3.2.25 \(\int \frac {(b \tan (e+f x))^{3/2}}{(a \sin (e+f x))^{5/2}} \, dx\) [125]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 145, 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 = {2673, 2681, 12, 2645, 335, 304, 209, 212} \begin {gather*} \frac {b^2 \sqrt {a \sin (e+f x)} \text {ArcTan}\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)}} \end {gather*}

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 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 2673

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

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Mathematica [A]
time = 0.40, size = 104, normalized size = 0.72 \begin {gather*} \frac {b \left (\text {ArcTan}\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 ) \cos ^2(e+f x)+2 \cos ^2(e+f x)^{3/4}\right ) \sqrt {b \tan (e+f x)}}{a^2 f \cos ^2(e+f x)^{3/4} \sqrt {a \sin (e+f x)}} \end {gather*}

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 = 0.32, size = 247, normalized size = 1.70

method	result	size
default	\(-\frac {\left (\cos \left (f x +e \right )-1\right ) \left (\cos \left (f x +e \right ) \ln \left (-\frac {2 \left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1\right )}{\sin \left (f x +e \right )^{2}}\right )+\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right ) \cos \left (f x +e \right )+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right ) \cos \left (f x +e \right ) \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}{2 f \left (a \sin \left (f x +e \right )\right )^{\frac {5}{2}} \sin \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\)	\(247\)

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,method=_RETURNVERBOSE)

[Out]

-1/2/f*(cos(f*x+e)-1)*(cos(f*x+e)*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*co
s(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))*cos(f*x+e)+4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)+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)/(a*sin(f*x+e))^(5/2)/sin(f*x+e)/(-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((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] Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (135) = 270\).
time = 0.64, size = 570, normalized size = 3.93 \begin {gather*} \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 {gather*}

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)] 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((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 = 0.00, 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((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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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