3.1435 \(\int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx\)

Optimal. Leaf size=424 \[ \frac {b^2 \sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{a^3 d^2 f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}+\frac {2 b \sqrt {g \cos (e+f x)}}{a^2 d^2 f g \sqrt {d \sin (e+f x)}}+\frac {2 \sqrt {2} b^3 \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{a^3 d^{5/2} f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {2} b^3 \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{a^3 d^{5/2} f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{3 a d^2 f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {g \cos (e+f x)}}{3 a d f g (d \sin (e+f x))^{3/2}} \]

[Out]

2*b^3*EllipticPi((d*sin(f*x+e))^(1/2)/d^(1/2)/(1+cos(f*x+e))^(1/2),-a/(b-(-a^2+b^2)^(1/2)),I)*2^(1/2)*cos(f*x+
e)^(1/2)/a^3/d^(5/2)/f/(-a^2+b^2)^(1/2)/(g*cos(f*x+e))^(1/2)-2*b^3*EllipticPi((d*sin(f*x+e))^(1/2)/d^(1/2)/(1+
cos(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)),I)*2^(1/2)*cos(f*x+e)^(1/2)/a^3/d^(5/2)/f/(-a^2+b^2)^(1/2)/(g*cos(f*
x+e))^(1/2)-2/3*(g*cos(f*x+e))^(1/2)/a/d/f/g/(d*sin(f*x+e))^(3/2)+2*b*(g*cos(f*x+e))^(1/2)/a^2/d^2/f/g/(d*sin(
f*x+e))^(1/2)-2/3*(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi+f*x)*EllipticF(cos(e+1/4*Pi+f*x),2^(1/2))*sin(2*f*x
+2*e)^(1/2)/a/d^2/f/(g*cos(f*x+e))^(1/2)/(d*sin(f*x+e))^(1/2)-b^2*(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi+f*x
)*EllipticF(cos(e+1/4*Pi+f*x),2^(1/2))*sin(2*f*x+2*e)^(1/2)/a^3/d^2/f/(g*cos(f*x+e))^(1/2)/(d*sin(f*x+e))^(1/2
)

________________________________________________________________________________________

Rubi [A]  time = 1.16, antiderivative size = 424, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {2910, 2570, 2573, 2641, 2563, 2908, 2907, 1218} \[ \frac {b^2 \sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{a^3 d^2 f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} b^3 \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{a^3 d^{5/2} f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {2} b^3 \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{a^3 d^{5/2} f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {2 b \sqrt {g \cos (e+f x)}}{a^2 d^2 f g \sqrt {d \sin (e+f x)}}+\frac {2 \sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{3 a d^2 f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {g \cos (e+f x)}}{3 a d f g (d \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[g*Cos[e + f*x]]*(d*Sin[e + f*x])^(5/2)*(a + b*Sin[e + f*x])),x]

[Out]

(2*Sqrt[2]*b^3*Sqrt[Cos[e + f*x]]*EllipticPi[-(a/(b - Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*Sin[e + f*x]]/(Sqrt[d]
*Sqrt[1 + Cos[e + f*x]])], -1])/(a^3*Sqrt[-a^2 + b^2]*d^(5/2)*f*Sqrt[g*Cos[e + f*x]]) - (2*Sqrt[2]*b^3*Sqrt[Co
s[e + f*x]]*EllipticPi[-(a/(b + Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*Sin[e + f*x]]/(Sqrt[d]*Sqrt[1 + Cos[e + f*x]
])], -1])/(a^3*Sqrt[-a^2 + b^2]*d^(5/2)*f*Sqrt[g*Cos[e + f*x]]) - (2*Sqrt[g*Cos[e + f*x]])/(3*a*d*f*g*(d*Sin[e
+ f*x])^(3/2)) + (2*b*Sqrt[g*Cos[e + f*x]])/(a^2*d^2*f*g*Sqrt[d*Sin[e + f*x]]) + (2*EllipticF[e - Pi/4 + f*x,
2]*Sqrt[Sin[2*e + 2*f*x]])/(3*a*d^2*f*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]]) + (b^2*EllipticF[e - Pi/4 +
f*x, 2]*Sqrt[Sin[2*e + 2*f*x]])/(a^3*d^2*f*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]])

Rule 1218

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(c/a), 4]}, Simp[(1*Ellipt
icPi[-(e/(d*q^2)), ArcSin[q*x], -1])/(d*Sqrt[a]*q), x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]

Rule 2563

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Simp[((a*Sin[e +
f*x])^(m + 1)*(b*Cos[e + f*x])^(n + 1))/(a*b*f*(m + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 2,
0] && NeQ[m, -1]

Rule 2570

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((b*Cos[e + f
*x])^(n + 1)*(a*Sin[e + f*x])^(m + 1))/(a*b*f*(m + 1)), x] + Dist[(m + n + 2)/(a^2*(m + 1)), Int[(b*Cos[e + f*
x])^n*(a*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2907

Int[Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[cos[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]))
, x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Dist[(2*Sqrt[2]*d*(b + q))/(f*q), Subst[Int[1/((d*(b + q) + a*x^2
)*Sqrt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]], x] - Dist[(2*Sqrt[2]*d*(b - q))/(f*
q), Subst[Int[1/((d*(b - q) + a*x^2)*Sqrt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]],
x]] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2908

Int[Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(
x_)])), x_Symbol] :> Dist[Sqrt[Cos[e + f*x]]/Sqrt[g*Cos[e + f*x]], Int[Sqrt[d*Sin[e + f*x]]/(Sqrt[Cos[e + f*x]
]*(a + b*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2910

