3.133 \(\int \frac{1}{(a \sin (e+f x))^{5/2} \sqrt{b \tan (e+f x)}} \, dx\)

Optimal. Leaf size=146 \[ -\frac{b}{2 a^2 f \sqrt{a \sin (e+f x)} (b \tan (e+f x))^{3/2}}+\frac{\sqrt{a \sin (e+f x)} \tan ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{4 a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}-\frac{\sqrt{a \sin (e+f x)} \tanh ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{4 a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.141696, antiderivative size = 146, 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 = {2599, 2601, 12, 2565, 329, 298, 203, 206} \[ -\frac{b}{2 a^2 f \sqrt{a \sin (e+f x)} (b \tan (e+f x))^{3/2}}+\frac{\sqrt{a \sin (e+f x)} \tan ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{4 a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}-\frac{\sqrt{a \sin (e+f x)} \tanh ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{4 a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2599

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*(m + n + 1)), x] + Dist[(m + 2)/(a^2*(m + n + 1)), Int[(a*Sin
[e + f*x])^(m + 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && NeQ[m + n + 1, 0]
&& 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{1}{(a \sin (e+f x))^{5/2} \sqrt{b \tan (e+f x)}} \, dx &=-\frac{b}{2 a^2 f \sqrt{a \sin (e+f x)} (b \tan (e+f x))^{3/2}}+\frac{\int \frac{1}{\sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}} \, dx}{4 a^2}\\ &=-\frac{b}{2 a^2 f \sqrt{a \sin (e+f x)} (b \tan (e+f x))^{3/2}}+\frac{\sqrt{a \sin (e+f x)} \int \frac{\sqrt{\cos (e+f x)} \csc (e+f x)}{a} \, dx}{4 a^2 \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=-\frac{b}{2 a^2 f \sqrt{a \sin (e+f x)} (b \tan (e+f x))^{3/2}}+\frac{\sqrt{a \sin (e+f x)} \int \sqrt{\cos (e+f x)} \csc (e+f x) \, dx}{4 a^3 \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=-\frac{b}{2 a^2 f \sqrt{a \sin (e+f x)} (b \tan (e+f x))^{3/2}}-\frac{\sqrt{a \sin (e+f x)} \operatorname{Subst}\left (\int \frac{\sqrt{x}}{1-x^2} \, dx,x,\cos (e+f x)\right )}{4 a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=-\frac{b}{2 a^2 f \sqrt{a \sin (e+f x)} (b \tan (e+f x))^{3/2}}-\frac{\sqrt{a \sin (e+f x)} \operatorname{Subst}\left (\int \frac{x^2}{1-x^4} \, dx,x,\sqrt{\cos (e+f x)}\right )}{2 a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=-\frac{b}{2 a^2 f \sqrt{a \sin (e+f x)} (b \tan (e+f x))^{3/2}}-\frac{\sqrt{a \sin (e+f x)} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{\cos (e+f x)}\right )}{4 a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}+\frac{\sqrt{a \sin (e+f x)} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{\cos (e+f x)}\right )}{4 a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ &=-\frac{b}{2 a^2 f \sqrt{a \sin (e+f x)} (b \tan (e+f x))^{3/2}}+\frac{\tan ^{-1}\left (\sqrt{\cos (e+f x)}\right ) \sqrt{a \sin (e+f x)}}{4 a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}-\frac{\tanh ^{-1}\left (\sqrt{\cos (e+f x)}\right ) \sqrt{a \sin (e+f x)}}{4 a^3 f \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.598743, size = 112, normalized size = 0.77 \[ \frac{-4 \cos ^2(e+f x)^{3/4} \cot (e+f x)+\sin (2 (e+f x)) \tan ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )-\sin (2 (e+f x)) \tanh ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )}{8 a^2 f \cos ^2(e+f x)^{3/4} \sqrt{a \sin (e+f x)} \sqrt{b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(Cos[e + f*x]^2)^(3/4)*Cot[e + f*x] + ArcTan[(Cos[e + f*x]^2)^(1/4)]*Sin[2*(e + f*x)] - ArcTanh[(Cos[e + f
*x]^2)^(1/4)]*Sin[2*(e + f*x)])/(8*a^2*f*(Cos[e + f*x]^2)^(3/4)*Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]])

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Maple [B]  time = 0.165, size = 319, normalized size = 2.2 \begin{align*} -{\frac{\sin \left ( fx+e \right ) }{8\,f} \left ( 4\,\cos \left ( fx+e \right ) \sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}+\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 ) -\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{5}{2}}}{\frac{1}{\sqrt{{\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}}}}{\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))^(5/2)/(b*tan(f*x+e))^(1/2),x)

[Out]

-1/8/f*(4*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+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)*a
rctan(1/2/(-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)/(a*sin(f*x+e))^(5/2)/(b*sin(f*x+e)/cos(f*x+e))^(1/2)/(-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{5}{2}} \sqrt{b \tan \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 4.46479, size = 1571, normalized size = 10.76 \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))^(5/2)/(b*tan(f*x+e))^(1/2),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)*log
(-(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)^2)/((a^3*b*f*cos(f
*x + e)^2 - a^3*b*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) - s
qrt(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*c
os(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))*cos(f*
x + e)^2)/((a^3*b*f*cos(f*x + e)^2 - a^3*b*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))**(5/2)/(b*tan(f*x+e))**(1/2),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{5}{2}} \sqrt{b \tan \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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