3.2.33 \(\int \frac {1}{(a \sin (e+f x))^{5/2} \sqrt {b \tan (e+f x)}} \, dx\) [133]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 146, 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 = {2679, 2681, 12, 2645, 335, 304, 209, 212} \begin {gather*} \frac {\sqrt {a \sin (e+f x)} \text {ArcTan}\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)}}-\frac {b}{2 a^2 f \sqrt {a \sin (e+f x)} (b \tan (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*b/(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 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 304

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 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 2679

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

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Mathematica [A]
time = 0.63, size = 112, normalized size = 0.77 \begin {gather*} \frac {-4 \cos ^2(e+f x)^{3/4} \cot (e+f x)+\text {ArcTan}\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 ) \sin (2 (e+f x))}{8 a^2 f \cos ^2(e+f x)^{3/4} \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(318\) vs. \(2(120)=240\).
time = 0.36, size = 319, normalized size = 2.18

method	result	size
default	\(-\frac {\left (4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )+\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right ) \cos \left (f x +e \right )+\cos \left (f x +e \right ) \ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right )-\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )-\ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right )\right ) \sin \left (f x +e \right )}{8 f \left (a \sin \left (f x +e \right )\right )^{\frac {5}{2}} \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\)	\(319\)

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,method=_RETURNVERBOSE)

[Out]

-1/8/f*(4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)+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)-arctan(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))*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 = 0.00, 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))^(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] Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (130) = 260\).
time = 0.63, size = 659, normalized size = 4.51 \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 )^{2}}{16 \, {\left (a^{3} b f \cos \left (f x + e\right )^{2} - a^{3} b 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 )^{2}}{16 \, {\left (a^{3} b f \cos \left (f x + e\right )^{2} - a^{3} b 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))^(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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, 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))^(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)

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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 )}^{5/2}\,\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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