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

Optimal. Leaf size=64 \[ \frac{6 \cos ^2(e+f x)^{7/6} (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{7/3} \, _2F_1\left (\frac{7}{6},\frac{23}{12};\frac{35}{12};\sin ^2(e+f x)\right )}{23 d f} \]

[Out]

(6*(Cos[e + f*x]^2)^(7/6)*Hypergeometric2F1[7/6, 23/12, 35/12, Sin[e + f*x]^2]*(b*Sin[e + f*x])^(3/2)*(d*Tan[e
+ f*x])^(7/3))/(23*d*f)

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Rubi [A]  time = 0.0986516, 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} \[ \frac{6 \cos ^2(e+f x)^{7/6} (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{7/3} \, _2F_1\left (\frac{7}{6},\frac{23}{12};\frac{35}{12};\sin ^2(e+f x)\right )}{23 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(Cos[e + f*x]^2)^(7/6)*Hypergeometric2F1[7/6, 23/12, 35/12, Sin[e + f*x]^2]*(b*Sin[e + f*x])^(3/2)*(d*Tan[e
+ f*x])^(7/3))/(23*d*f)

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 (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{4/3} \, dx &=\frac{\left (b \cos ^{\frac{7}{3}}(e+f x) (d \tan (e+f x))^{7/3}\right ) \int \frac{(b \sin (e+f x))^{17/6}}{\cos ^{\frac{4}{3}}(e+f x)} \, dx}{d (b \sin (e+f x))^{7/3}}\\ &=\frac{6 \cos ^2(e+f x)^{7/6} \, _2F_1\left (\frac{7}{6},\frac{23}{12};\frac{35}{12};\sin ^2(e+f x)\right ) (b \sin (e+f x))^{3/2} (d \tan (e+f x))^{7/3}}{23 d f}\\ \end{align*}

Mathematica [A]  time = 0.576644, size = 85, normalized size = 1.33 \[ \frac{3 d (b \sin (e+f x))^{3/2} \sqrt [3]{d \tan (e+f x)} \left (\sqrt [4]{\sec ^2(e+f x)}-\sec ^2(e+f x) \, _2F_1\left (\frac{11}{12},\frac{7}{4};\frac{23}{12};-\tan ^2(e+f x)\right )\right )}{f \sqrt [4]{\sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*d*(-(Hypergeometric2F1[11/12, 7/4, 23/12, -Tan[e + f*x]^2]*Sec[e + f*x]^2) + (Sec[e + f*x]^2)^(1/4))*(b*Sin
[e + f*x])^(3/2)*(d*Tan[e + f*x])^(1/3))/(f*(Sec[e + f*x]^2)^(1/4))

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Maple [F]  time = 0.111, size = 0, normalized size = 0. \begin{align*} \int \left ( b\sin \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{4}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b \sin \left (f x + e\right )} \left (d \tan \left (f x + e\right )\right )^{\frac{1}{3}} b d \sin \left (f x + e\right ) \tan \left (f x + e\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sin(f*x + e))*(d*tan(f*x + e))^(1/3)*b*d*sin(f*x + e)*tan(f*x + e), 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((b*sin(f*x+e))**(3/2)*(d*tan(f*x+e))**(4/3),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError