Integrand size = 23, antiderivative size = 261 \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=-\frac {3 b \left (3 a^2-b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}-\frac {\left (7 a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a \left (a^2-b^2\right )^2 d}+\frac {3 \left (5 a^4-2 a^2 b^2+b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^2 (a-b)^2 (a+b)^3 d}+\frac {b^2 \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {3 b^2 \left (3 a^2-b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \] Output:
-3/4*b*(3*a^2-b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/(a^2-b^2)^2/d -1/4*(7*a^2-b^2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a/(a^2-b^2)^2/d+3/ 4*(5*a^4-2*a^2*b^2+b^4)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/a ^2/(a-b)^2/(a+b)^3/d+1/2*b^2*cos(d*x+c)^(1/2)*sin(d*x+c)/a/(a^2-b^2)/d/(a+ b*cos(d*x+c))^2+3/4*b^2*(3*a^2-b^2)*cos(d*x+c)^(1/2)*sin(d*x+c)/a^2/(a^2-b ^2)^2/d/(a+b*cos(d*x+c))
Time = 3.14 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.15 \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\frac {\frac {4 b^2 \sqrt {\cos (c+d x)} \left (11 a^3-5 a b^2+\left (9 a^2 b-3 b^3\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {\frac {2 \left (16 a^4-19 a^2 b^2+9 b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {16 \left (-4 a^3+a b^2\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}-\frac {6 \left (3 a^2-b^2\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{16 a^2 d} \] Input:
Integrate[1/(Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^3),x]
Output:
((4*b^2*Sqrt[Cos[c + d*x]]*(11*a^3 - 5*a*b^2 + (9*a^2*b - 3*b^3)*Cos[c + d *x])*Sin[c + d*x])/((a^2 - b^2)^2*(a + b*Cos[c + d*x])^2) + ((2*(16*a^4 - 19*a^2*b^2 + 9*b^4)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + ( 16*(-4*a^3 + a*b^2)*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b )/(a + b), (c + d*x)/2, 2]))/(a + b) - (6*(3*a^2 - 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])/(a*Sqrt[Sin[c + d*x]^2]))/((a - b)^2*(a + b)^2))/(1 6*a^2*d)
Time = 1.77 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 3281, 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.
\(\displaystyle \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 3281 |
\(\displaystyle \frac {\int \frac {4 a^2-4 b \cos (c+d x) a-3 b^2+b^2 \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {4 a^2-4 b \cos (c+d x) a-3 b^2+b^2 \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}dx}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {4 a^2-4 b \sin \left (c+d x+\frac {\pi }{2}\right ) a-3 b^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\frac {\int \frac {8 a^4-5 b^2 a^2-4 b \left (4 a^2-b^2\right ) \cos (c+d x) a+3 b^4-3 b^2 \left (3 a^2-b^2\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a \left (a^2-b^2\right )}+\frac {3 b^2 \left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {8 a^4-5 b^2 a^2-4 b \left (4 a^2-b^2\right ) \cos (c+d x) a+3 b^4-3 b^2 \left (3 a^2-b^2\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{2 a \left (a^2-b^2\right )}+\frac {3 b^2 \left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {8 a^4-5 b^2 a^2-4 b \left (4 a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+3 b^4-3 b^2 \left (3 a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}+\frac {3 b^2 \left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {\frac {-3 b \left (3 a^2-b^2\right ) \int \sqrt {\cos (c+d x)}dx-\frac {\int -\frac {b \left (8 a^4-5 b^2 a^2+3 b^4\right )-a b^2 \left (7 a^2-b^2\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{2 a \left (a^2-b^2\right )}+\frac {3 b^2 \left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\frac {\int \frac {b \left (8 a^4-5 b^2 a^2+3 b^4\right )-a b^2 \left (7 a^2-b^2\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-3 b \left (3 a^2-b^2\right ) \int \sqrt {\cos (c+d x)}dx}{2 a \left (a^2-b^2\right )}+\frac {3 b^2 \left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\int \frac {b \left (8 a^4-5 b^2 a^2+3 b^4\right )-a b^2 \left (7 a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-3 b \left (3 a^2-b^2\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a \left (a^2-b^2\right )}+\frac {3 b^2 \left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {\frac {\int \frac {b \left (8 a^4-5 b^2 a^2+3 b^4\right )-a b^2 \left (7 a^2-b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {6 b \left (3 a^2-b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a \left (a^2-b^2\right )}+\frac {3 b^2 \left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {\frac {\frac {3 b \left (5 a^4-2 a^2 b^2+b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx-a