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

Optimal. Leaf size=98 \[ \frac{\tan (e+f x) (d \sin (e+f x))^m \cos ^2(e+f x)^{\frac{1}{2} (n p+1)} \left (b (c \tan (e+f x))^n\right )^p \text{Hypergeometric2F1}\left (\frac{1}{2} (n p+1),\frac{1}{2} (m+n p+1),\frac{1}{2} (m+n p+3),\sin ^2(e+f x)\right )}{f (m+n p+1)} \]

[Out]

((Cos[e + f*x]^2)^((1 + n*p)/2)*Hypergeometric2F1[(1 + n*p)/2, (1 + m + n*p)/2, (3 + m + n*p)/2, Sin[e + f*x]^
2]*(d*Sin[e + f*x])^m*Tan[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(1 + m + n*p))

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Rubi [A]  time = 0.18114, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3659, 2602, 2577} \[ \frac{\tan (e+f x) (d \sin (e+f x))^m \cos ^2(e+f x)^{\frac{1}{2} (n p+1)} \left (b (c \tan (e+f x))^n\right )^p \, _2F_1\left (\frac{1}{2} (n p+1),\frac{1}{2} (m+n p+1);\frac{1}{2} (m+n p+3);\sin ^2(e+f x)\right )}{f (m+n p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[e + f*x]^2)^((1 + n*p)/2)*Hypergeometric2F1[(1 + n*p)/2, (1 + m + n*p)/2, (3 + m + n*p)/2, Sin[e + f*x]^
2]*(d*Sin[e + f*x])^m*Tan[e + f*x]*(b*(c*Tan[e + f*x])^n)^p)/(f*(1 + m + 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 (d \sin (e+f x))^m \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 (d \sin (e+f x))^m (c \tan (e+f x))^{n p} \, dx\\ &=\left (d \cos ^{n p}(e+f x) \sin (e+f x) (d \sin (e+f x))^{-1-n p} \left (b (c \tan (e+f x))^n\right )^p\right ) \int \cos ^{-n p}(e+f x) (d \sin (e+f x))^{m+n p} \, 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} (1+m+n p);\frac{1}{2} (3+m+n p);\sin ^2(e+f x)\right ) (d \sin (e+f x))^m \tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1+m+n p)}\\ \end{align*}

Mathematica [C]  time = 2.01977, size = 295, normalized size = 3.01 \[ \frac{(m+n p+3) \sin (e+f x) (d \sin (e+f x))^m F_1\left (\frac{1}{2} (m+n p+1);n p,m+1;\frac{1}{2} (m+n p+3);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (m+n p+1) \left ((m+n p+3) F_1\left (\frac{1}{2} (m+n p+1);n p,m+1;\frac{1}{2} (m+n p+3);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-2 \tan ^2\left (\frac{1}{2} (e+f x)\right ) \left ((m+1) F_1\left (\frac{1}{2} (m+n p+3);n p,m+2;\frac{1}{2} (m+n p+5);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-n p F_1\left (\frac{1}{2} (m+n p+3);n p+1,m+1;\frac{1}{2} (m+n p+5);\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((3 + m + n*p)*AppellF1[(1 + m + n*p)/2, n*p, 1 + m, (3 + m + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]
*Sin[e + f*x]*(d*Sin[e + f*x])^m*(b*(c*Tan[e + f*x])^n)^p)/(f*(1 + m + n*p)*((3 + m + n*p)*AppellF1[(1 + m + n
*p)/2, n*p, 1 + m, (3 + m + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*((1 + m)*AppellF1[(3 + m + n*
p)/2, n*p, 2 + m, (5 + m + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - n*p*AppellF1[(3 + m + n*p)/2, 1
+ n*p, 1 + m, (5 + m + n*p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2))

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Maple [F]  time = 0.414, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sin \left ( fx+e \right ) \right ) ^{m} \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((d*sin(f*x+e))^m*(b*(c*tan(f*x+e))^n)^p,x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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} \left (d \sin \left (f x + e\right )\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))**m*(b*(c*tan(f*x+e))**n)**p,x)

[Out]

Timed out

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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} \left (d \sin \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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