3.129 \(\int (a+a \sin (e+f x))^m (g \tan (e+f x))^p \, dx\)

Optimal. Leaf size=111 \[ \frac{(1-\sin (e+f x))^{\frac{p+1}{2}} (a \sin (e+f x)+a)^m (g \tan (e+f x))^{p+1} (\sin (e+f x)+1)^{\frac{1}{2} (-2 m+p+1)} F_1\left (p+1;\frac{p+1}{2},\frac{1}{2} (-2 m+p+1);p+2;\sin (e+f x),-\sin (e+f x)\right )}{f g (p+1)} \]

[Out]

(AppellF1[1 + p, (1 + p)/2, (1 - 2*m + p)/2, 2 + p, Sin[e + f*x], -Sin[e + f*x]]*(1 - Sin[e + f*x])^((1 + p)/2
)*(1 + Sin[e + f*x])^((1 - 2*m + p)/2)*(a + a*Sin[e + f*x])^m*(g*Tan[e + f*x])^(1 + p))/(f*g*(1 + p))

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Rubi [A]  time = 0.12401, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2720, 135, 133} \[ \frac{(1-\sin (e+f x))^{\frac{p+1}{2}} (a \sin (e+f x)+a)^m (g \tan (e+f x))^{p+1} (\sin (e+f x)+1)^{\frac{1}{2} (-2 m+p+1)} F_1\left (p+1;\frac{p+1}{2},\frac{1}{2} (-2 m+p+1);p+2;\sin (e+f x),-\sin (e+f x)\right )}{f g (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m*(g*Tan[e + f*x])^p,x]

[Out]

(AppellF1[1 + p, (1 + p)/2, (1 - 2*m + p)/2, 2 + p, Sin[e + f*x], -Sin[e + f*x]]*(1 - Sin[e + f*x])^((1 + p)/2
)*(1 + Sin[e + f*x])^((1 - 2*m + p)/2)*(a + a*Sin[e + f*x])^m*(g*Tan[e + f*x])^(1 + p))/(f*g*(1 + p))

Rule 2720

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_), x_Symbol] :> Dist[((g*Ta
n[e + f*x])^(p + 1)*(a - b*Sin[e + f*x])^((p + 1)/2)*(a + b*Sin[e + f*x])^((p + 1)/2))/(f*g*(b*Sin[e + f*x])^(
p + 1)), Subst[Int[(x^p*(a + x)^(m - (p + 1)/2))/(a - x)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a,
b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !IntegerQ[p]

Rule 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^m (g \tan (e+f x))^p \, dx &=\frac{\left ((a \sin (e+f x))^{-1-p} (a-a \sin (e+f x))^{\frac{1+p}{2}} (a+a \sin (e+f x))^{\frac{1+p}{2}} (g \tan (e+f x))^{1+p}\right ) \operatorname{Subst}\left (\int (a-x)^{\frac{1}{2} (-1-p)} x^p (a+x)^{m+\frac{1}{2} (-1-p)} \, dx,x,a \sin (e+f x)\right )}{f g}\\ &=\frac{\left ((1-\sin (e+f x))^{\frac{1}{2}+\frac{p}{2}} (a \sin (e+f x))^{-1-p} (a-a \sin (e+f x))^{-\frac{1}{2}-\frac{p}{2}+\frac{1+p}{2}} (a+a \sin (e+f x))^{\frac{1+p}{2}} (g \tan (e+f x))^{1+p}\right ) \operatorname{Subst}\left (\int x^p (a+x)^{m+\frac{1}{2} (-1-p)} \left (1-\frac{x}{a}\right )^{\frac{1}{2} (-1-p)} \, dx,x,a \sin (e+f x)\right )}{f g}\\ &=\frac{\left ((1-\sin (e+f x))^{\frac{1}{2}+\frac{p}{2}} (a \sin (e+f x))^{-1-p} (1+\sin (e+f x))^{\frac{1}{2}-m+\frac{p}{2}} (a-a \sin (e+f x))^{-\frac{1}{2}-\frac{p}{2}+\frac{1+p}{2}} (a+a \sin (e+f x))^{-\frac{1}{2}+m-\frac{p}{2}+\frac{1+p}{2}} (g \tan (e+f x))^{1+p}\right ) \operatorname{Subst}\left (\int x^p \left (1-\frac{x}{a}\right )^{\frac{1}{2} (-1-p)} \left (1+\frac{x}{a}\right )^{m+\frac{1}{2} (-1-p)} \, dx,x,a \sin (e+f x)\right )}{f g}\\ &=\frac{F_1\left (1+p;\frac{1+p}{2},\frac{1}{2} (1-2 m+p);2+p;\sin (e+f x),-\sin (e+f x)\right ) (1-\sin (e+f x))^{\frac{1+p}{2}} (1+\sin (e+f x))^{\frac{1}{2} (1-2 m+p)} (a+a \sin (e+f x))^m (g \tan (e+f x))^{1+p}}{f g (1+p)}\\ \end{align*}

