3.125 \(\int \frac{(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=145 \[ \frac{b^2 \sqrt{a \sin (e+f x)} \tan ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}-\frac{b^2 \sqrt{a \sin (e+f x)} \tanh ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}+\frac{2 b \sqrt{b \tan (e+f x)}}{a^2 f \sqrt{a \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.150169, antiderivative size = 145, 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 = {2593, 2601, 12, 2565, 329, 298, 203, 206} \[ \frac{b^2 \sqrt{a \sin (e+f x)} \tan ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}-\frac{b^2 \sqrt{a \sin (e+f x)} \tanh ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}+\frac{2 b \sqrt{b \tan (e+f x)}}{a^2 f \sqrt{a \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x])^(3/2)/(a*Sin[e + f*x])^(5/2),x]

[Out]

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

Rule 2593

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

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 298

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

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])

Rubi steps

\begin{align*} \int \frac{(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{5/2}} \, dx &=\frac{2 b \sqrt{b \tan (e+f x)}}{a^2 f \sqrt{a \sin (e+f x)}}+\frac{b^2 \int \frac{1}{\sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}} \, dx}{a^2}\\ &=\frac{2 b \sqrt{b \tan (e+f x)}}{a^2 f \sqrt{a \sin (e+f x)}}+\frac{\left (b^2 \sqrt{a \sin (e+f x)}\right ) \int \frac{\sqrt{\cos (e+f x)} \csc (e+f x)}{a} \, dx}{a^2 \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=\frac{2 b \sqrt{b \tan (e+f x)}}{a^2 f \sqrt{a \sin (e+f x)}}+\frac{\left (b^2 \sqrt{a \sin (e+f x)}\right ) \int \sqrt{\cos (e+f x)} \csc (e+f x) \, dx}{a^3 \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=\frac{2 b \sqrt{b \tan (e+f x)}}{a^2 f \sqrt{a \sin (e+f x)}}-\frac{\left (b^2 \sqrt{a \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{1-x^2} \, dx,x,\cos (e+f x)\right )}{a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=\frac{2 b \sqrt{b \tan (e+f x)}}{a^2 f \sqrt{a \sin (e+f x)}}-\frac{\left (2 b^2 \sqrt{a \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-x^4} \, dx,x,\sqrt{\cos (e+f x)}\right )}{a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=\frac{2 b \sqrt{b \tan (e+f x)}}{a^2 f \sqrt{a \sin (e+f x)}}-\frac{\left (b^2 \sqrt{a \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{\cos (e+f x)}\right )}{a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}+\frac{\left (b^2 \sqrt{a \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{\cos (e+f x)}\right )}{a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=\frac{b^2 \tan ^{-1}\left (\sqrt{\cos (e+f x)}\right ) \sqrt{a \sin (e+f x)}}{a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}-\frac{b^2 \tanh ^{-1}\left (\sqrt{\cos (e+f x)}\right ) \sqrt{a \sin (e+f x)}}{a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}+\frac{2 b \sqrt{b \tan (e+f x)}}{a^2 f \sqrt{a \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.34573, size = 104, normalized size = 0.72 \[ \frac{b \sqrt{b \tan (e+f x)} \left (2 \cos ^2(e+f x)^{3/4}+\cos ^2(e+f x) \tan ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )-\cos ^2(e+f x) \tanh ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )\right )}{a^2 f \cos ^2(e+f x)^{3/4} \sqrt{a \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x])^(3/2)/(a*Sin[e + f*x])^(5/2),x]

[Out]

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

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Maple [A]  time = 0.148, size = 247, normalized size = 1.7 \begin{align*} -{\frac{ \left ( \cos \left ( fx+e \right ) -1 \right ) \cos \left ( fx+e \right ) }{2\,f\sin \left ( fx+e \right ) } \left ( \cos \left ( fx+e \right ) \ln \left ( -2\,{\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\,\cos \left ( fx+e \right ) \sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}+4\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}} \right ) \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}} \left ( a\sin \left ( fx+e \right ) \right ) ^{-{\frac{5}{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((b*tan(f*x+e))^(3/2)/(a*sin(f*x+e))^(5/2),x)

[Out]

-1/2/f*(cos(f*x+e)-1)*(cos(f*x+e)*ln(-2*(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)/(cos(f*x+e)+1)^2)^(1/2)+4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))*cos(f
*x+e)*(b*sin(f*x+e)/cos(f*x+e))^(3/2)/sin(f*x+e)/(a*sin(f*x+e))^(5/2)/(-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{\left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}{\left (a \sin \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(3/2)/(a*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e))^(3/2)/(a*sin(f*x + e))^(5/2), x)

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Fricas [B]  time = 5.00055, size = 1334, normalized size = 9.2 \begin{align*} \left [\frac{2 \, a b \sqrt{-\frac{b}{a}} \arctan \left (\frac{2 \, \sqrt{a \sin \left (f x + e\right )} \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt{-\frac{b}{a}} \cos \left (f x + e\right )}{{\left (b \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + a b \sqrt{-\frac{b}{a}} \log \left (-\frac{b \cos \left (f x + e\right )^{3} - 4 \, \sqrt{a \sin \left (f x + e\right )} \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt{-\frac{b}{a}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 5 \, b \cos \left (f x + e\right )^{2} - 5 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{3} + 3 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 8 \, \sqrt{a \sin \left (f x + e\right )} b \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{4 \, a^{3} f \sin \left (f x + e\right )}, -\frac{2 \, a b \sqrt{\frac{b}{a}} \arctan \left (\frac{2 \, \sqrt{a \sin \left (f x + e\right )} \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt{\frac{b}{a}} \cos \left (f x + e\right )}{{\left (b \cos \left (f x + e\right ) - b\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - a b \sqrt{\frac{b}{a}} \log \left (\frac{4 \,{\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right )} \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt{\frac{b}{a}} -{\left (b \cos \left (f x + e\right )^{2} + 6 \, b \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 8 \, \sqrt{a \sin \left (f x + e\right )} b \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{4 \, a^{3} f \sin \left (f x + e\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(3/2)/(a*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[1/4*(2*a*b*sqrt(-b/a)*arctan(2*sqrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(-b/a)*cos(f*x + e)
/((b*cos(f*x + e) + b)*sin(f*x + e)))*sin(f*x + e) + a*b*sqrt(-b/a)*log(-(b*cos(f*x + e)^3 - 4*sqrt(a*sin(f*x
+ e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(-b/a)*cos(f*x + e)*sin(f*x + e) - 5*b*cos(f*x + e)^2 - 5*b*cos(f*
x + e) + 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))*b*
sqrt(b*sin(f*x + e)/cos(f*x + e)))/(a^3*f*sin(f*x + e)), -1/4*(2*a*b*sqrt(b/a)*arctan(2*sqrt(a*sin(f*x + e))*s
qrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(b/a)*cos(f*x + e)/((b*cos(f*x + e) - b)*sin(f*x + e)))*sin(f*x + e) - a*
b*sqrt(b/a)*log((4*(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))*sqrt
(b/a) - (b*cos(f*x + e)^2 + 6*b*cos(f*x + e) + 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))*b*sqrt(b*sin(f*x + e)/cos(f*x + e)))/(a^3*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((b*tan(f*x+e))**(3/2)/(a*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}{\left (a \sin \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(3/2)/(a*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^(3/2)/(a*sin(f*x + e))^(5/2), x)