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

Optimal. Leaf size=141 \[ -\frac {a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \tan ^{-1}\left (\sqrt {\cos (e+f x)}\right )}{b^2 f \sqrt {a \sin (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \tanh ^{-1}\left (\sqrt {\cos (e+f x)}\right )}{b^2 f \sqrt {a \sin (e+f x)}}+\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 141, 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 = {2595, 2601, 12, 2565, 329, 212, 206, 203} \[ -\frac {a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \tan ^{-1}\left (\sqrt {\cos (e+f x)}\right )}{b^2 f \sqrt {a \sin (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \tanh ^{-1}\left (\sqrt {\cos (e+f x)}\right )}{b^2 f \sqrt {a \sin (e+f x)}}+\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 2595

Int[Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]/((b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Simp[(2*Sqrt[a*Sin
[e + f*x]])/(b*f*Sqrt[b*Tan[e + f*x]]), x] + Dist[a^2/b^2, Int[Sqrt[b*Tan[e + f*x]]/(a*Sin[e + f*x])^(3/2), x]
, x] /; FreeQ[{a, b, e, f}, x]

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])

Rubi steps

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

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Mathematica [A]  time = 0.34, size = 88, normalized size = 0.62 \[ \frac {\sqrt {a \sin (e+f x)} \left (2 \sqrt [4]{\cos ^2(e+f x)}-\tan ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )-\tanh ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )\right )}{b f \sqrt [4]{\cos ^2(e+f x)} \sqrt {b \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

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fricas [B]  time = 1.12, size = 529, normalized size = 3.75 \[ \left [\frac {2 \, b \sqrt {-\frac {a}{b}} \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 {a}{b}} \cos \left (f x + e\right )}{{\left (a \cos \left (f x + e\right ) + a\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + b \sqrt {-\frac {a}{b}} \log \left (-\frac {a \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 {a}{b}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 5 \, a \cos \left (f x + e\right )^{2} - 5 \, a \cos \left (f x + e\right ) + a}{\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 )}{4 \, b^{2} f \sin \left (f x + e\right )}, \frac {2 \, b \sqrt {\frac {a}{b}} \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 {a}{b}} \cos \left (f x + e\right )}{{\left (a \cos \left (f x + e\right ) - a\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + b \sqrt {\frac {a}{b}} \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 {a}{b}} - {\left (a \cos \left (f x + e\right )^{2} + 6 \, a \cos \left (f x + e\right ) + a\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 )}{4 \, b^{2} f \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \sin \left (f x + e\right )}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.54, size = 237, normalized size = 1.68 \[ -\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\right )-\ln \left (-\frac {2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+2 \cos \left (f x +e \right )-1}{\sin \left (f x +e \right )^{2}}\right )+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\right ) \sqrt {a \sin \left (f x +e \right )}}{2 f \sin \left (f x +e \right ) \cos \left (f x +e \right ) \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sqrt {-\frac {\cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \sin \left (f x + e\right )}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a \sin {\left (e + f x \right )}}}{\left (b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*sin(e + f*x))/(b*tan(e + f*x))**(3/2), x)

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