Mathematica [B]  time = 2.19787, size = 367, normalized size = 3.31 \[ -\frac{2 (p-3) \sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \cos ^3\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) (a (\sin (e+f x)+1))^m (g \tan (e+f x))^p F_1\left (\frac{1-p}{2};-p,m+1;\frac{3-p}{2};\cot ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )}{f (p-1) \left ((p-3) \cos ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) F_1\left (\frac{1-p}{2};-p,m+1;\frac{3-p}{2};\cot ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )+2 \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \left (p F_1\left (\frac{3-p}{2};1-p,m+1;\frac{5-p}{2};\cot ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )+(m+1) F_1\left (\frac{3-p}{2};-p,m+2;\frac{5-p}{2};\cot ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^m*(g*Tan[e + f*x])^p,x]

[Out]

(-2*(-3 + p)*AppellF1[(1 - p)/2, -p, 1 + m, (3 - p)/2, Cot[(2*e + Pi + 2*f*x)/4]^2, -Tan[(2*e - Pi + 2*f*x)/4]
^2]*Cos[(2*e - Pi + 2*f*x)/4]^3*(a*(1 + Sin[e + f*x]))^m*Sin[(2*e - Pi + 2*f*x)/4]*(g*Tan[e + f*x])^p)/(f*(-1
+ p)*((-3 + p)*AppellF1[(1 - p)/2, -p, 1 + m, (3 - p)/2, Cot[(2*e + Pi + 2*f*x)/4]^2, -Tan[(2*e - Pi + 2*f*x)/
4]^2]*Cos[(2*e - Pi + 2*f*x)/4]^2 + 2*(p*AppellF1[(3 - p)/2, 1 - p, 1 + m, (5 - p)/2, Cot[(2*e + Pi + 2*f*x)/4
]^2, -Tan[(2*e - Pi + 2*f*x)/4]^2] + (1 + m)*AppellF1[(3 - p)/2, -p, 2 + m, (5 - p)/2, Cot[(2*e + Pi + 2*f*x)/
4]^2, -Tan[(2*e - Pi + 2*f*x)/4]^2])*Sin[(2*e - Pi + 2*f*x)/4]^2))

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Maple [F]  time = 0.847, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( g\tan \left ( fx+e \right ) \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^m*(g*tan(f*x+e))^p,x)

[Out]

int((a+a*sin(f*x+e))^m*(g*tan(f*x+e))^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \left (g \tan \left (f x + e\right )\right )^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^m*(g*tan(f*x+e))^p,x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^m*(g*tan(f*x + e))^p, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^m*(g*tan(f*x+e))^p,x, algorithm="fricas")

[Out]

integral((a*sin(f*x + e) + a)^m*(g*tan(f*x + e))^p, 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((a+a*sin(f*x+e))**m*(g*tan(f*x+e))**p,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \left (g \tan \left (f x + e\right )\right )^{p}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^m*(g*tan(f*x+e))^p,x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^m*(g*tan(f*x + e))^p, x)