3.383 \(\int (b \csc (e+f x))^m (d \tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=79 \[ \frac{2 \cos ^2(e+f x)^{5/4} (d \tan (e+f x))^{5/2} (b \csc (e+f x))^m \, _2F_1\left (\frac{5}{4},\frac{1}{4} (5-2 m);\frac{1}{4} (9-2 m);\sin ^2(e+f x)\right )}{d f (5-2 m)} \]

[Out]

(2*(Cos[e + f*x]^2)^(5/4)*(b*Csc[e + f*x])^m*Hypergeometric2F1[5/4, (5 - 2*m)/4, (9 - 2*m)/4, Sin[e + f*x]^2]*
(d*Tan[e + f*x])^(5/2))/(d*f*(5 - 2*m))

________________________________________________________________________________________

Rubi [A]  time = 0.166997, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2618, 2602, 2577} \[ \frac{2 \cos ^2(e+f x)^{5/4} (d \tan (e+f x))^{5/2} (b \csc (e+f x))^m \, _2F_1\left (\frac{5}{4},\frac{1}{4} (5-2 m);\frac{1}{4} (9-2 m);\sin ^2(e+f x)\right )}{d f (5-2 m)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(Cos[e + f*x]^2)^(5/4)*(b*Csc[e + f*x])^m*Hypergeometric2F1[5/4, (5 - 2*m)/4, (9 - 2*m)/4, Sin[e + f*x]^2]*
(d*Tan[e + f*x])^(5/2))/(d*f*(5 - 2*m))

Rule 2618

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Csc[e + f*
x])^FracPart[m]*(Sin[e + f*x]/a)^FracPart[m], Int[(b*Tan[e + f*x])^n/(Sin[e + f*x]/a)^m, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]

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 \csc (e+f x))^m (d \tan (e+f x))^{3/2} \, dx &=\left ((b \csc (e+f x))^m \left (\frac{\sin (e+f x)}{b}\right )^m\right ) \int \left (\frac{\sin (e+f x)}{b}\right )^{-m} (d \tan (e+f x))^{3/2} \, dx\\ &=\frac{\left (\cos ^{\frac{5}{2}}(e+f x) (b \csc (e+f x))^{3+m} \left (\frac{\sin (e+f x)}{b}\right )^{\frac{1}{2}+m} (d \tan (e+f x))^{5/2}\right ) \int \frac{\left (\frac{\sin (e+f x)}{b}\right )^{\frac{3}{2}-m}}{\cos ^{\frac{3}{2}}(e+f x)} \, dx}{b d}\\ &=\frac{2 \cos ^2(e+f x)^{5/4} (b \csc (e+f x))^{3+m} \, _2F_1\left (\frac{5}{4},\frac{1}{4} (5-2 m);\frac{1}{4} (9-2 m);\sin ^2(e+f x)\right ) \sin ^3(e+f x) (d \tan (e+f x))^{5/2}}{b^3 d f (5-2 m)}\\ \end{align*}

Mathematica [A]  time = 5.68104, size = 87, normalized size = 1.1 \[ -\frac{2 (d \tan (e+f x))^{5/2} \sec ^2(e+f x)^{-m/2} (b \csc (e+f x))^m \, _2F_1\left (\frac{1}{4} (5-2 m),1-\frac{m}{2};\frac{1}{4} (9-2 m);-\tan ^2(e+f x)\right )}{d f (2 m-5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*Csc[e + f*x])^m*Hypergeometric2F1[(5 - 2*m)/4, 1 - m/2, (9 - 2*m)/4, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^
(5/2))/(d*f*(-5 + 2*m)*(Sec[e + f*x]^2)^(m/2))

________________________________________________________________________________________

Maple [F]  time = 0.174, size = 0, normalized size = 0. \begin{align*} \int \left ( b\csc \left ( fx+e \right ) \right ) ^{m} \left ( d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*tan(f*x + e))*(b*csc(f*x + e))^m*d*tan(f*x + e), x)

________________________________________________________________________________________

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*csc(f*x+e))**m*(d*tan(f*x+e))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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