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[1/a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] - Dist[b/(a*d), Int[((g*Cos
[e + f*x])^p*(d*Sin[e + f*x])^(n + 1))/(a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2
- b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[-1, p, 1] && LtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx &=\frac {\int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{5/2}} \, dx}{a}-\frac {b \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx}{a d}\\ &=-\frac {2 \sqrt {g \cos (e+f x)}}{3 a d f g (d \sin (e+f x))^{3/2}}+\frac {2 \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \, dx}{3 a d^2}+\frac {b^2 \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx}{a^2 d^2}-\frac {b \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2}} \, dx}{a^2 d}\\ &=-\frac {2 \sqrt {g \cos (e+f x)}}{3 a d f g (d \sin (e+f x))^{3/2}}+\frac {2 b \sqrt {g \cos (e+f x)}}{a^2 d^2 f g \sqrt {d \sin (e+f x)}}-\frac {b^3 \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a^3 d^3}+\frac {b^2 \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \, dx}{a^3 d^2}+\frac {\left (2 \sqrt {\sin (2 e+2 f x)}\right ) \int \frac {1}{\sqrt {\sin (2 e+2 f x)}} \, dx}{3 a d^2 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}\\ &=-\frac {2 \sqrt {g \cos (e+f x)}}{3 a d f g (d \sin (e+f x))^{3/2}}+\frac {2 b \sqrt {g \cos (e+f x)}}{a^2 d^2 f g \sqrt {d \sin (e+f x)}}+\frac {2 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{3 a d^2 f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}-\frac {\left (b^3 \sqrt {\cos (e+f x)}\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a^3 d^3 \sqrt {g \cos (e+f x)}}+\frac {\left (b^2 \sqrt {\sin (2 e+2 f x)}\right ) \int \frac {1}{\sqrt {\sin (2 e+2 f x)}} \, dx}{a^3 d^2 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}\\ &=-\frac {2 \sqrt {g \cos (e+f x)}}{3 a d f g (d \sin (e+f x))^{3/2}}+\frac {2 b \sqrt {g \cos (e+f x)}}{a^2 d^2 f g \sqrt {d \sin (e+f x)}}+\frac {2 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{3 a d^2 f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}+\frac {b^2 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{a^3 d^2 f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}-\frac {\left (2 \sqrt {2} b^3 \left (1-\frac {b}{\sqrt {-a^2+b^2}}\right ) \sqrt {\cos (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\left (b-\sqrt {-a^2+b^2}\right ) d+a x^2\right ) \sqrt {1-\frac {x^4}{d^2}}} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {1+\cos (e+f x)}}\right )}{a^3 d^2 f \sqrt {g \cos (e+f x)}}-\frac {\left (2 \sqrt {2} b^3 \left (1+\frac {b}{\sqrt {-a^2+b^2}}\right ) \sqrt {\cos (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\left (b+\sqrt {-a^2+b^2}\right ) d+a x^2\right ) \sqrt {1-\frac {x^4}{d^2}}} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {1+\cos (e+f x)}}\right )}{a^3 d^2 f \sqrt {g \cos (e+f x)}}\\ &=\frac {2 \sqrt {2} b^3 \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right )\right |-1\right )}{a^3 \sqrt {-a^2+b^2} d^{5/2} f \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {2} b^3 \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right )\right |-1\right )}{a^3 \sqrt {-a^2+b^2} d^{5/2} f \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {g \cos (e+f x)}}{3 a d f g (d \sin (e+f x))^{3/2}}+\frac {2 b \sqrt {g \cos (e+f x)}}{a^2 d^2 f g \sqrt {d \sin (e+f x)}}+\frac {2 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{3 a d^2 f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}+\frac {b^2 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{a^3 d^2 f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 20.23, size = 1140, normalized size = 2.69 \[ \frac {\cos (e+f x) \left (\frac {2 b \csc (e+f x)}{a^2}-\frac {2 \csc ^2(e+f x)}{3 a}\right ) \sin ^3(e+f x)}{f \sqrt {g \cos (e+f x)} (d \sin (e+f x))^{5/2}}+\frac {\sqrt {\cos (e+f x)} \left (\frac {4 a b \sqrt {\sin (e+f x)} \left (\frac {\sqrt {a} \left (-2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}}{\sqrt {a}}\right )+2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}}{\sqrt {a}}+1\right )+\log \left (-a+\sqrt {2} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)} \sqrt {a}-\sqrt {a^2-b^2} \tan (e+f x)\right )-\log \left (a+\sqrt {2} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)} \sqrt {a}+\sqrt {a^2-b^2} \tan (e+f x)\right )\right )}{4 \sqrt {2} \left (a^2-b^2\right )^{3/4}}-\frac {b F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};-\tan ^2(e+f x),\frac {\left (b^2-a^2\right ) \tan ^2(e+f x)}{a^2}\right ) \tan ^{\frac {5}{2}}(e+f x)}{5 a^2}\right ) \left (\sqrt {\tan ^2(e+f x)+1} a+b \tan (e+f x)\right )}{\cos ^{\frac {5}{2}}(e+f x) (a+b \sin (e+f x)) \sqrt {\tan (e+f x)} \left (\tan ^2(e+f x)+1\right )^{3/2}}-\frac {2 \left (2 a^2+3 b^2\right ) \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {5 a \left (a^2-b^2\right ) F_1\left (\frac {1}{4};\frac {3}{4},1;\frac {5}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right ) \sqrt {\cos (e+f x)}}{\left (1-\cos ^2(e+f x)\right )^{3/4} \left (\left (3 \left (a^2-b^2\right ) F_1\left (\frac {5}{4};\frac {7}{4},1;\frac {9}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )-4 b^2 F_1\left (\frac {5}{4};\frac {3}{4},2;\frac {9}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )\right ) \cos ^2(e+f x)+5 \left (a^2-b^2\right ) F_1\left (\frac {1}{4};\frac {3}{4},1;\frac {5}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )\right ) \left (a^2+b^2 \left (\cos ^2(e+f x)-1\right )\right )}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) b \left (2 \tan ^{-1}\left (1-\frac {(1+i) \sqrt {a} \sqrt {\cos (e+f x)}}{\sqrt [4]{b^2-a^2} \sqrt [4]{\cos ^2(e+f x)-1}}\right )-2 \tan ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\cos (e+f x)}}{\sqrt [4]{b^2-a^2} \sqrt [4]{\cos ^2(e+f x)-1}}+1\right )+\log \left (\frac {i a \cos (e+f x)}{\sqrt {\cos ^2(e+f x)-1}}-\frac {(1+i) \sqrt {a} \sqrt [4]{b^2-a^2} \sqrt {\cos (e+f x)}}{\sqrt [4]{\cos ^2(e+f x)-1}}+\sqrt {b^2-a^2}\right )-\log \left (\frac {i a \cos (e+f x)}{\sqrt {\cos ^2(e+f x)-1}}+\frac {(1+i) \sqrt {a} \sqrt [4]{b^2-a^2} \sqrt {\cos (e+f x)}}{\sqrt [4]{\cos ^2(e+f x)-1}}+\sqrt {b^2-a^2}\right )\right )}{\sqrt {a} \left (b^2-a^2\right )^{3/4}}\right ) \sqrt {\sin (e+f x)}}{\sqrt [4]{1-\cos ^2(e+f x)} (a+b \sin (e+f x))}\right ) \sin ^{\frac {5}{2}}(e+f x)}{3 a^2 f \sqrt {g \cos (e+f x)} (d \sin (e+f x))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sqrt[g*Cos[e + f*x]]*(d*Sin[e + f*x])^(5/2)*(a + b*Sin[e + f*x])),x]

