Integrand size = 35, antiderivative size = 211 \[ \int \frac {(a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {4 a^3 (5 A-9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 (5 A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {8 a^3 (10 A-3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \cos (c+d x))^3 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 A \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {2 (35 A-3 C) \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d} \] Output:
-4/5*a^3*(5*A-9*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/3*a^3*(5*A+3* C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d-8/15*a^3*(10*A-3*C)*cos(d*x+c) ^(1/2)*sin(d*x+c)/d+2/3*A*(a+a*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(3/2) +4*A*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)-2/15*(35*A-3*C )*cos(d*x+c)^(1/2)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.33 (sec) , antiderivative size = 909, normalized size of antiderivative = 4.31 \[ \int \frac {(a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:
Integrate[((a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/ 2),x]
Output:
Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(-1/40*((-2 5*A + 18*C + 5*A*Cos[2*c] + 18*C*Cos[2*c])*Csc[c]*Sec[c])/d + (C*Cos[d*x]* Sin[c])/(4*d) + (C*Cos[2*d*x]*Sin[2*c])/(40*d) + (C*Cos[c]*Sin[d*x])/(4*d) + (A*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(12*d) + (Sec[c]*Sec[c + d*x]*(A*Sin [c] + 9*A*Sin[d*x]))/(12*d) + (C*Cos[2*c]*Sin[2*d*x])/(40*d)) - (5*A*(a + a*Cos[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - Ar cTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - S in[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTa n[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(6*d*Sqrt[1 + Cot[c]^2]) - (C*(a + a*Cos[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, S in[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*Sec[d*x - ArcTan[Cot[c]]] *Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[ d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(2*d*Sqrt[1 + Cot[c]^2]) + (A*(a + a*Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((Hype rgeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[ d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[ c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos [c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c...
Time = 1.52 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.05, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.543, Rules used = {3042, 3523, 27, 3042, 3454, 27, 3042, 3455, 27, 3042, 3447, 3042, 3502, 27, 3042, 3227, 3042, 3119, 3120}
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.
| \(\displaystyle \int \frac {(a \cos (c+d x)+a)^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {2 \int \frac {(\cos (c+d x) a+a)^3 (6 a A-a (5 A-3 C) \cos (c+d x))}{2 \cos ^{\frac {3}{2}}(c+d x)}dx}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^3 (6 a A-a (5 A-3 C) \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)}dx}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (6 a A-a (5 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {2 \int \frac {(\cos (c+d x) a+a)^2 \left (a^2 (25 A+3 C)-a^2 (35 A-3 C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x)}}dx+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^2 \left (a^2 (25 A+3 C)-a^2 (35 A-3 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a^2 (25 A+3 C)-a^2 (35 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {2}{5} \int \frac {3 (\cos (c+d x) a+a) \left (3 a^3 (5 A+C)-2 a^3 (10 A-3 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx-\frac {2 (35 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {6}{5} \int \frac {(\cos (c+d x) a+a) \left (3 a^3 (5 A+C)-2 a^3 (10 A-3 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}}dx-\frac {2 (35 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {6}{5} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (3 a^3 (5 A+C)-2 a^3 (10 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (35 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {6}{5} \int \frac {-2 (10 A-3 C) \cos ^2(c+d x) a^4+3 (5 A+C) a^4+\left (3 a^4 (5 A+C)-2 a^4 (10 A-3 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {2 (35 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {6}{5} \int \frac {-2 (10 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^4+3 (5 A+C) a^4+\left (3 a^4 (5 A+C)-2 a^4 (10 A-3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (35 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {6}{5} \left (\frac {2}{3} \int \frac {5 a^4 (5 A+3 C)-3 a^4 (5 A-9 C) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx-\frac {4 a^4 (10 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (35 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {6}{5} \left (\frac {1}{3} \int \frac {5 a^4 (5 A+3 C)-3 a^4 (5 A-9 C) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx-\frac {4 a^4 (10 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (35 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {6}{5} \left (\frac {1}{3} \int \frac {5 a^4 (5 A+3 C)-3 a^4 (5 A-9 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {4 a^4 (10 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (35 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {6}{5} \left (\frac {1}{3} \left (5 a^4 (5 A+3 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^4 (5 A-9 C) \int \sqrt {\cos (c+d x)}dx\right )-\frac {4 a^4 (10 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (35 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {6}{5} \left (\frac {1}{3} \left (5 a^4 (5 A+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^4 (5 A-9 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {4 a^4 (10 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (35 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {6}{5} \left (\frac {1}{3} \left (5 a^4 (5 A+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a^4 (5 A-9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {4 a^4 (10 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-\frac {2 (35 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {-\frac {2 (35 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 \cos (c+d x)+a^4\right )}{5 d}+\frac {6}{5} \left (\frac {1}{3} \left (\frac {10 a^4 (5 A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a^4 (5 A-9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {4 a^4 (10 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {12 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{d \sqrt {\cos (c+d x)}}}{3 a}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^3}{3 d \cos ^{\frac {3}{2}}(c+d x)}\) |
Input:
Int[((a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/2),x]
Output:
(2*A*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((12* A*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]) - (2*(35 *A - 3*C)*Sqrt[Cos[c + d*x]]*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(5*d) + (6*(((-6*a^4*(5*A - 9*C)*EllipticE[(c + d*x)/2, 2])/d + (10*a^4*(5*A + 3 *C)*EllipticF[(c + d*x)/2, 2])/d)/3 - (4*a^4*(10*A - 3*C)*Sqrt[Cos[c + d*x ]]*Sin[c + d*x])/(3*d)))/5)/(3*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[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*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] :> Sim p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -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/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a *d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n + 2) + C* (c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(703\) vs. \(2(196)=392\).
