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

Optimal. Leaf size=63 \[ \frac{\tan ^3(e+f x) \text{Hypergeometric2F1}\left (2,\frac{1}{2} (n p+3),\frac{1}{2} (n p+5),-\tan ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+3)} \]

[Out]

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

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Rubi [A]  time = 0.112894, antiderivative size = 63, 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 = {3659, 2591, 364} \[ \frac{\tan ^3(e+f x) \, _2F_1\left (2,\frac{1}{2} (n p+3);\frac{1}{2} (n p+5);-\tan ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3659

Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[(b^IntPart[p]*(b*(c*Tan[e + f*x
])^n)^FracPart[p])/(c*Tan[e + f*x])^(n*FracPart[p]), Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; Fre
eQ[{b, c, e, f, n, p}, x] &&  !IntegerQ[p] &&  !IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x]
)^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])

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) \left (b (c \tan (e+f x))^n\right )^p \, dx &=\left ((c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p\right ) \int \sin ^2(e+f x) (c \tan (e+f x))^{n p} \, dx\\ &=\frac{\left (c (c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p\right ) \operatorname{Subst}\left (\int \frac{x^{2+n p}}{\left (c^2+x^2\right )^2} \, dx,x,c \tan (e+f x)\right )}{f}\\ &=\frac{\, _2F_1\left (2,\frac{1}{2} (3+n p);\frac{1}{2} (5+n p);-\tan ^2(e+f x)\right ) \tan ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (3+n p)}\\ \end{align*}

Mathematica [C]  time = 2.48457, size = 517, normalized size = 8.21 \[ \frac{8 (2 n p+6) \sin ^3\left (\frac{1}{2} (e+f x)\right ) \cos ^5\left (\frac{1}{2} (e+f x)\right ) \left (F_1\left (\frac{1}{2} (n p+1);n p,2;\frac{1}{2} (n p+3);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-F_1\left (\frac{1}{2} (n p+1);n p,3;\frac{1}{2} (n p+3);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+1) \left (2 (n p+3) \cos ^2\left (\frac{1}{2} (e+f x)\right ) F_1\left (\frac{1}{2} (n p+1);n p,2;\frac{1}{2} (n p+3);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-2 (n p+3) \cos ^2\left (\frac{1}{2} (e+f x)\right ) F_1\left (\frac{1}{2} (n p+1);n p,3;\frac{1}{2} (n p+3);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+2 (\cos (e+f x)-1) \left (2 F_1\left (\frac{1}{2} (n p+3);n p,3;\frac{1}{2} (n p+5);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-3 F_1\left (\frac{1}{2} (n p+3);n p,4;\frac{1}{2} (n p+5);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+n p \left (F_1\left (\frac{1}{2} (n p+3);n p+1,3;\frac{1}{2} (n p+5);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-F_1\left (\frac{1}{2} (n p+3);n p+1,2;\frac{1}{2} (n p+5);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(8*(6 + 2*n*p)*(AppellF1[(1 + n*p)/2, n*p, 2, (3 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - AppellF1
[(1 + n*p)/2, n*p, 3, (3 + n*p)/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*(c*Tan[e + f*x])^n)^p)/(f*(1 + n*p)*(2*(3 + n*p)*AppellF1[(1 + n*p)/2, n*p, 2, (3 + n*p)/2, Tan[(e + f
*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 - 2*(3 + n*p)*AppellF1[(1 + n*p)/2, n*p, 3, (3 + n*p)/2, Tan
[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 + 2*(2*AppellF1[(3 + n*p)/2, n*p, 3, (5 + n*p)/2, Tan
[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 3*AppellF1[(3 + n*p)/2, n*p, 4, (5 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[
(e + f*x)/2]^2] + n*p*(-AppellF1[(3 + n*p)/2, 1 + n*p, 2, (5 + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2
] + AppellF1[(3 + n*p)/2, 1 + n*p, 3, (5 + n*p)/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 = 7.694, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( fx+e \right ) \right ) ^{2} \left ( b \left ( c\tan \left ( fx+e \right ) \right ) ^{n} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \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*(c*tan(f*x+e))^n)^p,x, algorithm="maxima")

[Out]

integrate(((c*tan(f*x + e))^n*b)^p*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 (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \left (c \tan{\left (e + f x \right )}\right )^{n}\right )^{p} \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*(c*tan(f*x+e))**n)**p,x)

[Out]

Integral((b*(c*tan(e + f*x))**n)**p*sin(e + f*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \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*(c*tan(f*x+e))^n)^p,x, algorithm="giac")

[Out]

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