Integrand size = 23, antiderivative size = 264 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {3 a \left (a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^2 \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^4-5 a^2 b^2+8 b^4\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {3 a \left (a^4-2 a^2 b^2+5 b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 (a-b)^2 b^3 (a+b)^3 d}-\frac {a^2 \sqrt {\cos (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {3 a \left (a^2-3 b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:
-3/4*a*(a^2-3*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b^2/(a^2-b^2)^2/d +1/4*(3*a^4-5*a^2*b^2+8*b^4)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/b^3/(a ^2-b^2)^2/d-3/4*a*(a^4-2*a^2*b^2+5*b^4)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/ (a+b),2^(1/2))/(a-b)^2/b^3/(a+b)^3/d-1/2*a^2*cos(d*x+c)^(1/2)*sin(d*x+c)/b /(a^2-b^2)/d/(a+b*cos(d*x+c))^2+3/4*a*(a^2-3*b^2)*cos(d*x+c)^(1/2)*sin(d*x +c)/b/(a^2-b^2)^2/d/(a+b*cos(d*x+c))
Time = 2.44 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\frac {4 a \sqrt {\cos (c+d x)} \left (a^3-7 a b^2+3 b \left (a^2-3 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}-\frac {\frac {2 \left (a^3+5 a b^2\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}-\frac {16 \left (a^2+2 b^2\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {6 \left (a^2-3 b^2\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{16 b d} \] Input:
Integrate[Cos[c + d*x]^(5/2)/(a + b*Cos[c + d*x])^3,x]
Output:
((4*a*Sqrt[Cos[c + d*x]]*(a^3 - 7*a*b^2 + 3*b*(a^2 - 3*b^2)*Cos[c + d*x])* Sin[c + d*x])/((a^2 - b^2)^2*(a + b*Cos[c + d*x])^2) - ((2*(a^3 + 5*a*b^2) *EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) - (16*(a^2 + 2*b^2)*(( a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2 , 2]))/(a + b) + (6*(a^2 - 3*b^2)*(-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)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x]) /(b^2*Sqrt[Sin[c + d*x]^2]))/((a - b)^2*(a + b)^2))/(16*b*d)
Time = 1.69 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.03, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 3271, 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.
\(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 3271 |
\(\displaystyle -\frac {\int \frac {a^2-4 b \cos (c+d x) a-\left (3 a^2-4 b^2\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}dx}{2 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {a^2-4 b \cos (c+d x) a-\left (3 a^2-4 b^2\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}dx}{4 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {a^2-4 b \sin \left (c+d x+\frac {\pi }{2}\right ) a+\left (4 b^2-3 a^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{4 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle -\frac {\frac {\int -\frac {-3 \left (a^2-3 b^2\right ) \cos ^2(c+d x) a^2+\left (a^2-7 b^2\right ) a^2+4 b \left (a^2+2 b^2\right ) \cos (c+d x) a}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a \left (a^2-b^2\right )}-\frac {3 a \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {-\frac {\int \frac {-3 \left (a^2-3 b^2\right ) \cos ^2(c+d x) a^2+\left (a^2-7 b^2\right ) a^2+4 b \left (a^2+2 b^2\right ) \cos (c+d x) a}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{2 a \left (a^2-b^2\right )}-\frac {3 a \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {-\frac {\int \frac {-3 \left (a^2-3 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^2+\left (a^2-7 b^2\right ) a^2+4 b \left (a^2+2 b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}-\frac {3 a \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle -\frac {-\frac {-\frac {3 a^2 \left (a^2-3 b^2\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {b \left (a^2-7 b^2\right ) a^2+\left (3 a^4-5 b^2 a^2+8 b^4\right ) \cos (c+d x) a}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{2 a \left (a^2-b^2\right )}-\frac {3 a \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {-\frac {\frac {\int \frac {b \left (a^2-7 b^2\right ) a^2+\left (3 a^4-5 b^2 a^2+8 b^4\right ) \cos (c+d x) a}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {3 a^2 \left (a^2-3 b^2\right ) \int \sqrt {\cos (c+d x)}dx}{b}}{2 a \left (a^2-b^2\right )}-\frac {3 a \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {-\frac {\frac {\int \frac {b \left (a^2-7 b^2\right ) a^2+\left (3 a^4-5 b^2 a^2+8 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {3 a^2 \left (a^2-3 b^2\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{2 a \left (a^2-b^2\right )}-\frac {3 a \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle -\frac {-\frac {\frac {\int \frac {b \left (a^2-7 b^2\right ) a^2+\left (3 a^4-5 