3.1431 \(\int \frac {(d \sin (e+f x))^{3/2}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\)

Optimal. Leaf size=508 \[ \frac {2 \sqrt {2} a d^{3/2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{b f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {2} a d^{3/2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{b f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {g \cos (e+f x)}}\right )}{\sqrt {2} b f \sqrt {g}}+\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {g \cos (e+f x)}}+1\right )}{\sqrt {2} b f \sqrt {g}}+\frac {d^{3/2} \log \left (-\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}+\sqrt {d} \tan (e+f x)+\sqrt {d}\right )}{2 \sqrt {2} b f \sqrt {g}}-\frac {d^{3/2} \log \left (\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}+\sqrt {d} \tan (e+f x)+\sqrt {d}\right )}{2 \sqrt {2} b f \sqrt {g}} \]

[Out]

-1/2*d^(3/2)*arctan(1-2^(1/2)*g^(1/2)*(d*sin(f*x+e))^(1/2)/d^(1/2)/(g*cos(f*x+e))^(1/2))/b/f*2^(1/2)/g^(1/2)+1
/2*d^(3/2)*arctan(1+2^(1/2)*g^(1/2)*(d*sin(f*x+e))^(1/2)/d^(1/2)/(g*cos(f*x+e))^(1/2))/b/f*2^(1/2)/g^(1/2)+1/4
*d^(3/2)*ln(d^(1/2)-2^(1/2)*g^(1/2)*(d*sin(f*x+e))^(1/2)/(g*cos(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/b/f*2^(1/2)/
g^(1/2)-1/4*d^(3/2)*ln(d^(1/2)+2^(1/2)*g^(1/2)*(d*sin(f*x+e))^(1/2)/(g*cos(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/b
/f*2^(1/2)/g^(1/2)+2*a*d^(3/2)*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)*cos(f*x+e)^(1/2)/b/f/(-a^2+b^2)^(1/2)/(g*cos(f*x+e))^(1/2)-2*a*d^(3/2)*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)*cos(f*x+e)^(1/2)/b/f/(-a^2+b^2)^(1/2
)/(g*cos(f*x+e))^(1/2)

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Rubi [A]  time = 0.77, antiderivative size = 508, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.297, Rules used = {2909, 2574, 297, 1162, 617, 204, 1165, 628, 2908, 2907, 1218} \[ \frac {2 \sqrt {2} a d^{3/2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{b f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {2} a d^{3/2} \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {b^2-a^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{b f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {g \cos (e+f x)}}\right )}{\sqrt {2} b f \sqrt {g}}+\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {g \cos (e+f x)}}+1\right )}{\sqrt {2} b f \sqrt {g}}+\frac {d^{3/2} \log \left (-\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}+\sqrt {d} \tan (e+f x)+\sqrt {d}\right )}{2 \sqrt {2} b f \sqrt {g}}-\frac {d^{3/2} \log \left (\frac {\sqrt {2} \sqrt {g} \sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)}}+\sqrt {d} \tan (e+f x)+\sqrt {d}\right )}{2 \sqrt {2} b f \sqrt {g}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[g]*Sqrt[d*Sin[e + f*x]])/(Sqrt[d]*Sqrt[g*Cos[e + f*x]])])/(Sqrt[2]*b*f*Sqr
t[g])) + (d^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[g]*Sqrt[d*Sin[e + f*x]])/(Sqrt[d]*Sqrt[g*Cos[e + f*x]])])/(Sqrt[2]*
b*f*Sqrt[g]) + (2*Sqrt[2]*a*d^(3/2)*Sqrt[Cos[e + f*x]]*EllipticPi[-(a/(b - Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*S
in[e + f*x]]/(Sqrt[d]*Sqrt[1 + Cos[e + f*x]])], -1])/(b*Sqrt[-a^2 + b^2]*f*Sqrt[g*Cos[e + f*x]]) - (2*Sqrt[2]*
a*d^(3/2)*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])/(b*Sqrt[-a^2 + b^2]*f*Sqrt[g*Cos[e + f*x]]) + (d^(3/2)*Log[Sqrt[d] - (Sqrt[2]*Sqrt[
g]*Sqrt[d*Sin[e + f*x]])/Sqrt[g*Cos[e + f*x]] + Sqrt[d]*Tan[e + f*x]])/(2*Sqrt[2]*b*f*Sqrt[g]) - (d^(3/2)*Log[
Sqrt[d] + (Sqrt[2]*Sqrt[g]*Sqrt[d*Sin[e + f*x]])/Sqrt[g*Cos[e + f*x]] + Sqrt[d]*Tan[e + f*x]])/(2*Sqrt[2]*b*f*
Sqrt[g])

