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

Optimal. Leaf size=76 \[ \frac{\sin (e+f x) \cos ^2(e+f x)^{\frac{n+1}{2}} (b \tan (e+f x))^{n+1} \, _2F_1\left (\frac{n+1}{2},\frac{n+2}{2};\frac{n+4}{2};\sin ^2(e+f x)\right )}{b f (n+2)} \]

[Out]

((Cos[e + f*x]^2)^((1 + n)/2)*Hypergeometric2F1[(1 + n)/2, (2 + n)/2, (4 + n)/2, Sin[e + f*x]^2]*Sin[e + f*x]*
(b*Tan[e + f*x])^(1 + n))/(b*f*(2 + n))

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Rubi [A]  time = 0.0702338, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2602, 2577} \[ \frac{\sin (e+f x) \cos ^2(e+f x)^{\frac{n+1}{2}} (b \tan (e+f x))^{n+1} \, _2F_1\left (\frac{n+1}{2},\frac{n+2}{2};\frac{n+4}{2};\sin ^2(e+f x)\right )}{b f (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[e + f*x]^2)^((1 + n)/2)*Hypergeometric2F1[(1 + n)/2, (2 + n)/2, (4 + n)/2, Sin[e + f*x]^2]*Sin[e + f*x]*
(b*Tan[e + f*x])^(1 + n))/(b*f*(2 + n))

Rule 2602

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[e + f
*x]^(n + 1)*(b*Tan[e + f*x])^(n + 1))/(b*(a*Sin[e + f*x])^(n + 1)), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^
n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[n]

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps

\begin{align*} \int \sin (e+f x) (b \tan (e+f x))^n \, dx &=\frac{\left (\cos ^{1+n}(e+f x) \sin ^{-1-n}(e+f x) (b \tan (e+f x))^{1+n}\right ) \int \cos ^{-n}(e+f x) \sin ^{1+n}(e+f x) \, dx}{b}\\ &=\frac{\cos ^2(e+f x)^{\frac{1+n}{2}} \, _2F_1\left (\frac{1+n}{2},\frac{2+n}{2};\frac{4+n}{2};\sin ^2(e+f x)\right ) \sin (e+f x) (b \tan (e+f x))^{1+n}}{b f (2+n)}\\ \end{align*}

Mathematica [C]  time = 1.04791, size = 252, normalized size = 3.32 \[ \frac{8 (n+4) \sin ^2\left (\frac{1}{2} (e+f x)\right ) \cos ^4\left (\frac{1}{2} (e+f x)\right ) F_1\left (\frac{n}{2}+1;n,2;\frac{n}{2}+2;\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right ) (b \tan (e+f x))^n}{f (n+2) \left (2 (\cos (e+f x)-1) \left (2 F_1\left (\frac{n}{2}+2;n,3;\frac{n}{2}+3;\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-n F_1\left (\frac{n}{2}+2;n+1,2;\frac{n}{2}+3;\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )+(n+4) (\cos (e+f x)+1) F_1\left (\frac{n}{2}+1;n,2;\frac{n}{2}+2;\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(8*(4 + n)*AppellF1[1 + n/2, n, 2, 2 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^4*Sin[(e
+ f*x)/2]^2*(b*Tan[e + f*x])^n)/(f*(2 + n)*(2*(2*AppellF1[2 + n/2, n, 3, 3 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e
+ f*x)/2]^2] - n*AppellF1[2 + n/2, 1 + n, 2, 3 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*(-1 + Cos[e +
f*x]) + (4 + n)*AppellF1[1 + n/2, n, 2, 2 + 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 = 0.664, size = 0, normalized size = 0. \begin{align*} \int \sin \left ( fx+e \right ) \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)*(b*tan(f*x+e))^n,x)

[Out]

int(sin(f*x+e)*(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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (b \tan \left (f x + e\right )\right )^{n} \sin \left (f x + e\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*tan(f*x + e))^n*sin(f*x + e), 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{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*tan(e + f*x))**n*sin(e + f*x), 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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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