Integrand size = 35, antiderivative size = 363 \[ \int (a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=-\frac {\left (3 b^2 (11 A-16 C)+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{24 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {a \left (8 a^2 (2 A+3 C)+b^2 (59 A+96 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{24 d \sqrt {a+b \cos (c+d x)}}+\frac {5 b \left (A b^2+4 a^2 (A+2 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{8 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (15 A b^2+8 a^2 (2 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{24 d}+\frac {5 A b (a+b \cos (c+d x))^{3/2} \sec (c+d x) \tan (c+d x)}{12 d}+\frac {A (a+b \cos (c+d x))^{5/2} \sec ^2(c+d x) \tan (c+d x)}{3 d} \]
-1/24*(3*b^2*(11*A-16*C)+8*a^2*(2*A+3*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos (1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*(a+b *cos(d*x+c))^(1/2)/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+1/24*a*(8*a^2*(2*A+3*C )+b^2*(59*A+96*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti cF(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1 /2)/d/(a+b*cos(d*x+c))^(1/2)+5/8*b*(A*b^2+4*a^2*(A+2*C))*(cos(1/2*d*x+1/2* c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(b/ (a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/d/(a+b*cos(d*x+c))^(1/2)+5/12 *A*b*(a+b*cos(d*x+c))^(3/2)*sec(d*x+c)*tan(d*x+c)/d+1/3*A*(a+b*cos(d*x+c)) ^(5/2)*sec(d*x+c)^2*tan(d*x+c)/d+1/24*(15*A*b^2+8*a^2*(2*A+3*C))*(a+b*cos( d*x+c))^(1/2)*tan(d*x+c)/d
Result contains complex when optimal does not.
Time = 6.73 (sec) , antiderivative size = 477, normalized size of antiderivative = 1.31 \[ \int (a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {\frac {8 a b^2 (13 A+72 C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 b \left (-3 b^2 (A-16 C)+8 a^2 (13 A+27 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i \left (3 b^2 (11 A-16 C)+8 a^2 (2 A+3 C)\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {-\frac {b (1+\cos (c+d x))}{a-b}} \csc (c+d x) \left (-2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (-2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )+b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right )}{a b \sqrt {-\frac {1}{a+b}}}+4 \sqrt {a+b \cos (c+d x)} \sec ^2(c+d x) \left (26 a A b \sin (c+d x)+\left (\frac {33 A b^2}{2}+4 a^2 (2 A+3 C)\right ) \sin (2 (c+d x))+8 a^2 A \tan (c+d x)\right )}{96 d} \]
((8*a*b^2*(13*A + 72*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + (2*b*(-3*b^2*(A - 16*C) + 8*a^2*(13*A + 27*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, ( c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] - ((2*I)*(3*b^2*(11*A - 16*C) + 8*a^2*(2*A + 3*C))*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqr t[-((b*(1 + Cos[c + d*x]))/(a - b))]*Csc[c + d*x]*(-2*a*(a - b)*EllipticE[ I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*(-2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]] ], (a + b)/(a - b)] + b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1) ]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)])))/(a*b*Sqrt[-(a + b)^(-1)]) + 4*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^2*(26*a*A*b*Sin[c + d*x] + ((33 *A*b^2)/2 + 4*a^2*(2*A + 3*C))*Sin[2*(c + d*x)] + 8*a^2*A*Tan[c + d*x]))/( 96*d)
Time = 3.28 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.04, number of steps used = 24, number of rules used = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.686, Rules used = {3042, 3527, 27, 3042, 3526, 27, 3042, 3526, 27, 3042, 3538, 25, 3042, 3134, 3042, 3132, 3481, 3042, 3142, 3042, 3140, 3286, 3042, 3284}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sec ^4(c+d x) (a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3527 |
| \(\displaystyle \frac {1}{3} \int \frac {1}{2} (a+b \cos (c+d x))^{3/2} \left (-b (A-6 C) \cos ^2(c+d x)+2 a (2 A+3 C) \cos (c+d x)+5 A b\right ) \sec ^3(c+d x)dx+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \int (a+b \cos (c+d x))^{3/2} \left (-b (A-6 C) \cos ^2(c+d x)+2 a (2 A+3 C) \cos (c+d x)+5 A b\right ) \sec ^3(c+d x)dx+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (-b (A-6 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 a (2 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 A b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{2} \int \frac {1}{2} \sqrt {a+b \cos (c+d x)} \left (8 (2 A+3 C) a^2+2 b (11 A+24 C) \cos (c+d x) a+15 A b^2-3 b^2 (3 A-8 C) \cos ^2(c+d x)\right ) \sec ^2(c+d x)dx+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \int \sqrt {a+b \cos (c+d x)} \left (8 (2 A+3 C) a^2+2 b (11 A+24 C) \cos (c+d x) a+15 A b^2-3 b^2 (3 A-8 C) \cos ^2(c+d x)\right ) \sec ^2(c+d x)dx+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \int \frac {\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (8 (2 A+3 C) a^2+2 b (11 A+24 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+15 A b^2-3 b^2 (3 A-8 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\int \frac {\left (2 a (13 A+72 C) \cos (c+d x) b^2-\left (8 (2 A+3 C) a^2+3 b^2 (11 A-16 C)\right ) \cos ^2(c+d x) b+15 \left (4 (A+2 C) a^2+A b^2\right ) b\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {\left (2 a (13 A+72 C) \cos (c+d x) b^2-\left (8 (2 A+3 C) a^2+3 b^2 (11 A-16 C)\right ) \cos ^2(c+d x) b+15 \left (4 (A+2 C) a^2+A b^2\right ) b\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {2 a (13 A+72 C) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2-\left (8 (2 A+3 C) a^2+3 b^2 (11 A-16 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2 b+15 \left (4 (A+2 C) a^2+A b^2\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {1}{2} \left (-\left (\left (8 a^2 (2 A+3 C)+3 b^2 (11 A-16 C)\right ) \int \sqrt {a+b \cos (c+d x)}dx\right )-\frac {\int -\frac {\left (15 \left (4 (A+2 C) a^2+A b^2\right ) b^2+a \left (8 (2 A+3 C) a^2+b^2 (59 A+96 C)\right ) \cos (c+d x) b\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}\right )+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {\left (15 \left (4 (A+2 C) a^2+A b^2\right ) b^2+a \left (8 (2 A+3 C) a^2+b^2 (59 A+96 C)\right ) \cos (c+d x) b\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\left (8 a^2 (2 A+3 C)+3 b^2 (11 A-16 C)\right ) \int \sqrt {a+b \cos (c+d x)}dx\right )+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {15 \left (4 (A+2 C) a^2+A b^2\right ) b^2+a \left (8 (2 A+3 C) a^2+b^2 (59 A+96 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\left (8 a^2 (2 A+3 C)+3 b^2 (11 A-16 C)\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {15 \left (4 (A+2 C) a^2+A b^2\right ) b^2+a \left (8 (2 A+3 C) a^2+b^2 (59 A+96 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (8 a^2 (2 A+3 C)+3 b^2 (11 A-16 C)\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {15 \left (4 (A+2 C) a^2+A b^2\right ) b^2+a \left (8 (2 A+3 C) a^2+b^2 (59 A+96 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (8 a^2 (2 A+3 C)+3 b^2 (11 A-16 C)\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {15 \left (4 (A+2 C) a^2+A b^2\right ) b^2+a \left (8 (2 A+3 C) a^2+b^2 (59 A+96 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (8 a^2 (2 A+3 C)+3 b^2 (11 A-16 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {a b \left (8 a^2 (2 A+3 C)+b^2 (59 A+96 C)\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx+15 b^2 \left (4 a^2 (A+2 C)+A b^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {2 \left (8 a^2 (2 A+3 C)+3 b^2 (11 A-16 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {15 b^2 \left (4 a^2 (A+2 C)+A b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a b \left (8 a^2 (2 A+3 C)+b^2 (59 A+96 C)\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (8 a^2 (2 A+3 C)+3 b^2 (11 A-16 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {15 b^2 \left (4 a^2 (A+2 C)+A b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a b \left (8 a^2 (2 A+3 C)+b^2 (59 A+96 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (8 a^2 (2 A+3 C)+3 b^2 (11 A-16 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {15 b^2 \left (4 a^2 (A+2 C)+A b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a b \left (8 a^2 (2 A+3 C)+b^2 (59 A+96 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (8 a^2 (2 A+3 C)+3 b^2 (11 A-16 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {15 b^2 \left (4 a^2 (A+2 C)+A b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a b \left (8 a^2 (2 A+3 C)+b^2 (59 A+96 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (8 a^2 (2 A+3 C)+3 b^2 (11 A-16 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\frac {15 b^2 \left (4 a^2 (A+2 C)+A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 a b \left (8 a^2 (2 A+3 C)+b^2 (59 A+96 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (8 a^2 (2 A+3 C)+3 b^2 (11 A-16 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\frac {15 b^2 \left (4 a^2 (A+2 C)+A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 a b \left (8 a^2 (2 A+3 C)+b^2 (59 A+96 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (8 a^2 (2 A+3 C)+3 b^2 (11 A-16 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )+\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {\left (8 a^2 (2 A+3 C)+15 A b^2\right ) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}+\frac {1}{2} \left (\frac {\frac {2 a b \left (8 a^2 (2 A+3 C)+b^2 (59 A+96 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {30 b^2 \left (4 a^2 (A+2 C)+A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 \left (8 a^2 (2 A+3 C)+3 b^2 (11 A-16 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\right )\right )+\frac {5 A b \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^{3/2}}{2 d}\right )+\frac {A \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))^{5/2}}{3 d}\) |
(A*(a + b*Cos[c + d*x])^(5/2)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((5*A*b *(a + b*Cos[c + d*x])^(3/2)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (((-2*(3*b^ 2*(11*A - 16*C) + 8*a^2*(2*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + ((2*a* b*(8*a^2*(2*A + 3*C) + b^2*(59*A + 96*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b )]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]) + ( 30*b^2*(A*b^2 + 4*a^2*(A + 2*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Ellipt icPi[2, (c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]))/b)/2 + ((15*A*b^2 + 8*a^2*(2*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*Tan[c + d*x])/d)/ 4)/6
3.7.44.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)] Int[Sqrt[a/(a + b) + ( b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 , 0] && !GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]] Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && !GtQ[a + b, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt [c + d*Sin[e + f*x]] Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !GtQ[c + d, 0]
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f *x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d , 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^ 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A* d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b *c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( A*d^2*(m + n + 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(2672\) vs. \(2(420)=840\).
Time = 197.11 (sec) , antiderivative size = 2673, normalized size of antiderivative = 7.36
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(2673\) |
| parts | \(\text {Expression too large to display}\) | \(2704\) |
-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*C*b^3*( sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/( -2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos( 1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+6*C*a*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)* ((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a +b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^ (1/2))-2*C*b^2*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c) ^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2 )^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1/ 2*d*x+1/2*c),(-2*b/(a-b))^(1/2)))+2*A*a^3*(-1/3*cos(1/2*d*x+1/2*c)/a*(-2*s in(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2 *c)^2-1)^3+5/12*b/a^2*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)* sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^2-1/24*(16*a^2+15*b ^2)/a^3*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/ 2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)+5/48*b^2/a^2*(sin(1/2*d*x+1/2*c)^ 2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2* c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2* b/(a-b))^(1/2))+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^ 2+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2) ^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/3*(sin(1/2*d*...
Timed out. \[ \int (a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \]
Timed out. \[ \int (a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \]
\[ \int (a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{4} \,d x } \]
\[ \int (a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{4} \,d x } \]
Timed out. \[ \int (a+b \cos (c+d x))^{5/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^4} \,d x \]