Integrand size = 25, antiderivative size = 566 \[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\sqrt {\sec (c+d x)}} \, dx=-\frac {(a-b) \sqrt {a+b} \left (33 a^2+16 b^2\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{24 a d \sqrt {\sec (c+d x)}}+\frac {\sqrt {a+b} \left (33 a^2+26 a b+16 b^2\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{24 d \sqrt {\sec (c+d x)}}-\frac {5 a \sqrt {a+b} \left (a^2+4 b^2\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{8 b d \sqrt {\sec (c+d x)}}+\frac {b^2 \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {13 a b \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{12 d \sqrt {\sec (c+d x)}}+\frac {\left (33 a^2+16 b^2\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{24 d} \] Output:
-1/24*(a-b)*(a+b)^(1/2)*(33*a^2+16*b^2)*cos(d*x+c)^(1/2)*csc(d*x+c)*Ellipt icE((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/ 2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/d/sec( d*x+c)^(1/2)+1/24*(a+b)^(1/2)*(33*a^2+26*a*b+16*b^2)*cos(d*x+c)^(1/2)*csc( d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+ b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^( 1/2)/d/sec(d*x+c)^(1/2)-5/8*a*(a+b)^(1/2)*(a^2+4*b^2)*cos(d*x+c)^(1/2)*csc (d*x+c)*EllipticPi((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(a+ b)/b,(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c) )/(a-b))^(1/2)/b/d/sec(d*x+c)^(1/2)+1/3*b^2*(a+b*cos(d*x+c))^(1/2)*sin(d*x +c)/d/sec(d*x+c)^(3/2)+13/12*a*b*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/d/sec(d *x+c)^(1/2)+1/24*(33*a^2+16*b^2)*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)*s in(d*x+c)/d
Time = 15.14 (sec) , antiderivative size = 970, normalized size of antiderivative = 1.71 \[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\sqrt {\sec (c+d x)}} \, dx =\text {Too large to display} \] Input:
Integrate[(a + b*Cos[c + d*x])^(5/2)/Sqrt[Sec[c + d*x]],x]
Output:
(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((b^2*Sin[c + d*x])/12 + (13* a*b*Sin[2*(c + d*x)])/24 + (b^2*Sin[3*(c + d*x)])/12))/d + (Sqrt[(1 - Tan[ (c + d*x)/2]^2)^(-1)]*(33*a^3*Tan[(c + d*x)/2] + 33*a^2*b*Tan[(c + d*x)/2] + 16*a*b^2*Tan[(c + d*x)/2] + 16*b^3*Tan[(c + d*x)/2] - 66*a^2*b*Tan[(c + d*x)/2]^3 - 32*b^3*Tan[(c + d*x)/2]^3 - 33*a^3*Tan[(c + d*x)/2]^5 + 33*a^ 2*b*Tan[(c + d*x)/2]^5 - 16*a*b^2*Tan[(c + d*x)/2]^5 + 16*b^3*Tan[(c + d*x )/2]^5 + 30*a^3*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] *Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[( c + d*x)/2]^2)/(a + b)] + 120*a*b^2*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2] ], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 30*a^3*EllipticPi[-1, ArcSin [Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 120*a*b^2*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b )]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + (33*a^3 + 33*a^2*b + 16*a*b ^2 + 16*b^3)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b + a*Tan[(c + d* x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a*(24*a^2 - 13*a*b + 38*b^2)* EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c +...
Time = 2.67 (sec) , antiderivative size = 544, normalized size of antiderivative = 0.96, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.760, Rules used = {3042, 4710, 3042, 3272, 27, 3042, 3528, 27, 3042, 3540, 25, 3042, 3532, 3042, 3288, 3477, 3042, 3295, 3473}
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.
