3.2.38 \(\int \frac {1}{(a \sin (e+f x))^{3/2} (b \tan (e+f x))^{3/2}} \, dx\) [138]

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

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2677, 2681, 12, 2645, 335, 218, 212, 209} \begin {gather*} \frac {\sqrt {\cos (e+f x)} \text {ArcTan}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {b \tan (e+f x)}}{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)}} \end {gather*}

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*1/(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*Ta
n[e + f*x]])/(4*a*b^2*f*Sqrt[a*Sin[e + f*x]])

Rule 12

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

Rule 209

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

Rule 212

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 335

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 2645

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 2677

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 2681

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])

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.36, size = 103, normalized size = 0.68 \begin {gather*} \frac {\left (\text {ArcTan}\left (\sqrt [4]{\cos ^2(e+f x)}\right )+\tanh ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )-2 \sqrt [4]{\cos ^2(e+f x)} \csc ^2(e+f x)\right ) \sin ^2(e+f x)}{4 b f \sqrt [4]{\cos ^2(e+f x)} (a \sin (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(319\) vs. \(2(125)=250\).
time = 0.35, size = 320, normalized size = 2.12

method	result	size
default	\(\frac {\left (\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right ) \cos \left (f x +e \right )-\cos \left (f x +e \right ) \ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right )-\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )+\ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right )-4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right ) \sin \left (f x +e \right )}{8 f \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \cos \left (f x +e \right )}\)	\(320\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (135) = 270\).
time = 0.69, size = 662, normalized size = 4.38 \begin {gather*} \left [-\frac {2 \, \sqrt {-a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \arctan \left (\frac {2 \, \sqrt {-a b} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{{\left (a b \cos \left (f x + e\right ) + a b\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + \sqrt {-a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (-\frac {a b \cos \left (f x + e\right )^{3} - 5 \, a b \cos \left (f x + e\right )^{2} + 4 \, \sqrt {-a b} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 5 \, a b \cos \left (f x + e\right ) + a 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 )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{16 \, {\left (a^{2} b^{2} f \cos \left (f x + e\right )^{2} - a^{2} b^{2} f\right )} \sin \left (f x + e\right )}, -\frac {2 \, \sqrt {a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \arctan \left (\frac {2 \, \sqrt {a b} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{{\left (a b \cos \left (f x + e\right ) - a b\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - \sqrt {a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (-\frac {4 \, \sqrt {a b} {\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 )}} + {\left (a b \cos \left (f x + e\right )^{2} + 6 \, a b \cos \left (f x + e\right ) + a 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 )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{16 \, {\left (a^{2} b^{2} f \cos \left (f x + e\right )^{2} - a^{2} b^{2} f\right )} \sin \left (f x + e\right )}\right ] \end {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 5010 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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