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

Optimal. Leaf size=91 \[ \frac{\sin (e+f x) \tan (e+f x) \cos ^2(e+f x)^{\frac{1}{2} (n p+1)} \text{Hypergeometric2F1}\left (\frac{1}{2} (n p+1),\frac{1}{2} (n p+2),\frac{1}{2} (n p+4),\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+2)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.107261, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3659, 2602, 2577} \[ \frac{\sin (e+f x) \tan (e+f x) \cos ^2(e+f x)^{\frac{1}{2} (n p+1)} \, _2F_1\left (\frac{1}{2} (n p+1),\frac{1}{2} (n p+2);\frac{1}{2} (n p+4);\sin ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[e + f*x]^2)^((1 + n*p)/2)*Hypergeometric2F1[(1 + n*p)/2, (2 + n*p)/2, (4 + n*p)/2, Sin[e + f*x]^2]*Sin[e
+ f*x]*Tan[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(2 + 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 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) \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 (e+f x) (c \tan (e+f x))^{n p} \, dx\\ &=\left (\cos ^{n p}(e+f x) \sin ^{-n p}(e+f x) \left (b (c \tan (e+f x))^n\right )^p\right ) \int \cos ^{-n p}(e+f x) \sin ^{1+n p}(e+f x) \, dx\\ &=\frac{\cos ^2(e+f x)^{\frac{1}{2} (1+n p)} \, _2F_1\left (\frac{1}{2} (1+n p),\frac{1}{2} (2+n p);\frac{1}{2} (4+n p);\sin ^2(e+f x)\right ) \sin (e+f x) \tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (2+n p)}\\ \end{align*}

Mathematica [C]  time = 1.30609, size = 284, normalized size = 3.12 \[ \frac{8 (n p+4) \sin ^2\left (\frac{1}{2} (e+f x)\right ) \cos ^4\left (\frac{1}{2} (e+f x)\right ) F_1\left (\frac{n p}{2}+1;n p,2;\frac{n p}{2}+2;\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+2) \left (2 (n p+4) \cos ^2\left (\frac{1}{2} (e+f x)\right ) F_1\left (\frac{n p}{2}+1;n p,2;\frac{n p}{2}+2;\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+2 (\cos (e+f x)-1) \left (2 F_1\left (\frac{n p}{2}+2;n p,3;\frac{n p}{2}+3;\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-n p F_1\left (\frac{n p}{2}+2;n p+1,2;\frac{n p}{2}+3;\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(8*(4 + n*p)*AppellF1[1 + (n*p)/2, n*p, 2, 2 + (n*p)/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*(c*Tan[e + f*x])^n)^p)/(f*(2 + n*p)*(2*(4 + n*p)*AppellF1[1 + (n*p)/2, n*p, 2, 2 +
(n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2 + 2*(2*AppellF1[2 + (n*p)/2, n*p, 3, 3 +
(n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - n*p*AppellF1[2 + (n*p)/2, 1 + n*p, 2, 3 + (n*p)/2, Tan[(e
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2])*(-1 + Cos[e + f*x])))

________________________________________________________________________________________

Maple [F]  time = 5.46, size = 0, normalized size = 0. \begin{align*} \int \sin \left ( fx+e \right ) \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)*(b*(c*tan(f*x+e))^n)^p,x)

[Out]

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

________________________________________________________________________________________

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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)*(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), x)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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{\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*(c*tan(f*x+e))**n)**p,x)

[Out]

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

________________________________________________________________________________________

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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)*(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), x)