3.358 \(\int (b \sec (e+f x))^m \tan ^2(e+f x) \, dx\)

Optimal. Leaf size=63 \[ \frac {\tan ^3(e+f x) \cos ^2(e+f x)^{\frac {m+3}{2}} (b \sec (e+f x))^m \, _2F_1\left (\frac {3}{2},\frac {m+3}{2};\frac {5}{2};\sin ^2(e+f x)\right )}{3 f} \]

[Out]

1/3*(cos(f*x+e)^2)^(3/2+1/2*m)*hypergeom([3/2, 3/2+1/2*m],[5/2],sin(f*x+e)^2)*(b*sec(f*x+e))^m*tan(f*x+e)^3/f

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Rubi [A]  time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2617} \[ \frac {\tan ^3(e+f x) \cos ^2(e+f x)^{\frac {m+3}{2}} (b \sec (e+f x))^m \, _2F_1\left (\frac {3}{2},\frac {m+3}{2};\frac {5}{2};\sin ^2(e+f x)\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^m*Tan[e + f*x]^2,x]

[Out]

((Cos[e + f*x]^2)^((3 + m)/2)*Hypergeometric2F1[3/2, (3 + m)/2, 5/2, Sin[e + f*x]^2]*(b*Sec[e + f*x])^m*Tan[e
+ f*x]^3)/(3*f)

Rule 2617

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1)*(Cos[e + f*x]^2)^((m + n + 1)/2)*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2,
(n + 3)/2, Sin[e + f*x]^2])/(b*f*(n + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rubi steps

\begin {align*} \int (b \sec (e+f x))^m \tan ^2(e+f x) \, dx &=\frac {\cos ^2(e+f x)^{\frac {3+m}{2}} \, _2F_1\left (\frac {3}{2},\frac {3+m}{2};\frac {5}{2};\sin ^2(e+f x)\right ) (b \sec (e+f x))^m \tan ^3(e+f x)}{3 f}\\ \end {align*}

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Mathematica [C]  time = 25.16, size = 6612, normalized size = 104.95 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(b*Sec[e + f*x])^m*Tan[e + f*x]^2,x]

[Out]

Result too large to show

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (b \sec \left (f x + e\right )\right )^{m} \tan \left (f x + e\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^m*tan(f*x+e)^2,x, algorithm="fricas")

[Out]

integral((b*sec(f*x + e))^m*tan(f*x + e)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^m*tan(f*x+e)^2,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^m*tan(f*x + e)^2, x)

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maple [F]  time = 0.56, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x +e \right )\right )^{m} \left (\tan ^{2}\left (f x +e \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^m*tan(f*x+e)^2,x)

[Out]

int((b*sec(f*x+e))^m*tan(f*x+e)^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^m*tan(f*x+e)^2,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^m*tan(f*x + e)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^2\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^2*(b/cos(e + f*x))^m,x)

[Out]

int(tan(e + f*x)^2*(b/cos(e + f*x))^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))**m*tan(f*x+e)**2,x)

[Out]

Integral((b*sec(e + f*x))**m*tan(e + f*x)**2, x)

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