3.176 \(\int \sin ^2(e+f x) (b \tan (e+f x))^n \, dx\)

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

[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))

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Rubi [A]  time = 0.0531362, antiderivative size = 50, normalized size of antiderivative = 1., 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 = {2591, 364} \[ \frac{(b \tan (e+f x))^{n+3} \, _2F_1\left (2,\frac{n+3}{2};\frac{n+5}{2};-\tan ^2(e+f x)\right )}{b^3 f (n+3)} \]

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 2591

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])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \sin ^2(e+f x) (b \tan (e+f x))^n \, dx &=\frac{b \operatorname{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*}

Mathematica [C]  time = 2.11771, size = 450, normalized size = 9. \[ \frac{16 (n+3) \sin ^3\left (\frac{1}{2} (e+f x)\right ) \cos ^5\left (\frac{1}{2} (e+f x)\right ) \left (F_1\left (\frac{n+1}{2};n,2;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-F_1\left (\frac{n+1}{2};n,3;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right ) (b \tan (e+f x))^n}{f (n+1) \left (-2 (n+3) \cos ^2\left (\frac{1}{2} (e+f x)\right ) F_1\left (\frac{n+1}{2};n,3;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+2 (\cos (e+f x)-1) \left (2 F_1\left (\frac{n+3}{2};n,3;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-3 F_1\left (\frac{n+3}{2};n,4;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+n \left (F_1\left (\frac{n+3}{2};n+1,3;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-F_1\left (\frac{n+3}{2};n+1,2;\frac{n+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )\right )+(n+3) (\cos (e+f x)+1) F_1\left (\frac{n+1}{2};n,2;\frac{n+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )} \]

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 = 0.71, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( fx+e \right ) \right ) ^{2} \left ( b\tan \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{2}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (\cos \left (f x + e\right )^{2} - 1\right )} \left (b \tan \left (f x + e\right )\right )^{n}, x\right ) \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan{\left (e + f x \right )}\right )^{n} \sin ^{2}{\left (e + f x \right )}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{2}\,{d x} \end{align*}

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)