3.1421 \(\int \frac {(g \cos (e+f x))^{3/2}}{(d \sin (e+f x))^{7/2} (a+b \sin (e+f x))} \, dx\)

Optimal. Leaf size=525 \[ -\frac {2 b g^2 \sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{3 a^2 d^3 f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}+\frac {2 b g \sqrt {g \cos (e+f x)}}{3 a^2 d^2 f (d \sin (e+f x))^{3/2}}-\frac {2 \sqrt {2} b^2 g^2 \sqrt {b^2-a^2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{a^4 d^{7/2} f \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} b^2 g^2 \sqrt {b^2-a^2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{a^4 d^{7/2} f \sqrt {g \cos (e+f x)}}+\frac {b g^2 \left (a^2-b^2\right ) \sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{a^4 d^3 f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}+\frac {2 g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}{a^3 d^3 f \sqrt {d \sin (e+f x)}}-\frac {8 g \sqrt {g \cos (e+f x)}}{5 a d^3 f \sqrt {d \sin (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{5 a d f (d \sin (e+f x))^{5/2}} \]

[Out]

-2*b^2*g^2*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)*(-a
^2+b^2)^(1/2)*cos(f*x+e)^(1/2)/a^4/d^(7/2)/f/(g*cos(f*x+e))^(1/2)+2*b^2*g^2*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)*(-a^2+b^2)^(1/2)*cos(f*x+e)^(1/2)/a^4/d^(7/2)/f/
(g*cos(f*x+e))^(1/2)-2/5*g*(g*cos(f*x+e))^(1/2)/a/d/f/(d*sin(f*x+e))^(5/2)+2/3*b*g*(g*cos(f*x+e))^(1/2)/a^2/d^
2/f/(d*sin(f*x+e))^(3/2)-8/5*g*(g*cos(f*x+e))^(1/2)/a/d^3/f/(d*sin(f*x+e))^(1/2)+2*(a^2-b^2)*g*(g*cos(f*x+e))^
(1/2)/a^3/d^3/f/(d*sin(f*x+e))^(1/2)+2/3*b*g^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^2/d^3/f/(g*cos(f*x+e))^(1/2)/(d*sin(f*x+e))^(1/2)-b*(a^2-b^2)*g^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^4/d^3
/f/(g*cos(f*x+e))^(1/2)/(d*sin(f*x+e))^(1/2)

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Rubi [A]  time = 1.35, antiderivative size = 525, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.243, Rules used = {2899, 2570, 2563, 2573, 2641, 2910, 2908, 2907, 1218} \[ \frac {b g^2 \left (a^2-b^2\right ) \sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{a^4 d^3 f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {2} b^2 g^2 \sqrt {b^2-a^2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{a^4 d^{7/2} f \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} b^2 g^2 \sqrt {b^2-a^2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{a^4 d^{7/2} f \sqrt {g \cos (e+f x)}}+\frac {2 g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}{a^3 d^3 f \sqrt {d \sin (e+f x)}}-\frac {2 b g^2 \sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{3 a^2 d^3 f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}+\frac {2 b g \sqrt {g \cos (e+f x)}}{3 a^2 d^2 f (d \sin (e+f x))^{3/2}}-\frac {8 g \sqrt {g \cos (e+f x)}}{5 a d^3 f \sqrt {d \sin (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{5 a d f (d \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(g*Cos[e + f*x])^(3/2)/((d*Sin[e + f*x])^(7/2)*(a + b*Sin[e + f*x])),x]

[Out]

(-2*Sqrt[2]*b^2*Sqrt[-a^2 + b^2]*g^2*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^4*d^(7/2)*f*Sqrt[g*Cos[e + f*x]]) + (2*Sqrt[2]*b^2*Sq
rt[-a^2 + b^2]*g^2*Sqrt[Cos[e + f*x]]*EllipticPi[-(a/(b + Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*Sin[e + f*x]]/(Sqr
t[d]*Sqrt[1 + Cos[e + f*x]])], -1])/(a^4*d^(7/2)*f*Sqrt[g*Cos[e + f*x]]) - (2*g*Sqrt[g*Cos[e + f*x]])/(5*a*d*f
*(d*Sin[e + f*x])^(5/2)) + (2*b*g*Sqrt[g*Cos[e + f*x]])/(3*a^2*d^2*f*(d*Sin[e + f*x])^(3/2)) - (8*g*Sqrt[g*Cos
[e + f*x]])/(5*a*d^3*f*Sqrt[d*Sin[e + f*x]]) + (2*(a^2 - b^2)*g*Sqrt[g*Cos[e + f*x]])/(a^3*d^3*f*Sqrt[d*Sin[e
+ f*x]]) - (2*b*g^2*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[Sin[2*e + 2*f*x]])/(3*a^2*d^3*f*Sqrt[g*Cos[e + f*x]]*Sqr
t[d*Sin[e + f*x]]) + (b*(a^2 - b^2)*g^2*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[Sin[2*e + 2*f*x]])/(a^4*d^3*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 2899