b^2 a^2+8 b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {6 a^2 \left (a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}-\frac {3 a \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle -\frac {-\frac {\frac {\frac {a \left (3 a^4-5 a^2 b^2+8 b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}-\frac {3 a^2 \left (a^4-2 a^2 b^2+5 b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{b}-\frac {6 a^2 \left (a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}-\frac {3 a \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {-\frac {\frac {\frac {a \left (3 a^4-5 a^2 b^2+8 b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {3 a^2 \left (a^4-2 a^2 b^2+5 b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}-\frac {6 a^2 \left (a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}-\frac {3 a \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {-\frac {\frac {\frac {2 a \left (3 a^4-5 a^2 b^2+8 b^4\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {3 a^2 \left (a^4-2 a^2 b^2+5 b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}-\frac {6 a^2 \left (a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}-\frac {3 a \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 b \left (a^2-b^2\right )}-\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle -\frac {a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {-\frac {3 a \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\frac {2 a \left (3 a^4-5 a^2 b^2+8 b^4\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {6 a^2 \left (a^4-2 a^2 b^2+5 b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}}{b}-\frac {6 a^2 \left (a^2-3 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}}{4 b \left (a^2-b^2\right )}\) |
Input:
Int[Cos[c + d*x]^(5/2)/(a + b*Cos[c + d*x])^3,x]
Output:
-1/2*(a^2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - (-1/2*((-6*a^2*(a^2 - 3*b^2)*EllipticE[(c + d*x)/2, 2])/(b*d) + ((2*a*(3*a^4 - 5*a^2*b^2 + 8*b^4)*EllipticF[(c + d*x)/2, 2])/(b*d) - (6* a^2*(a^4 - 2*a^2*b^2 + 5*b^4)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/( b*(a + b)*d))/b)/(a*(a^2 - b^2)) - (3*a*(a^2 - 3*b^2)*Sqrt[Cos[c + d*x]]*S in[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x])))/(4*b*(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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co s[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((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 - 3)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^ 2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, 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[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
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_)]), 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[(-(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. \(1913\) vs. \(2(255)=510\).
Time = 80.74 (sec) , antiderivative size = 1914, normalized size of antiderivative = 7.25
Input:
int(cos(d*x+c)^(5/2)/(a+cos(d*x+c)*b)^3,x,method=_RETURNVERBOSE)
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*(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))+12 /b^2*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+1/2*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))+6/b^3*a^2*(-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(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+s in(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)*EllipticPi(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)*EllipticP i(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))-2/b^3*a^3*(-1/2*b^2/a/(a^2-b^...
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(5/2)/(a+b*cos(d*x+c))**3,x)
Output:
Timed out
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:
integrate(cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")
Output:
integrate(cos(d*x + c)^(5/2)/(b*cos(d*x + c) + a)^3, x)
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:
integrate(cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")
Output:
integrate(cos(d*x + c)^(5/2)/(b*cos(d*x + c) + a)^3, x)
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \] Input:
int(cos(c + d*x)^(5/2)/(a + b*cos(c + d*x))^3,x)
Output:
int(cos(c + d*x)^(5/2)/(a + b*cos(c + d*x))^3, x)
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\int \frac {\sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{3} b^{3}+3 \cos \left (d x +c \right )^{2} a \,b^{2}+3 \cos \left (d x +c \right ) a^{2} b +a^{3}}d x \] Input:
int(cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^3,x)
Output:
int((sqrt(cos(c + d*x))*cos(c + d*x)**2)/(cos(c + d*x)**3*b**3 + 3*cos(c + d*x)**2*a*b**2 + 3*cos(c + d*x)*a**2*b + a**3),x)