[Out]

(Cos[e + f*x]*((2*b*Csc[e + f*x])/a^2 - (2*Csc[e + f*x]^2)/(3*a))*Sin[e + f*x]^3)/(f*Sqrt[g*Cos[e + f*x]]*(d*S
in[e + f*x])^(5/2)) + (Sqrt[Cos[e + f*x]]*Sin[e + f*x]^(5/2)*((-2*(2*a^2 + 3*b^2)*(a + b*Sqrt[1 - Cos[e + f*x]
^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 3/4, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[
e + f*x]])/((1 - Cos[e + f*x]^2)^(3/4)*(5*(a^2 - b^2)*AppellF1[1/4, 3/4, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e +
f*x]^2)/(-a^2 + b^2)] + (-4*b^2*AppellF1[5/4, 3/4, 2, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]
+ 3*(a^2 - b^2)*AppellF1[5/4, 7/4, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^2)
*(a^2 + b^2*(-1 + Cos[e + f*x]^2))) - ((1/8 - I/8)*b*(2*ArcTan[1 - ((1 + I)*Sqrt[a]*Sqrt[Cos[e + f*x]])/((-a^2
+ b^2)^(1/4)*(-1 + Cos[e + f*x]^2)^(1/4))] - 2*ArcTan[1 + ((1 + I)*Sqrt[a]*Sqrt[Cos[e + f*x]])/((-a^2 + b^2)^
(1/4)*(-1 + Cos[e + f*x]^2)^(1/4))] + Log[Sqrt[-a^2 + b^2] + (I*a*Cos[e + f*x])/Sqrt[-1 + Cos[e + f*x]^2] - ((
1 + I)*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]])/(-1 + Cos[e + f*x]^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] + (I
*a*Cos[e + f*x])/Sqrt[-1 + Cos[e + f*x]^2] + ((1 + I)*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]])/(-1 + Cos
[e + f*x]^2)^(1/4)]))/(Sqrt[a]*(-a^2 + b^2)^(3/4)))*Sqrt[Sin[e + f*x]])/((1 - Cos[e + f*x]^2)^(1/4)*(a + b*Sin
[e + f*x])) + (4*a*b*Sqrt[Sin[e + f*x]]*((Sqrt[a]*(-2*ArcTan[1 - (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]]
)/Sqrt[a]] + 2*ArcTan[1 + (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a]] + Log[-a + Sqrt[2]*Sqrt[a]*(
a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]] - Sqrt[a^2 - b^2]*Tan[e + f*x]] - Log[a + Sqrt[2]*Sqrt[a]*(a^2 - b^2)^(1/4
)*Sqrt[Tan[e + f*x]] + Sqrt[a^2 - b^2]*Tan[e + f*x]]))/(4*Sqrt[2]*(a^2 - b^2)^(3/4)) - (b*AppellF1[5/4, 1/2, 1
, 9/4, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2)/a^2]*Tan[e + f*x]^(5/2))/(5*a^2))*(b*Tan[e + f*x] + a*Sq
rt[1 + Tan[e + f*x]^2]))/(Cos[e + f*x]^(5/2)*(a + b*Sin[e + f*x])*Sqrt[Tan[e + f*x]]*(1 + Tan[e + f*x]^2)^(3/2
))))/(3*a^2*f*Sqrt[g*Cos[e + f*x]]*(d*Sin[e + f*x])^(5/2))

