3.152 \(\int \frac{(d \tan (e+f x))^{3/2}}{(b \sin (e+f x))^{4/3}} \, dx\)

Optimal. Leaf size=64 \[ \frac{6 \cos ^2(e+f x)^{5/4} (d \tan (e+f x))^{5/2} \, _2F_1\left (\frac{7}{12},\frac{5}{4};\frac{19}{12};\sin ^2(e+f x)\right )}{7 d f (b \sin (e+f x))^{4/3}} \]

[Out]

(6*(Cos[e + f*x]^2)^(5/4)*Hypergeometric2F1[7/12, 5/4, 19/12, Sin[e + f*x]^2]*(d*Tan[e + f*x])^(5/2))/(7*d*f*(
b*Sin[e + f*x])^(4/3))

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Rubi [A]  time = 0.0941235, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2602, 2577} \[ \frac{6 \cos ^2(e+f x)^{5/4} (d \tan (e+f x))^{5/2} \, _2F_1\left (\frac{7}{12},\frac{5}{4};\frac{19}{12};\sin ^2(e+f x)\right )}{7 d f (b \sin (e+f x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^(3/2)/(b*Sin[e + f*x])^(4/3),x]

[Out]

(6*(Cos[e + f*x]^2)^(5/4)*Hypergeometric2F1[7/12, 5/4, 19/12, Sin[e + f*x]^2]*(d*Tan[e + f*x])^(5/2))/(7*d*f*(
b*Sin[e + f*x])^(4/3))

Rule 2602

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[e + f
*x]^(n + 1)*(b*Tan[e + f*x])^(n + 1))/(b*(a*Sin[e + f*x])^(n + 1)), 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]

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps

\begin{align*} \int \frac{(d \tan (e+f x))^{3/2}}{(b \sin (e+f x))^{4/3}} \, dx &=\frac{\left (b \cos ^{\frac{5}{2}}(e+f x) (d \tan (e+f x))^{5/2}\right ) \int \frac{\sqrt [6]{b \sin (e+f x)}}{\cos ^{\frac{3}{2}}(e+f x)} \, dx}{d (b \sin (e+f x))^{5/2}}\\ &=\frac{6 \cos ^2(e+f x)^{5/4} \, _2F_1\left (\frac{7}{12},\frac{5}{4};\frac{19}{12};\sin ^2(e+f x)\right ) (d \tan (e+f x))^{5/2}}{7 d f (b \sin (e+f x))^{4/3}}\\ \end{align*}

Mathematica [A]  time = 0.478017, size = 69, normalized size = 1.08 \[ -\frac{2 d (b \sin (e+f x))^{2/3} \sqrt{d \tan (e+f x)} \left (4 \sqrt [4]{\cos ^2(e+f x)} \, _2F_1\left (\frac{1}{4},\frac{7}{12};\frac{19}{12};\sin ^2(e+f x)\right )-7\right )}{7 b^2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[e + f*x])^(3/2)/(b*Sin[e + f*x])^(4/3),x]

[Out]

(-2*d*(-7 + 4*(Cos[e + f*x]^2)^(1/4)*Hypergeometric2F1[1/4, 7/12, 19/12, Sin[e + f*x]^2])*(b*Sin[e + f*x])^(2/
3)*Sqrt[d*Tan[e + f*x]])/(7*b^2*f)

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Maple [F]  time = 0.116, size = 0, normalized size = 0. \begin{align*} \int{ \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \left ( b\sin \left ( fx+e \right ) \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^(3/2)/(b*sin(f*x+e))^(4/3),x)

[Out]

int((d*tan(f*x+e))^(3/2)/(b*sin(f*x+e))^(4/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \tan \left (f x + e\right )\right )^{\frac{3}{2}}}{\left (b \sin \left (f x + e\right )\right )^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(3/2)/(b*sin(f*x+e))^(4/3),x, algorithm="maxima")

[Out]

integrate((d*tan(f*x + e))^(3/2)/(b*sin(f*x + e))^(4/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\left (b \sin \left (f x + e\right )\right )^{\frac{2}{3}} \sqrt{d \tan \left (f x + e\right )} d \tan \left (f x + e\right )}{b^{2} \cos \left (f x + e\right )^{2} - b^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(3/2)/(b*sin(f*x+e))^(4/3),x, algorithm="fricas")

[Out]

integral(-(b*sin(f*x + e))^(2/3)*sqrt(d*tan(f*x + e))*d*tan(f*x + e)/(b^2*cos(f*x + e)^2 - b^2), x)

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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((d*tan(f*x+e))**(3/2)/(b*sin(f*x+e))**(4/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \tan \left (f x + e\right )\right )^{\frac{3}{2}}}{\left (b \sin \left (f x + e\right )\right )^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^(3/2)/(b*sin(f*x+e))^(4/3),x, algorithm="giac")

[Out]

integrate((d*tan(f*x + e))^(3/2)/(b*sin(f*x + e))^(4/3), x)