3.4.58 \(\int (b \sec (e+f x))^m \tan ^2(e+f x) \, dx\) [358]

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

[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 = 0.03, 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 = {2697} \begin {gather*} \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} \end {gather*}

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 2697

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)/(b*f*(n + 1)))*Hypergeometric2F1[(n + 1)/2,
(m + n + 1)/2, (n + 3)/2, Sin[e + f*x]^2], 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] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 25.32, size = 6726, normalized size = 106.76 \begin {gather*} \text {Result too large to show} \end {gather*}

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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Maple [F]
time = 0.08, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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

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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