Integrand size = 42, antiderivative size = 256 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=-\frac {\left (2 a^2 B-3 b^2 B+a b C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}+\frac {(b B-a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a \left (a^2-b^2\right ) d}-\frac {\left (5 a^2 b B-3 b^3 B-3 a^3 C+a b^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 (a-b) (a+b)^2 d}+\frac {\left (2 a^2 B-3 b^2 B+a b C\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {b (b B-a C) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \] Output:
-(2*B*a^2-3*B*b^2+C*a*b)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/(a^2-b^ 2)/d+(B*b-C*a)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a/(a^2-b^2)/d-(5*B*a ^2*b-3*B*b^3-3*C*a^3+C*a*b^2)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1 /2))/a^2/(a-b)/(a+b)^2/d+(2*B*a^2-3*B*b^2+C*a*b)*sin(d*x+c)/a^2/(a^2-b^2)/ d/cos(d*x+c)^(1/2)+b*(B*b-C*a)*sin(d*x+c)/a/(a^2-b^2)/d/cos(d*x+c)^(1/2)/( a+b*cos(d*x+c))
Time = 5.00 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.23 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\frac {-\frac {\frac {2 \left (-10 a^2 b B+9 b^3 B+4 a^3 C-3 a b^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}-\frac {8 a \left (a^2 B-2 b^2 B+a b C\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{b (a+b)}-\frac {2 \left (2 a^2 B-3 b^2 B+a b C\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{(-a+b) (a+b)}+4 \sqrt {\cos (c+d x)} \left (\frac {b^2 (b B-a C) \sin (c+d x)}{\left (-a^2+b^2\right ) (a+b \cos (c+d x))}+2 B \tan (c+d x)\right )}{4 a^2 d} \] Input:
Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + b*C os[c + d*x])^2),x]
Output:
(-(((2*(-10*a^2*b*B + 9*b^3*B + 4*a^3*C - 3*a*b^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) - (8*a*(a^2*B - 2*b^2*B + a*b*C)*((a + b)*El lipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(b *(a + b)) - (2*(2*a^2*B - 3*b^2*B + a*b*C)*(-2*a*b*EllipticE[ArcSin[Sqrt[C os[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[ c + d*x])/(a*b*Sqrt[Sin[c + d*x]^2]))/((-a + b)*(a + b))) + 4*Sqrt[Cos[c + d*x]]*((b^2*(b*B - a*C)*Sin[c + d*x])/((-a^2 + b^2)*(a + b*Cos[c + d*x])) + 2*B*Tan[c + d*x]))/(4*a^2*d)
Time = 2.09 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.95, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.405, Rules used = {3042, 3508, 3042, 3479, 27, 3042, 3534, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 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.
\(\displaystyle \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3508 |
\(\displaystyle \int \frac {B+C \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B+C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3479 |
\(\displaystyle \frac {\int \frac {2 B a^2+b C a-2 (b B-a C) \cos (c+d x) a+b (b B-a C) \cos ^2(c+d x)-3 b^2 B}{2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {2 B a^2+b C a-2 (b B-a C) \cos (c+d x) a+b (b B-a C) \cos ^2(c+d x)-3 b^2 B}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {2 B a^2+b C a-2 (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+b (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2-3 b^2 B}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\frac {2 \int -\frac {-2 C a^3+4 b B a^2+b^2 C a+2 \left (B a^2+b C a-2 b^2 B\right ) \cos (c+d x) a+b \left (2 B a^2+b C a-3 b^2 B\right ) \cos ^2(c+d x)-3 b^3 B}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}+\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {-2 C a^3+4 b B a^2+b^2 C a+2 \left (B a^2+b C a-2 b^2 B\right ) \cos (c+d x) a+b \left (2 B a^2+b C a-3 b^2 B\right ) \cos ^2(c+d x)-3 b^3 B}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {-2 C a^3+4 b B a^2+b^2 C a+2 \left (B a^2+b C a-2 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+b \left (2 B a^2+b C a-3 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-3 b^3 B}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\left (2 a^2 B+a b C-3 b^2 B\right ) \int \sqrt {\cos (c+d x)}dx-\frac {\int -\frac {b \left (-2 C a^3+4 b B a^2+b^2 C a-3 b^3 B\right )-a b^2 (b B-a C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\left (2 a^2 B+a b C-3 b^2 B\right ) \int \sqrt {\cos (c+d x)}dx+\frac {\int \frac {b \left (-2 C a^3+4 b B a^2+b^2 C a-3 b^3 B\right )-a b^2 (b B-a C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\left (2 a^2 B+a b C-3 b^2 B\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\int \frac {b \left (-2 C a^3+4 b B a^2+b^2 C a-3 b^3 B\right )-a b^2 (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {\int \frac {b \left (-2 C a^3+4 b B a^2+b^2 C a-3 b^3 B\right )-a b^2 (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {b \left (-3 a^3 C+5 a^2 b B+a b^2 C-3 b^3 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx-a b (b B-a C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}+\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {b \left (-3 a^3 C+5 a^2 b B+a b^2 C-3 b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx-a b (b B-a C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {b \left (-3 a^3 C+5 a^2 b B+a b^2 C-3 b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx-\frac {2 a b (b B-a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b}+\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{2 a \left (a^2-b^2\right )}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}+\frac {\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {2 \left (2 a^2 B+a b C-3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {\frac {2 b \left (-3 a^3 C+5 a^2 b B+a b^2 C-3 b^3 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}-\frac {2 a b (b B-a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b}}{a}}{2 a \left (a^2-b^2\right )}\) |
Input:
Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])^2),x]
Output:
(b*(b*B - a*C)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(a + b*Co s[c + d*x])) + (-(((2*(2*a^2*B - 3*b^2*B + a*b*C)*EllipticE[(c + d*x)/2, 2 ])/d + ((-2*a*b*(b*B - a*C)*EllipticF[(c + d*x)/2, 2])/d + (2*b*(5*a^2*b*B - 3*b^3*B - 3*a^3*C + a*b^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]) /((a + b)*d))/b)/a) + (2*(2*a^2*B - 3*b^2*B + a*b*C)*Sin[c + d*x])/(a*d*Sq rt[Cos[c + d*x]]))/(2*a*(a^2 - b^2))
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[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[((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. \(855\) vs. \(2(257)=514\).
Time = 5.10 (sec) , antiderivative size = 856, normalized size of antiderivative = 3.34
Input:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x,me thod=_RETURNVERBOSE)
Output:
-(-(1-2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B/a^2/sin(1/2 *d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2* d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(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)))-2*(B*b-C*a)/a*(-1/a*b^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*s in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*b*cos(1/2*d*x+1/2*c)^2+ a-b)-1/2/(a+b)/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*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)*EllipticF(cos(1/ 2*d*x+1/2*c),2^(1/2))-1/2/a*b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2* 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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2/a*b/(a^2-b^2)*(sin(1/2*d* x+1/2*c)^2)^(1/2)*(1-2*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a ^2-b^2)/(-2*a*b+2*b^2)*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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Elliptic Pi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+2*b^2)*b^3 *(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*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)*EllipticPi(cos(1/2*d*x+1/2*c),-2 *b/(a-b),2^(1/2)))+4*B*b^2/a^2/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2) *(1-2*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+...
Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^ 2,x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2)/(a+b*cos(d*x+c) )**2,x)
Output:
Timed out
Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^ 2,x, algorithm="maxima")
Output:
Timed out
\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^ 2,x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/((b*cos(d*x + c) + a)^2*cos( d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \] Input:
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + b*cos(c + d*x))^2),x)
Output:
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + b*cos(c + d*x))^2), x)
\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4} b^{2}+2 \cos \left (d x +c \right )^{3} a b +\cos \left (d x +c \right )^{2} a^{2}}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3} b^{2}+2 \cos \left (d x +c \right )^{2} a b +\cos \left (d x +c \right ) a^{2}}d x \right ) c \] Input:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x)
Output:
int(sqrt(cos(c + d*x))/(cos(c + d*x)**4*b**2 + 2*cos(c + d*x)**3*a*b + cos (c + d*x)**2*a**2),x)*b + int(sqrt(cos(c + d*x))/(cos(c + d*x)**3*b**2 + 2 *cos(c + d*x)**2*a*b + cos(c + d*x)*a**2),x)*c