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] + (-Dist[(b*g^2)/(a
^2*d), Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] - Dist[(g^2*(a^2 - b^2))/(a^2*d^2), Int[(
(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 2))/(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] && GtQ[p, 1] && (LeQ[n, -2] || (EqQ[n, -3/2] && EqQ[p, 3/2]))

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 {(g \cos (e+f x))^{3/2}}{(d \sin (e+f x))^{7/2} (a+b \sin (e+f x))} \, dx &=\frac {g^2 \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{7/2}} \, dx}{a}-\frac {\left (\left (a^2-b^2\right ) g^2\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx}{a^2 d^2}-\frac {\left (b g^2\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{5/2}} \, dx}{a^2 d}\\ &=-\frac {2 g \sqrt {g \cos (e+f x)}}{5 a d f (d \sin (e+f x))^{5/2}}+\frac {2 b g \sqrt {g \cos (e+f x)}}{3 a^2 d^2 f (d \sin (e+f x))^{3/2}}-\frac {\left (2 b g^2\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \, dx}{3 a^2 d^3}+\frac {\left (b \left (a^2-b^2\right ) g^2\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx}{a^3 d^3}+\frac {\left (4 g^2\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2}} \, dx}{5 a d^2}-\frac {\left (\left (a^2-b^2\right ) g^2\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2}} \, dx}{a^3 d^2}\\ &=-\frac {2 g \sqrt {g \cos (e+f x)}}{5 a d f (d \sin (e+f x))^{5/2}}+\frac {2 b g \sqrt {g \cos (e+f x)}}{3 a^2 d^2 f (d \sin (e+f x))^{3/2}}-\frac {8 g \sqrt {g \cos (e+f x)}}{5 a d^3 f \sqrt {d \sin (e+f x)}}+\frac {2 \left (a^2-b^2\right ) g \sqrt {g \cos (e+f x)}}{a^3 d^3 f \sqrt {d \sin (e+f x)}}-\frac {\left (b^2 \left (a^2-b^2\right ) g^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a^4 d^4}+\frac {\left (b \left (a^2-b^2\right ) g^2\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \, dx}{a^4 d^3}-\frac {\left (2 b g^2 \sqrt {\sin (2 e+2 f x)}\right ) \int \frac {1}{\sqrt {\sin (2 e+2 f x)}} \, dx}{3 a^2 d^3 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}\\ &=-\frac {2 g \sqrt {g \cos (e+f x)}}{5 a d f (d \sin (e+f x))^{5/2}}+\frac {2 b g \sqrt {g \cos (e+f x)}}{3 a^2 d^2 f (d \sin (e+f x))^{3/2}}-\frac {8 g \sqrt {g \cos (e+f x)}}{5 a d^3 f \sqrt {d \sin (e+f x)}}+\frac {2 \left (a^2-b^2\right ) g \sqrt {g \cos (e+f x)}}{a^3 d^3 f \sqrt {d \sin (e+f x)}}-\frac {2 b g^2 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{3 a^2 d^3 f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}-\frac {\left (b^2 \left (a^2-b^2\right ) g^2 \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^4 d^4 \sqrt {g \cos (e+f x)}}+\frac {\left (b \left (a^2-b^2\right ) g^2 \sqrt {\sin (2 e+2 f x)}\right ) \int \frac {1}{\sqrt {\sin (2 e+2 f x)}} \, dx}{a^4 d^3 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}\\ &=-\frac {2 g \sqrt {g \cos (e+f x)}}{5 a d f (d \sin (e+f x))^{5/2}}+\frac {2 b g \sqrt {g \cos (e+f x)}}{3 a^2 d^2 f (d \sin (e+f x))^{3/2}}-\frac {8 g \sqrt {g \cos (e+f x)}}{5 a d^3 f \sqrt {d \sin (e+f x)}}+\frac {2 \left (a^2-b^2\right ) g \sqrt {g \cos (e+f x)}}{a^3 d^3 f \sqrt {d \sin (e+f x)}}-\frac {2 b g^2 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{3 a^2 d^3 f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}+\frac {b \left (a^2-b^2\right ) g^2 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{a^4 d^3 f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}-\frac {\left (2 \sqrt {2} b^2 \left (a^2-b^2\right ) \left (1-\frac {b}{\sqrt {-a^2+b^2}}\right ) g^2 \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^4 d^3 f \sqrt {g \cos (e+f x)}}-\frac {\left (2 \sqrt {2} b^2 \left (a^2-b^2\right ) \left (1+\frac {b}{\sqrt {-a^2+b^2}}\right ) g^2 \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^4 d^3 f \sqrt {g \cos (e+f x)}}\\ &=-\frac {2 \sqrt {2} b^2 \sqrt {-a^2+b^2} g^2 \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^4 d^{7/2} f \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} b^2 \sqrt {-a^2+b^2} g^2 \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^4 d^{7/2} f \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{5 a d f (d \sin (e+f x))^{5/2}}+\frac {2 b g \sqrt {g \cos (e+f x)}}{3 a^2 d^2 f (d \sin (e+f x))^{3/2}}-\frac {8 g \sqrt {g \cos (e+f x)}}{5 a d^3 f \sqrt {d \sin (e+f x)}}+\frac {2 \left (a^2-b^2\right ) g \sqrt {g \cos (e+f x)}}{a^3 d^3 f \sqrt {d \sin (e+f x)}}-\frac {2 b g^2 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{3 a^2 d^3 f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}+\frac {b \left (a^2-b^2\right ) g^2 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{a^4 d^3 f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 21.56, size = 1165, normalized size = 2.22 \[ \frac {b (g \cos (e+f x))^{3/2} \left (-\frac {2 \left (a^2-3 b^2\right ) \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \sqrt {\sin (e+f x)} \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 [4]{1-\cos ^2(e+f x)} (a+b \sin (e+f x))}-\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}}\right ) \sin ^{\frac {7}{2}}(e+f x)}{3 a^3 f \cos ^{\frac {3}{2}}(e+f x) (d \sin (e+f x))^{7/2}}+\frac {(g \cos (e+f x))^{3/2} \left (-\frac {2 \csc ^3(e+f x)}{5 a}+\frac {2 b \csc ^2(e+f x)}{3 a^2}+\frac {2 \left (a^2-5 b^2\right ) \csc (e+f x)}{5 a^3}\right ) \tan (e+f x) \sin ^3(e+f x)}{f (d \sin (e+f x))^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(g*Cos[e + f*x])^(3/2)/((d*Sin[e + f*x])^(7/2)*(a + b*Sin[e + f*x])),x]