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(g*cos(f*x + e))*(b*sin(f*x + e) + a)*(d*sin(f*x + e))^(5/2)), x)

________________________________________________________________________________________

maple [B]  time = 0.78, size = 2987, normalized size = 7.04 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x)

[Out]

-1/3/f*(3*cos(f*x+e)*sin(f*x+e)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),a/(a-b+(-a^2+b^2)^(1
/2)),1/2*2^(1/2))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)
*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*(-a^2+b^2)^(1/2)*b^3+3*cos(f*x+e)*sin(f*x+e)*EllipticPi((-(-1+cos(f*x+e)-s
in(f*x+e))/sin(f*x+e))^(1/2),a/(a-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1
/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*a*b^3-3*cos(f*x+e)*sin(f*
x+e)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),a/(a-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*(-(-1+cos
(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e
))^(1/2)*b^4+3*cos(f*x+e)*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e)
)/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/
2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*b^3-3*cos(f*x+e)*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*
x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*Ellip
ticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*a*b^3+3*cos(f*x+e)
*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+
cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)
-a),1/2*2^(1/2))*b^4+4*cos(f*x+e)*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+si
n(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+
e))^(1/2),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a^3-4*cos(f*x+e)*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(
1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+
e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a^2*b+6*cos(f*x+e)*sin(f*x+e)*(-(-1+cos(f*x+e)-
sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)
*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a*b^2-6*cos(f*x+e)*sin
(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(
f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-a^2+b^2)^(1/
2)*b^3+3*sin(f*x+e)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),a/(a-b+(-a^2+b^2)^(1/2)),1/2*2^(
1/2))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*
x+e))/sin(f*x+e))^(1/2)*(-a^2+b^2)^(1/2)*b^3+3*sin(f*x+e)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^
(1/2),a/(a-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin
(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*a*b^3-3*sin(f*x+e)*EllipticPi((-(-1+cos(f*x+e)-s
in(f*x+e))/sin(f*x+e))^(1/2),a/(a-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1
/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*b^4+3*sin(f*x+e)*(-(-1+co
s(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+
e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*(-a
^2+b^2)^(1/2)*b^3-3*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(
f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/
(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*a*b^3+3*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos
(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e
))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*b^4+4*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(
f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1
+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a^3-4*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(
f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*Ell
ipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a^2*b+6*sin(f*x+e)*(-(-1+c
os(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x
+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a*b^2-6*sin(
f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f
*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-a^2+b^2)^(1/2
)*b^3+6*2^(1/2)*(-a^2+b^2)^(1/2)*cos(f*x+e)*sin(f*x+e)*a^2*b-6*cos(f*x+e)*sin(f*x+e)*2^(1/2)*(-a^2+b^2)^(1/2)*
a*b^2-2*(-a^2+b^2)^(1/2)*cos(f*x+e)*2^(1/2)*a^3+2*cos(f*x+e)*2^(1/2)*(-a^2+b^2)^(1/2)*a^2*b)*sin(f*x+e)/(d*sin
(f*x+e))^(5/2)/(g*cos(f*x+e))^(1/2)*2^(1/2)/(-a^2+b^2)^(1/2)/(a-b+(-a^2+b^2)^(1/2))/(b+(-a^2+b^2)^(1/2)-a)/a^2

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(g*cos(f*x + e))*(b*sin(f*x + e) + a)*(d*sin(f*x + e))^(5/2)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{\sqrt {g\,\cos \left (e+f\,x\right )}\,{\left (d\,\sin \left (e+f\,x\right )\right )}^{5/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((g*cos(e + f*x))^(1/2)*(d*sin(e + f*x))^(5/2)*(a + b*sin(e + f*x))),x)

[Out]

int(1/((g*cos(e + f*x))^(1/2)*(d*sin(e + f*x))^(5/2)*(a + b*sin(e + f*x))), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sin(f*x+e))**(5/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________