\(\displaystyle \int \frac {(a+b \cos (c+d x))^{5/2}}{\sqrt {\sec (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4710 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{5/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx\) |
\(\Big \downarrow \) 3272 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \int \frac {\sqrt {\cos (c+d x)} \left (13 a b^2 \cos ^2(c+d x)+2 b \left (9 a^2+2 b^2\right ) \cos (c+d x)+3 a \left (2 a^2+b^2\right )\right )}{2 \sqrt {a+b \cos (c+d x)}}dx+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{6} \int \frac {\sqrt {\cos (c+d x)} \left (13 a b^2 \cos ^2(c+d x)+2 b \left (9 a^2+2 b^2\right ) \cos (c+d x)+3 a \left (2 a^2+b^2\right )\right )}{\sqrt {a+b \cos (c+d x)}}dx+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{6} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (13 a b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b \left (9 a^2+2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a \left (2 a^2+b^2\right )\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
\(\Big \downarrow \) 3528 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{6} \left (\frac {\int \frac {13 a^2 b^2+\left (33 a^2+16 b^2\right ) \cos ^2(c+d x) b^2+2 a \left (12 a^2+19 b^2\right ) \cos (c+d x) b}{2 \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{2 b}+\frac {13 a b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{6} \left (\frac {\int \frac {13 a^2 b^2+\left (33 a^2+16 b^2\right ) \cos ^2(c+d x) b^2+2 a \left (12 a^2+19 b^2\right ) \cos (c+d x) b}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{4 b}+\frac {13 a b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{6} \left (\frac {\int \frac {13 a^2 b^2+\left (33 a^2+16 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2 b^2+2 a \left (12 a^2+19 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 b}+\frac {13 a b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
\(\Big \downarrow \) 3540 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{6} \left (\frac {\frac {\int -\frac {-26 a^2 \cos (c+d x) b^3-15 a \left (a^2+4 b^2\right ) \cos ^2(c+d x) b^2+a \left (33 a^2+16 b^2\right ) b^2}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}+\frac {b \left (33 a^2+16 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}}{4 b}+\frac {13 a b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{6} \left (\frac {\frac {b \left (33 a^2+16 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {\int \frac {-26 a^2 \cos (c+d x) b^3-15 a \left (a^2+4 b^2\right ) \cos ^2(c+d x) b^2+a \left (33 a^2+16 b^2\right ) b^2}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}}{4 b}+\frac {13 a b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{6} \left (\frac {\frac {b \left (33 a^2+16 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {\int \frac {-26 a^2 \sin \left (c+d x+\frac {\pi }{2}\right ) b^3-15 a \left (a^2+4 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2 b^2+a \left (33 a^2+16 b^2\right ) b^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}}{4 b}+\frac {13 a b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
\(\Big \downarrow \) 3532 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{6} \left (\frac {\frac {b \left (33 a^2+16 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a b^2 \left (33 a^2+16 b^2\right )-26 a^2 b^3 \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-15 a b^2 \left (a^2+4 b^2\right ) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx}{2 b}}{4 b}+\frac {13 a b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{6} \left (\frac {\frac {b \left (33 a^2+16 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a b^2 \left (33 a^2+16 b^2\right )-26 a^2 b^3 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-15 a b^2 \left (a^2+4 b^2\right ) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}}{4 b}+\frac {13 a b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
\(\Big \downarrow \) 3288 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{6} \left (\frac {\frac {b \left (33 a^2+16 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a b^2 \left (33 a^2+16 b^2\right )-26 a^2 b^3 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {30 a b \sqrt {a+b} \left (a^2+4 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}}{4 b}+\frac {13 a b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
\(\Big \downarrow \) 3477 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{6} \left (\frac {\frac {b \left (33 a^2+16 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {a b^2 \left (33 a^2+16 b^2\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-a b^2 \left (33 a^2+26 a b+16 b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+\frac {30 a b \sqrt {a+b} \left (a^2+4 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}}{4 b}+\frac {13 a b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{6} \left (\frac {\frac {b \left (33 a^2+16 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {-a b^2 \left (33 a^2+26 a b+16 b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a b^2 \left (33 a^2+16 b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {30 a b \sqrt {a+b} \left (a^2+4 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}}{4 b}+\frac {13 a b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
\(\Big \downarrow \) 3295 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{6} \left (\frac {\frac {b \left (33 a^2+16 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {a b^2 \left (33 a^2+16 b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b^2 \sqrt {a+b} \left (33 a^2+26 a b+16 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}+\frac {30 a b \sqrt {a+b} \left (a^2+4 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}}{4 b}+\frac {13 a b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
\(\Big \downarrow \) 3473 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{6} \left (\frac {\frac {b \left (33 a^2+16 b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}-\frac {-\frac {2 b^2 \sqrt {a+b} \left (33 a^2+26 a b+16 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}+\frac {2 b^2 (a-b) \sqrt {a+b} \left (33 a^2+16 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}+\frac {30 a b \sqrt {a+b} \left (a^2+4 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}}{4 b}+\frac {13 a b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\right )+\frac {b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}\right )\) |