[Out]

((g*Cos[e + f*x])^(3/2)*((2*(a^2 - 5*b^2)*Csc[e + f*x])/(5*a^3) + (2*b*Csc[e + f*x]^2)/(3*a^2) - (2*Csc[e + f*
x]^3)/(5*a))*Sin[e + f*x]^3*Tan[e + f*x])/(f*(d*Sin[e + f*x])^(7/2)) + (b*(g*Cos[e + f*x])^(3/2)*Sin[e + f*x]^
(7/2)*((-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)*App
ellF1[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*A
rcTan[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)*Sqr
t[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*Arc
Tan[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*Sqrt[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^3*f*Cos[e + f*x]^(3/2)*(d*Sin[e + f*x])^(7/2))

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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((g*cos(f*x+e))^(3/2)/(d*sin(f*x+e))^(7/2)/(a+b*sin(f*x+e)),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}}}{{\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(3/2)/(d*sin(f*x+e))^(7/2)/(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate((g*cos(f*x + e))^(3/2)/((b*sin(f*x + e) + a)*(d*sin(f*x + e))^(7/2)), x)

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maple [B]  time = 0.72, size = 5828, normalized size = 11.10 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*cos(f*x+e))^(3/2)/(d*sin(f*x+e))^(7/2)/(a+b*sin(f*x+e)),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}}}{{\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(3/2)/(d*sin(f*x+e))^(7/2)/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((g*cos(f*x + e))^(3/2)/((b*sin(f*x + e) + a)*(d*sin(f*x + e))^(7/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}}{{\left (d\,\sin \left (e+f\,x\right )\right )}^{7/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*cos(e + f*x))^(3/2)/((d*sin(e + f*x))^(7/2)*(a + b*sin(e + f*x))),x)

[Out]

int((g*cos(e + f*x))^(3/2)/((d*sin(e + f*x))^(7/2)*(a + b*sin(e + f*x))), x)

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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((g*cos(f*x+e))**(3/2)/(d*sin(f*x+e))**(7/2)/(a+b*sin(f*x+e)),x)

[Out]

Timed out

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