Integrand size = 42, antiderivative size = 299 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {(3 b B-4 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 a^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {(b B-4 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a d \sqrt {a+b \cos (c+d x)}}+\frac {\left (4 a^2 B+3 b^2 B-4 a b C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 a^2 d \sqrt {a+b \cos (c+d x)}}-\frac {(3 b B-4 a C) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 a^2 d}+\frac {B \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{2 a d} \] Output:
1/4*(3*B*b-4*C*a)*(a+b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1 /2)*(b/(a+b))^(1/2))/a^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-1/4*(B*b-4*C*a)* ((a+b*cos(d*x+c))/(a+b))^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2)*(b/(a +b))^(1/2))/a/d/(a+b*cos(d*x+c))^(1/2)+1/4*(4*B*a^2+3*B*b^2-4*C*a*b)*((a+b *cos(d*x+c))/(a+b))^(1/2)*EllipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(b/(a+b) )^(1/2))/a^2/d/(a+b*cos(d*x+c))^(1/2)-1/4*(3*B*b-4*C*a)*(a+b*cos(d*x+c))^( 1/2)*tan(d*x+c)/a^2/d+1/2*B*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)*tan(d*x+c)/a /d
Result contains complex when optimal does not.
Time = 6.40 (sec) , antiderivative size = 420, normalized size of antiderivative = 1.40 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {\frac {8 a b B \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (8 a^2 B+9 b^2 B-12 a b C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 i (3 b B-4 a C) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} \csc (c+d x) \left (-2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (-2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )+b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right )}{a b \sqrt {-\frac {1}{a+b}}}+4 \sqrt {a+b \cos (c+d x)} (2 a B+(-3 b B+4 a C) \cos (c+d x)) \sec (c+d x) \tan (c+d x)}{16 a^2 d} \] Input:
Integrate[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/Sqrt[a + b* Cos[c + d*x]],x]
Output:
((8*a*b*B*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*(8*a^2*B + 9*b^2*B - 12*a*b*C)*Sqr t[(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*B - 4*a*C)*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[(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*S qrt[-(a + b)^(-1)]) + 4*Sqrt[a + b*Cos[c + d*x]]*(2*a*B + (-3*b*B + 4*a*C) *Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(16*a^2*d)
Time = 2.72 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.02, number of steps used = 23, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.548, Rules used = {3042, 3508, 3042, 3479, 27, 3042, 3534, 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.
\(\displaystyle \int \frac {\sec ^4(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3508 |
\(\displaystyle \int \frac {\sec ^3(c+d x) (B+C \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B+C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3479 |
\(\displaystyle \frac {\int -\frac {\left (-b B \cos ^2(c+d x)-2 a B \cos (c+d x)+3 b B-4 a C\right ) \sec ^2(c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{2 a}+\frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\int \frac {\left (-b B \cos ^2(c+d x)-2 a B \cos (c+d x)+3 b B-4 a C\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\int \frac {-b B \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a B \sin \left (c+d x+\frac {\pi }{2}\right )+3 b B-4 a C}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 a}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {\int -\frac {\left (4 B a^2-4 b C a+2 b B \cos (c+d x) a+b (3 b B-4 a C) \cos ^2(c+d x)+3 b^2 B\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{a}+\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}}{4 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\int \frac {\left (4 B a^2-4 b C a+2 b B \cos (c+d x) a+b (3 b B-4 a C) \cos ^2(c+d x)+3 b^2 B\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\int \frac {4 B a^2-4 b C a+2 b B \sin \left (c+d x+\frac {\pi }{2}\right ) a+b (3 b B-4 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+3 b^2 B}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {(3 b B-4 a C) \int \sqrt {a+b \cos (c+d x)}dx-\frac {\int -\frac {\left (b \left (4 B a^2-4 b C a+3 b^2 B\right )-a b (b B-4 a C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{2 a}}{4 a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {\left (b \left (4 B a^2-4 b C a+3 b^2 B\right )-a b (b B-4 a C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}+(3 b B-4 a C) \int \sqrt {a+b \cos (c+d x)}dx}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {b \left (4 B a^2-4 b C a+3 b^2 B\right )-a b (b B-4 a C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+(3 b B-4 a C) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3134 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {b \left (4 B a^2-4 b C a+3 b^2 B\right )-a b (b B-4 a C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {(3 b B-4 a C) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {b \left (4 B a^2-4 b C a+3 b^2 B\right )-a b (b B-4 a C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {(3 b B-4 a C) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3132 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {b \left (4 B a^2-4 b C a+3 b^2 B\right )-a b (b B-4 a C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 (3 b B-4 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {b \left (4 a^2 B-4 a b C+3 b^2 B\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx-a b (b B-4 a C) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}+\frac {2 (3 b B-4 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {b \left (4 a^2 B-4 a b C+3 b^2 B\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a b (b B-4 a C) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 (3 b B-4 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3142 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {b \left (4 a^2 B-4 a b C+3 b^2 B\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a b (b B-4 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 (3 b B-4 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {b \left (4 a^2 B-4 a b C+3 b^2 B\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a b (b B-4 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 (3 b B-4 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3140 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {b \left (4 a^2 B-4 a b C+3 b^2 B\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a b (b B-4 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 (3 b B-4 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3286 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\frac {b \left (4 a^2 B-4 a b C+3 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 a b (b B-4 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 (3 b B-4 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\frac {b \left (4 a^2 B-4 a b C+3 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 a b (b B-4 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 (3 b B-4 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 a d}-\frac {\frac {(3 b B-4 a C) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\frac {2 b \left (4 a^2 B-4 a b C+3 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {2 a b (b B-4 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 (3 b B-4 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}}{4 a}\) |
Input:
Int[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/Sqrt[a + b*Cos[c + d*x]],x]
Output:
(B*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(2*a*d) - (-1/2*((2 *(3*b*B - 4*a*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*(b*B - 4*a*C)*Sqr t[(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]]) + (2*b*(4*a^2*B + 3*b^2*B - 4*a*b*C)*Sqrt[(a + b *Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt [a + b*Cos[c + d*x]]))/b)/a + ((3*b*B - 4*a*C)*Sqrt[a + b*Cos[c + d*x]]*Ta n[c + d*x])/(a*d))/(4*a)
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_)])^(n_), x_Symbol] :> Si mp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin [e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) *(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n }, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(I ntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 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[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
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. \(1181\) vs. \(2(287)=574\).
Time = 2.10 (sec) , antiderivative size = 1182, normalized size of antiderivative = 3.95
method | result | size |
default | \(\text {Expression too large to display}\) | \(1182\) |
parts | \(\text {Expression too large to display}\) | \(1244\) |
Input:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2),x,me thod=_RETURNVERBOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B*(-1/2* cos(1/2*d*x+1/2*c)/a*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2 )^(1/2)/(-1+2*cos(1/2*d*x+1/2*c)^2)^2+3/4*b/a^2*cos(1/2*d*x+1/2*c)*(-2*b*s in(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1+2*cos(1/2*d*x+1/ 2*c)^2)-1/8*b/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a- b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(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))+3/8/a*(sin(1/2*d*x+1/2 *c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d* x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c ),(-2*b/(a-b))^(1/2))-3/8*b^2/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2 *d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2 *d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/2* (sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/ (-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(co s(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))-3/8/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2 )*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+ (a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a -b))^(1/2))*b^2)+2*C*(-cos(1/2*d*x+1/2*c)/a*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+ b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1+2*cos(1/2*d*x+1/2*c)^2)+1/2*(sin(1/2*d* x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*si...
Timed out. \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2 ),x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+b*cos(d*x+c))**( 1/2),x)
Output:
Timed out
\[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sec \left (d x + c\right )^{4}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2 ),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*sec(d*x + c)^4/sqrt(b*cos(d* x + c) + a), x)
\[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sec \left (d x + c\right )^{4}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2 ),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*sec(d*x + c)^4/sqrt(b*cos(d* x + c) + a), x)
Timed out. \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^4\,\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \] Input:
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + b*cos(c + d*x ))^(1/2)),x)
Output:
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + b*cos(c + d*x ))^(1/2)), x)
\[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{\cos \left (d x +c \right ) b +a}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}}{\cos \left (d x +c \right ) b +a}d x \right ) c \] Input:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c))^(1/2),x)
Output:
int((sqrt(cos(c + d*x)*b + a)*cos(c + d*x)*sec(c + d*x)**4)/(cos(c + d*x)* b + a),x)*b + int((sqrt(cos(c + d*x)*b + a)*cos(c + d*x)**2*sec(c + d*x)** 4)/(cos(c + d*x)*b + a),x)*c