Time = 4.33 (sec) , antiderivative size = 704, normalized size of antiderivative = 3.34
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(704\) |
| parts | \(\text {Expression too large to display}\) | \(963\) |
Input:
int((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x,method=_RETUR NVERBOSE)
Output:
-4/15*(24*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d *x+1/2*c)*sin(1/2*d*x+1/2*c)^8-96*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1 /2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+6*(-2*sin(1/2*d*x+1 /2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(15*A+13*C)*cos(1/2*d*x+1/2*c)*sin(1/2 *d*x+1/2*c)^4-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(25*A +9*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-2*(sin(1/2*d*x+1/2*c)^2)^(1/ 2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1 /2*c)^2)^(1/2)*(25*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+15*A*EllipticE( cos(1/2*d*x+1/2*c),2^(1/2))+15*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-27* C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^2+25*A*(sin(1/ 2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d *x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+15 *A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c )^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2 ^(1/2))+15*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2 *d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d* x+1/2*c),2^(1/2))-27*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2 )*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE( cos(1/2*d*x+1/2*c),2^(1/2)))*a^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2* c)^2)^(1/2)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(3/2)/sin(1/2*d*x+1/2*c)/d
Result contains complex when optimal does not.
Time = 0.18 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.17 \[ \int \frac {(a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, A - 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, A - 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (3 \, C a^{3} \cos \left (d x + c\right )^{3} + 15 \, C a^{3} \cos \left (d x + c\right )^{2} + 45 \, A a^{3} \cos \left (d x + c\right ) + 5 \, A a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algori thm="fricas")
Output:
-2/15*(5*I*sqrt(2)*(5*A + 3*C)*a^3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(5*A + 3*C)*a^3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt( 2)*(5*A - 9*C)*a^3*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInver se(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*(5*A - 9*C)*a^3*co s(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c ) - I*sin(d*x + c))) - (3*C*a^3*cos(d*x + c)^3 + 15*C*a^3*cos(d*x + c)^2 + 45*A*a^3*cos(d*x + c) + 5*A*a^3)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos( d*x + c)^2)
Timed out. \[ \int \frac {(a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))**3*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2),x)
Output:
Timed out
\[ \int \frac {(a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algori thm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^3/cos(d*x + c)^(5/2) , x)
\[ \int \frac {(a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algori thm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^3/cos(d*x + c)^(5/2) , x)
Time = 0.79 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.12 \[ \int \frac {(a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2\,\left (A\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+3\,A\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{d}+\frac {6\,C\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a^3\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{d}+\frac {6\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,a^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:
int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^3)/cos(c + d*x)^(5/2),x)
Output:
(2*(A*a^3*ellipticE(c/2 + (d*x)/2, 2) + 3*A*a^3*ellipticF(c/2 + (d*x)/2, 2 )))/d + (6*C*a^3*ellipticE(c/2 + (d*x)/2, 2))/d + (4*C*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (2*C*a^3*cos(c + d*x)^(1/2)*sin(c + d*x))/d + (6*A*a^3*s in(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^( 1/2)*(sin(c + d*x)^2)^(1/2)) + (2*A*a^3*sin(c + d*x)*hypergeom([-3/4, 1/2] , 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) - (2*C*a^3*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2))
\[ \int \frac {(a+a \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=a^{3} \left (3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) a +3 \left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) c +3 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) c \right ) \] Input:
int((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x)
Output:
a**3*(3*int(sqrt(cos(c + d*x))/cos(c + d*x),x)*a + int(sqrt(cos(c + d*x))/ cos(c + d*x),x)*c + int(sqrt(cos(c + d*x))/cos(c + d*x)**3,x)*a + 3*int(sq rt(cos(c + d*x))/cos(c + d*x)**2,x)*a + int(sqrt(cos(c + d*x)),x)*a + 3*in t(sqrt(cos(c + d*x)),x)*c + 3*int(sqrt(cos(c + d*x))*cos(c + d*x),x)*c + i nt(sqrt(cos(c + d*x))*cos(c + d*x)**2,x)*c)