Input:
Int[(a + b*Cos[c + d*x])^(5/2)/Sqrt[Sec[c + d*x]],x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((b^2*Cos[c + d*x]^(3/2)*Sqrt[a + b* Cos[c + d*x]]*Sin[c + d*x])/(3*d) + ((13*a*b*Sqrt[Cos[c + d*x]]*Sqrt[a + b *Cos[c + d*x]]*Sin[c + d*x])/(2*d) + (-1/2*((2*(a - b)*b^2*Sqrt[a + b]*(33 *a^2 + 16*b^2)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqr t[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d* x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a*d) - (2*b^2*Sqrt[a + b]*(33*a^2 + 26*a*b + 16*b^2)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*C os[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[( a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/d + ( 30*a*b*Sqrt[a + b]*(a^2 + 4*b^2)*Cot[c + d*x]*EllipticPi[(a + b)/b, ArcSin [Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/d)/b + (b*(33*a^2 + 16*b^2)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x]) /(d*Sqrt[Cos[c + d*x]]))/(4*b))/6)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d *(m + n) + b^2*(b*c*(m - 2) + a*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m ] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[2*b*(Tan[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c *((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*Ellipti cPi[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/b]
Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f _.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] ], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) ^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* (c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] )/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(c + d)/b]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S imp[(A - B)/(a - b) Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* x]]), x], x] - Simp[(A*b - a*B)/(a - b) Int[(1 + Sin[e + f*x])/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e , f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A, B]
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)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[C/b^2 Int[Sqrt[a + b*Sin[e + f*x]] /Sqrt[c + d*Sin[e + f*x]], x], x] + Simp[1/b^2 Int[(A*b^2 - a^2*C + b*(b* B - 2*a*C)*Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*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]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(Sqrt[c + d*Sin[e + f *x]]/(d*f*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[1/(2*d) Int[(1/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]))*Simp[2*a*A*d - C*(b*c - a*d) - 2*(a*c*C - d*(A*b + a*B))*Sin[e + f*x] + (2*b*B*d - C*(b*c + a*d))*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]
Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(1053\) vs. \(2(497)=994\).
Time = 14.85 (sec) , antiderivative size = 1054, normalized size of antiderivative = 1.86
Input:
int((a+cos(d*x+c)*b)^(5/2)/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/24/d*(a+cos(d*x+c)*b)^(1/2)/(b*cos(d*x+c)^2+a*cos(d*x+c)+cos(d*x+c)*b+a) /sec(d*x+c)^(1/2)*((cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(co s(d*x+c)+1)/(a+b))^(1/2)*a^3*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,(-(a-b)/( a+b))^(1/2))*(-30*cos(d*x+c)-60-30*sec(d*x+c))+(cos(d*x+c)/(cos(d*x+c)+1)) ^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a*b^2*EllipticPi(cot( d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*(-120*cos(d*x+c)-240-120*sec(d* x+c))+(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/( a+b))^(1/2)*a^3*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(-33 *cos(d*x+c)-66-33*sec(d*x+c))+(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d* x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a^2*b*EllipticE(cot(d*x+c)-csc(d*x+c), (-(a-b)/(a+b))^(1/2))*(-33*cos(d*x+c)-66-33*sec(d*x+c))+(cos(d*x+c)/(cos(d *x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a*b^2*Ellipt icE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(-16*cos(d*x+c)-32-16*sec( d*x+c))+(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1) /(a+b))^(1/2)*b^3*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(- 16*cos(d*x+c)-32-16*sec(d*x+c))+(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos( d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a^3*EllipticF(cot(d*x+c)-csc(d*x+c), (-(a-b)/(a+b))^(1/2))*(48*cos(d*x+c)+96+48*sec(d*x+c))+(cos(d*x+c)/(cos(d* x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a^2*b*Ellipti cF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(-26*cos(d*x+c)-52-26*se...
\[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x, algorithm="fricas")
Output:
integral((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*sqrt(b*cos(d*x + c) + a)/sqrt(sec(d*x + c)), x)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**(5/2)/sec(d*x+c)**(1/2),x)
Output:
Timed out
\[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x, algorithm="maxima")
Output:
integrate((b*cos(d*x + c) + a)^(5/2)/sqrt(sec(d*x + c)), x)
\[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x, algorithm="giac")
Output:
integrate((b*cos(d*x + c) + a)^(5/2)/sqrt(sec(d*x + c)), x)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \] Input:
int((a + b*cos(c + d*x))^(5/2)/(1/cos(c + d*x))^(1/2),x)
Output:
int((a + b*cos(c + d*x))^(5/2)/(1/cos(c + d*x))^(1/2), x)
\[ \int \frac {(a+b \cos (c+d x))^{5/2}}{\sqrt {\sec (c+d x)}} \, dx=2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )}{\sec \left (d x +c \right )}d x \right ) a b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2}}{\sec \left (d x +c \right )}d x \right ) b^{2}+\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}}{\sec \left (d x +c \right )}d x \right ) a^{2} \] Input:
int((a+b*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x)
Output:
2*int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x)*b + a)*cos(c + d*x))/sec(c + d *x),x)*a*b + int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x)*b + a)*cos(c + d*x) **2)/sec(c + d*x),x)*b**2 + int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x)*b + a))/sec(c + d*x),x)*a**2