b \left (7 a^2-b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}-\frac {6 b \left (3 a^2-b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a \left (a^2-b^2\right )}+\frac {3 b^2 \left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {3 b \left (5 a^4-2 a^2 b^2+b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx-a b \left (7 a^2-b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {6 b \left (3 a^2-b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a \left (a^2-b^2\right )}+\frac {3 b^2 \left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {\frac {3 b \left (5 a^4-2 a^2 b^2+b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx-\frac {2 a b \left (7 a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b}-\frac {6 b \left (3 a^2-b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a \left (a^2-b^2\right )}+\frac {3 b^2 \left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{4 a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {b^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {\frac {3 b^2 \left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {\frac {6 b \left (5 a^4-2 a^2 b^2+b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}-\frac {2 a b \left (7 a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b}-\frac {6 b \left (3 a^2-b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\) |
Input:
Int[1/(Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^3),x]
Output:
(b^2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d* x])^2) + (((-6*b*(3*a^2 - b^2)*EllipticE[(c + d*x)/2, 2])/d + ((-2*a*b*(7* a^2 - b^2)*EllipticF[(c + d*x)/2, 2])/d + (6*b*(5*a^4 - 2*a^2*b^2 + b^4)*E llipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a + b)*d))/b)/(2*a*(a^2 - b^2) ) + (3*b^2*(3*a^2 - b^2)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(a*(a^2 - b^2)*d *(a + b*Cos[c + d*x])))/(4*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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*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[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2* n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 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[(((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. \(1175\) vs. \(2(252)=504\).
Time = 3.11 (sec) , antiderivative size = 1176, normalized size of antiderivative = 4.51
Input:
int(1/cos(d*x+c)^(1/2)/(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)*(-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)^2-3/2*b^2*(3*a^2-b^2)/a^2/(a^2-b^2)^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*c os(1/2*d*x+1/2*c)^2+a-b)-7/4/(a+b)/(a^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2/(a+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))*b+3/4/(a+b)/(a^2-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+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 llipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-9/4*b/(a^2-b^2)^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+s in(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3/4*b^3/a ^2/(a^2-b^2)^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)*EllipticF(cos(1/2 *d*x+1/2*c),2^(1/2))+9/4*b/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co s(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))-3/4*b^3/a^2/(a^2-b^2)^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...
Timed out. \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate(1/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate(1/cos(d*x+c)**(1/2)/(a+b*cos(d*x+c))**3,x)
Output:
Timed out
Timed out. \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate(1/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")
Output:
Timed out
\[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sqrt {\cos \left (d x + c\right )}} \,d x } \] Input:
integrate(1/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")
Output:
integrate(1/((b*cos(d*x + c) + a)^3*sqrt(cos(d*x + c))), x)
Timed out. \[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\int \frac {1}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \] Input:
int(1/(cos(c + d*x)^(1/2)*(a + b*cos(c + d*x))^3),x)
Output:
int(1/(cos(c + d*x)^(1/2)*(a + b*cos(c + d*x))^3), x)
\[ \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4} b^{3}+3 \cos \left (d x +c \right )^{3} a \,b^{2}+3 \cos \left (d x +c \right )^{2} a^{2} b +\cos \left (d x +c \right ) a^{3}}d x \] Input:
int(1/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^3,x)
Output:
int(sqrt(cos(c + d*x))/(cos(c + d*x)**4*b**3 + 3*cos(c + d*x)**3*a*b**2 + 3*cos(c + d*x)**2*a**2*b + cos(c + d*x)*a**3),x)