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

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

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 2574

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[{k = Denomina
tor[m]}, Dist[(k*a*b)/f, Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Sin[e + f*x])^(1/k)/(b*Cos
[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

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 2909

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

Rubi steps

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

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Mathematica [C]  time = 16.45, size = 518, normalized size = 1.02 \[ \frac {10 \left (a^2-b^2\right ) \cot (e+f x) (d \sin (e+f x))^{3/2} \left (a+b \sqrt {\sin ^2(e+f x)}\right ) \left (\frac {a F_1\left (\frac {1}{4};-\frac {1}{4},1;\frac {5}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )}{\cos ^2(e+f x) \left (\left (b^2-a^2\right ) F_1\left (\frac {5}{4};\frac {3}{4},1;\frac {9}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )-4 b^2 F_1\left (\frac {5}{4};-\frac {1}{4},2;\frac {9}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )\right )+5 \left (a^2-b^2\right ) F_1\left (\frac {1}{4};-\frac {1}{4},1;\frac {5}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )}+\frac {b \sqrt {\sin ^2(e+f x)} F_1\left (\frac {1}{4};-\frac {3}{4},1;\frac {5}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )}{\cos ^2(e+f x) \left (4 b^2 F_1\left (\frac {5}{4};-\frac {3}{4},2;\frac {9}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )+3 \left (a^2-b^2\right ) F_1\left (\frac {5}{4};\frac {1}{4},1;\frac {9}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )\right )-5 \left (a^2-b^2\right ) F_1\left (\frac {1}{4};-\frac {3}{4},1;\frac {5}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )}\right )}{f \sqrt {g \cos (e+f x)} (b \sin (e+f x)-a) (a+b \sin (e+f x))^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(10*(a^2 - b^2)*Cot[e + f*x]*(d*Sin[e + f*x])^(3/2)*(a + b*Sqrt[Sin[e + f*x]^2])*((a*AppellF1[1/4, -1/4, 1, 5/
4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])/(5*(a^2 - b^2)*AppellF1[1/4, -1/4, 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, -1/4, 2, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]
^2)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/4, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]
)*Cos[e + f*x]^2) + (b*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[Sin
[e + f*x]^2])/(-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)*AppellF
1[5/4, 1/4, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^2)))/(f*Sqrt[g*Cos[e + f*
x]]*(-a + b*Sin[e + f*x])*(a + b*Sin[e + f*x])^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((d*sin(f*x+e))^(3/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))^(3/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.87, size = 941, normalized size = 1.85 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sin(f*x+e))^(3/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x)

[Out]

1/f*(I*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a*(-a^2+b^2)^(1/2)-I*E
llipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))*b*(-a^2+b^2)^(1/2)-I*EllipticP
i((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))*a*(-a^2+b^2)^(1/2)+I*EllipticPi((-(-1+
cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))*b*(-a^2+b^2)^(1/2)-EllipticPi((-(-1+cos(f*x+e)
-sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a+EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e
))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))*b*(-a^2+b^2)^(1/2)-EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x
+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a+EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)
,1/2+1/2*I,1/2*2^(1/2))*b*(-a^2+b^2)^(1/2)+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))*(-a^2+b^2)^(1/2)*a+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)*a+a^2*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))-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))*a*b-a^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))+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)*(d*sin(f*x+e))^(3/2)*((-1+cos(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))/sin(f*x+e))^(1/2)/(-1+cos(f*x+e))/(g*cos(f*x+e))^(1/2)*2^(1/2)*a/b/(-a^2
+b^2)^(1/2)/(a-b+(-a^2+b^2)^(1/2))/(b+(-a^2+b^2)^(1/2)-a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))^(3/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sin(e + f*x))^(3/2)/((g*cos(e + f*x))^(1/2)*(a + b*sin(e + f*x))),x)

[Out]

int((d*sin(e + f*x))^(3/2)/((g*cos(e + f*x))^(1/2)*(a + b*sin(e + f*x))), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))**(3/2)/(a+b*sin(f*x+e))/(g*cos(f*x+e))**(1/2),x)

[Out]

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

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