Integrand size = 33, antiderivative size = 224 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=-\frac {\left (a A b-3 a^2 B+2 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 \left (a^2-b^2\right ) d}+\frac {\left (a^2 A b-2 A b^3-3 a^3 B+4 a b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^3 \left (a^2-b^2\right ) d}-\frac {a \left (a^2 A b-3 A b^3-3 a^3 B+5 a b^2 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^3 (a+b)^2 d}+\frac {a (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:
-(A*a*b-3*B*a^2+2*B*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b^2/(a^2-b^ 2)/d+(A*a^2*b-2*A*b^3-3*B*a^3+4*B*a*b^2)*InverseJacobiAM(1/2*d*x+1/2*c,2^( 1/2))/b^3/(a^2-b^2)/d-a*(A*a^2*b-3*A*b^3-3*B*a^3+5*B*a*b^2)*EllipticPi(sin (1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/(a-b)/b^3/(a+b)^2/d+a*(A*b-B*a)*cos(d*x +c)^(1/2)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))
Time = 3.27 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\frac {-\frac {4 a (-A b+a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {2 \left (a A b+a^2 B-2 b^2 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 (-A b+a B) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {2 \left (-a A b+3 a^2 B-2 b^2 B\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}}{4 b d} \] Input:
Integrate[(Cos[c + d*x]^(3/2)*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^2 ,x]
Output:
((-4*a*(-(A*b) + a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*(a + b *Cos[c + d*x])) + ((2*(a*A*b + a^2*B - 2*b^2*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*(-(A*b) + a*B)*((a + b)*EllipticF[(c + d*x)/ 2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(a + b) + (2*(-(a*A* b) + 3*a^2*B - 2*b^2*B)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*E llipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sq rt[Sin[c + d*x]^2]))/((a - b)*(a + b)))/(4*b*d)
Time = 1.34 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.98, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3468, 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.
\(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3468 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int -\frac {-\left (\left (-3 B a^2+A b a+2 b^2 B\right ) \cos ^2(c+d x)\right )-2 b (A b-a B) \cos (c+d x)+a (A b-a B)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {-\left (\left (-3 B a^2+A b a+2 b^2 B\right ) \cos ^2(c+d x)\right )-2 b (A b-a B) \cos (c+d x)+a (A b-a B)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\left (3 B a^2-A b a-2 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )+a (A b-a B)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {-\frac {\left (-3 a^2 B+a A b+2 b^2 B\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {a b (A b-a B)+\left (-3 B a^3+A b a^2+4 b^2 B a-2 A b^3\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {a b (A b-a B)+\left (-3 B a^3+A b a^2+4 b^2 B a-2 A b^3\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {\left (-3 a^2 B+a A b+2 b^2 B\right ) \int \sqrt {\cos (c+d x)}dx}{b}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {a b (A b-a B)+\left (-3 B a^3+A b a^2+4 b^2 B a-2 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {\left (-3 a^2 B+a A b+2 b^2 B\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {\int \frac {a b (A b-a B)+\left (-3 B a^3+A b a^2+4 b^2 B a-2 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {\frac {\frac {\left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}-\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {\frac {2 \left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}+\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {\frac {2 \left (-3 a^3 B+a^2 A b+4 a b^2 B-2 A b^3\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {2 a \left (-3 a^3 B+a^2 A b+5 a b^2 B-3 A b^3\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}}{b}-\frac {2 \left (-3 a^2 B+a A b+2 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}\) |
Input:
Int[(Cos[c + d*x]^(3/2)*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^2,x]
Output:
((-2*(a*A*b - 3*a^2*B + 2*b^2*B)*EllipticE[(c + d*x)/2, 2])/(b*d) + ((2*(a ^2*A*b - 2*A*b^3 - 3*a^3*B + 4*a*b^2*B)*EllipticF[(c + d*x)/2, 2])/(b*d) - (2*a*(a^2*A*b - 3*A*b^3 - 3*a^3*B + 5*a*b^2*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d))/b)/(2*b*(a^2 - b^2)) + (a*(A*b - a*B)*Sqrt [Cos[c + d*x]]*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))
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[(-(b*c - a*d))*(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)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a *B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2, 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] && GtQ[m, 1] && LtQ[n, -1]
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_)] + (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. \(848\) vs. \(2(227)=454\).
Time = 8.07 (sec) , antiderivative size = 849, normalized size of antiderivative = 3.79
Input:
int(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^2,x,method=_RETURNV ERBOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/b^3/(-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)*(A*b*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) )-2*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a-B*b*EllipticE(cos(1/2*d*x+1/ 2*c),2^(1/2)))+2*a^2*(A*b-B*a)/b^3*(-b^2/a/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(- 2*sin(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/(a+b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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)*EllipticF(co s(1/2*d*x+1/2*c),2^(1/2))-1/2*b/(a^2-b^2)/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*( -2*cos(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*b/(a^2-b^2)/a*(sin(1 /2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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)*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)*(-2*cos(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)*E llipticPi(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)*(-2*cos(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)*EllipticPi(cos(1/2*d*x+1 /2*c),-2*b/(a-b),2^(1/2)))+4*a/b^2*(2*A*b-3*B*a)/(-2*a*b+2*b^2)*(sin(1/2*d *x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/...
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algorith m="fricas")
Output:
Timed out
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(3/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))**2,x)
Output:
Timed out
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algorith m="maxima")
Output:
integrate((B*cos(d*x + c) + A)*cos(d*x + c)^(3/2)/(b*cos(d*x + c) + a)^2, x)
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x, algorith m="giac")
Output:
integrate((B*cos(d*x + c) + A)*cos(d*x + c)^(3/2)/(b*cos(d*x + c) + a)^2, x)
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \] Input:
int((cos(c + d*x)^(3/2)*(A + B*cos(c + d*x)))/(a + b*cos(c + d*x))^2,x)
Output:
int((cos(c + d*x)^(3/2)*(A + B*cos(c + d*x)))/(a + b*cos(c + d*x))^2, x)
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx=\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right ) b +a}d x \] Input:
int(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^2,x)
Output:
int((sqrt(cos(c + d*x))*cos(c + d*x))/(cos(c + d*x)*b + a),x)