3.2.76 \(\int \sin ^2(e+f x) (b \tan (e+f x))^n \, dx\) [176]

Optimal. Leaf size=50 \[ \frac {\, _2F_1\left (2,\frac {3+n}{2};\frac {5+n}{2};-\tan ^2(e+f x)\right ) (b \tan (e+f x))^{3+n}}{b^3 f (3+n)} \]

[Out]

hypergeom([2, 3/2+1/2*n],[5/2+1/2*n],-tan(f*x+e)^2)*(b*tan(f*x+e))^(3+n)/b^3/f/(3+n)

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Rubi [A]
time = 0.04, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2671, 371} \begin {gather*} \frac {(b \tan (e+f x))^{n+3} \, _2F_1\left (2,\frac {n+3}{2};\frac {n+5}{2};-\tan ^2(e+f x)\right )}{b^3 f (n+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^2*(b*Tan[e + f*x])^n,x]

[Out]

(Hypergeometric2F1[2, (3 + n)/2, (5 + n)/2, -Tan[e + f*x]^2]*(b*Tan[e + f*x])^(3 + n))/(b^3*f*(3 + n))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2671

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[b*(ff/f), Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff
)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 2.28, size = 450, normalized size = 9.00 \begin {gather*} \frac {16 (3+n) \left (F_1\left (\frac {1+n}{2};n,2;\frac {3+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-F_1\left (\frac {1+n}{2};n,3;\frac {3+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \cos ^5\left (\frac {1}{2} (e+f x)\right ) \sin ^3\left (\frac {1}{2} (e+f x)\right ) (b \tan (e+f x))^n}{f (1+n) \left (-2 (3+n) F_1\left (\frac {1+n}{2};n,3;\frac {3+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right )+2 \left (2 F_1\left (\frac {3+n}{2};n,3;\frac {5+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-3 F_1\left (\frac {3+n}{2};n,4;\frac {5+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+n \left (-F_1\left (\frac {3+n}{2};1+n,2;\frac {5+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+F_1\left (\frac {3+n}{2};1+n,3;\frac {5+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )\right ) (-1+\cos (e+f x))+(3+n) F_1\left (\frac {1+n}{2};n,2;\frac {3+n}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1+\cos (e+f x))\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sin[e + f*x]^2*(b*Tan[e + f*x])^n,x]

[Out]

(16*(3 + n)*(AppellF1[(1 + n)/2, n, 2, (3 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - AppellF1[(1 + n)/
2, n, 3, (3 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Cos[(e + f*x)/2]^5*Sin[(e + f*x)/2]^3*(b*Tan[e +
f*x])^n)/(f*(1 + n)*(-2*(3 + n)*AppellF1[(1 + n)/2, n, 3, (3 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]
*Cos[(e + f*x)/2]^2 + 2*(2*AppellF1[(3 + n)/2, n, 3, (5 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 3*A
ppellF1[(3 + n)/2, n, 4, (5 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + n*(-AppellF1[(3 + n)/2, 1 + n,
2, (5 + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + AppellF1[(3 + n)/2, 1 + n, 3, (5 + n)/2, Tan[(e + f*x
)/2]^2, -Tan[(e + f*x)/2]^2]))*(-1 + Cos[e + f*x]) + (3 + n)*AppellF1[(1 + n)/2, n, 2, (3 + n)/2, Tan[(e + f*x
)/2]^2, -Tan[(e + f*x)/2]^2]*(1 + Cos[e + f*x])))

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Maple [F]
time = 0.49, size = 0, normalized size = 0.00 \[\int \left (\sin ^{2}\left (f x +e \right )\right ) \left (b \tan \left (f x +e \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^2*(b*tan(f*x+e))^n,x)

[Out]

int(sin(f*x+e)^2*(b*tan(f*x+e))^n,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(sin(f*x+e)^2*(b*tan(f*x+e))^n,x, algorithm="maxima")

[Out]

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

[Out]

integral(-(cos(f*x + e)^2 - 1)*(b*tan(f*x + e))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**2*(b*tan(f*x+e))**n,x)

[Out]

Integral((b*tan(e + f*x))**n*sin(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(sin(f*x+e)^2*(b*tan(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^n*sin(f*x + e)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^2*(b*tan(e + f*x))^n,x)

[Out]

int(sin(e + f*x)^2*(b*tan(e + f*x))^n, x)

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