3.138 \(\int \frac{1}{(a \sin (e+f x))^{3/2} (b \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=151 \[ \frac{\sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)} \tan ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{4 a b^2 f \sqrt{a \sin (e+f x)}}+\frac{\sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)} \tanh ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{4 a b^2 f \sqrt{a \sin (e+f x)}}-\frac{1}{2 b f (a \sin (e+f x))^{3/2} \sqrt{b \tan (e+f x)}} \]

[Out]

-1/(2*b*f*(a*Sin[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]]) + (ArcTan[Sqrt[Cos[e + f*x]]]*Sqrt[Cos[e + f*x]]*Sqrt[b
*Tan[e + f*x]])/(4*a*b^2*f*Sqrt[a*Sin[e + f*x]]) + (ArcTanh[Sqrt[Cos[e + f*x]]]*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[
e + f*x]])/(4*a*b^2*f*Sqrt[a*Sin[e + f*x]])

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Rubi [A]  time = 0.154978, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {2597, 2601, 12, 2565, 329, 212, 206, 203} \[ \frac{\sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)} \tan ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{4 a b^2 f \sqrt{a \sin (e+f x)}}+\frac{\sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)} \tanh ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{4 a b^2 f \sqrt{a \sin (e+f x)}}-\frac{1}{2 b f (a \sin (e+f x))^{3/2} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a*Sin[e + f*x])^(3/2)*(b*Tan[e + f*x])^(3/2)),x]

[Out]

-1/(2*b*f*(a*Sin[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]]) + (ArcTan[Sqrt[Cos[e + f*x]]]*Sqrt[Cos[e + f*x]]*Sqrt[b
*Tan[e + f*x]])/(4*a*b^2*f*Sqrt[a*Sin[e + f*x]]) + (ArcTanh[Sqrt[Cos[e + f*x]]]*Sqrt[Cos[e + f*x]]*Sqrt[b*Tan[
e + f*x]])/(4*a*b^2*f*Sqrt[a*Sin[e + f*x]])

Rule 2597

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sin[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n + 1)), x] - Dist[(n + 1)/(b^2*(m + n + 1)), Int[(a*Sin[e + f*x])
^m*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && NeQ[m + n + 1, 0] && Integer
sQ[2*m, 2*n] &&  !(EqQ[n, -3/2] && EqQ[m, 1])

Rule 2601

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(Cos[e + f*x
]^n*(b*Tan[e + f*x])^n)/(a*Sin[e + f*x])^n, Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 212

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

Rule 206

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

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 0.330586, size = 103, normalized size = 0.68 \[ \frac{\sin ^2(e+f x) \left (\tan ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )-2 \sqrt [4]{\cos ^2(e+f x)} \csc ^2(e+f x)+\tanh ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )\right )}{4 b f \sqrt [4]{\cos ^2(e+f x)} (a \sin (e+f x))^{3/2} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a*Sin[e + f*x])^(3/2)*(b*Tan[e + f*x])^(3/2)),x]

[Out]

((ArcTan[(Cos[e + f*x]^2)^(1/4)] + ArcTanh[(Cos[e + f*x]^2)^(1/4)] - 2*(Cos[e + f*x]^2)^(1/4)*Csc[e + f*x]^2)*
Sin[e + f*x]^2)/(4*b*f*(Cos[e + f*x]^2)^(1/4)*(a*Sin[e + f*x])^(3/2)*Sqrt[b*Tan[e + f*x]])

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Maple [B]  time = 0.138, size = 320, normalized size = 2.1 \begin{align*} -{\frac{\sin \left ( fx+e \right ) }{8\,f\cos \left ( fx+e \right ) } \left ( \cos \left ( fx+e \right ) \ln \left ( -{\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( 2\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}} \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}+2\,\cos \left ( fx+e \right ) -1 \right ) } \right ) -\cos \left ( fx+e \right ) \arctan \left ({\frac{1}{2}{\frac{1}{\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}}}} \right ) +4\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}-\ln \left ( -{\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( 2\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}} \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}+2\,\cos \left ( fx+e \right ) -1 \right ) } \right ) +\arctan \left ({\frac{1}{2}{\frac{1}{\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}}}} \right ) \right ) \left ( a\sin \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}} \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/f*(cos(f*x+e)*ln(-(2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^2-cos(f*x+e)^2-2*(-cos(f*x+e)/(cos(f
*x+e)+1)^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)-cos(f*x+e)*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))+4*
(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-ln(-(2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^2-cos(f*x+e)^2-2*(
-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)+arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1
/2)))*sin(f*x+e)/cos(f*x+e)/(a*sin(f*x+e))^(3/2)/(b*sin(f*x+e)/cos(f*x+e))^(3/2)/(-cos(f*x+e)/(cos(f*x+e)+1)^2
)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \sin \left (f x + e\right )\right )^{\frac{3}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((a*sin(f*x + e))^(3/2)*(b*tan(f*x + e))^(3/2)), x)

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Fricas [B]  time = 4.49621, size = 1580, normalized size = 10.46 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sin(f*x+e))^(3/2)/(b*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(2*sqrt(-a*b)*(cos(f*x + e)^2 - 1)*arctan(2*sqrt(-a*b)*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x
+ e))*cos(f*x + e)/((a*b*cos(f*x + e) + a*b)*sin(f*x + e)))*sin(f*x + e) + sqrt(-a*b)*(cos(f*x + e)^2 - 1)*lo
g(-(a*b*cos(f*x + e)^3 - 5*a*b*cos(f*x + e)^2 + 4*sqrt(-a*b)*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x
+ e))*cos(f*x + e)*sin(f*x + e) - 5*a*b*cos(f*x + e) + a*b)/(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + 3*cos(f*x + e
) + 1))*sin(f*x + e) - 8*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e))/((a^2*b^2*f*cos(
f*x + e)^2 - a^2*b^2*f)*sin(f*x + e)), -1/16*(2*sqrt(a*b)*(cos(f*x + e)^2 - 1)*arctan(2*sqrt(a*b)*sqrt(a*sin(f
*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e)/((a*b*cos(f*x + e) - a*b)*sin(f*x + e)))*sin(f*x + e)
- sqrt(a*b)*(cos(f*x + e)^2 - 1)*log(-(4*sqrt(a*b)*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(a*sin(f*x + e))*sqrt(b
*sin(f*x + e)/cos(f*x + e)) + (a*b*cos(f*x + e)^2 + 6*a*b*cos(f*x + e) + a*b)*sin(f*x + e))/((cos(f*x + e)^2 -
2*cos(f*x + e) + 1)*sin(f*x + e)))*sin(f*x + e) - 8*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*co
s(f*x + e))/((a^2*b^2*f*cos(f*x + e)^2 - a^2*b^2*f)*sin(f*x + e))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \sin \left (f x + e\right )\right )^{\frac{3}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((a*sin(f*x + e))^(3/2)*(b*tan(f*x + e))^(3/2)), x)