3.911 \(\int (g \sec (e+f x))^p (d \sin (e+f x))^n (a+a \sin (e+f x))^m \, dx\)

Optimal. Leaf size=127 \[ \frac{\sec (e+f x) (1-\sin (e+f x))^{\frac{p+1}{2}} (a \sin (e+f x)+a)^m (d \sin (e+f x))^{n+1} (g \sec (e+f x))^p (\sin (e+f x)+1)^{\frac{1}{2} (-2 m+p+1)} F_1\left (n+1;\frac{p+1}{2},\frac{1}{2} (-2 m+p+1);n+2;\sin (e+f x),-\sin (e+f x)\right )}{d f (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.349434, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {2926, 2886, 135, 133} \[ \frac{\sec (e+f x) (1-\sin (e+f x))^{\frac{p+1}{2}} (a \sin (e+f x)+a)^m (d \sin (e+f x))^{n+1} (g \sec (e+f x))^p (\sin (e+f x)+1)^{\frac{1}{2} (-2 m+p+1)} F_1\left (n+1;\frac{p+1}{2},\frac{1}{2} (-2 m+p+1);n+2;\sin (e+f x),-\sin (e+f x)\right )}{d f (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2926

Int[((g_.)*sec[(e_.) + (f_.)*(x_)])^(p_)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(2*IntPart[p])*(g*Cos[e + f*x])^FracPart[p]*(g*Sec[e + f*x])^FracP
art[p], Int[((a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e
, f, g, m, n, p}, x] &&  !IntegerQ[p]

Rule 2886

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

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

Mathematica [B]  time = 3.26272, size = 347, normalized size = 2.73 \[ \frac{g (p-3) (a (\sin (e+f x)+1))^m (d \sin (e+f x))^n (g \sec (e+f x))^{p-1} F_1\left (\frac{1-p}{2};-n,m+n-p+1;\frac{3-p}{2};\cot ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )}{f (p-1) \left (2 \tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \left (n F_1\left (\frac{3-p}{2};1-n,m+n-p+1;\frac{5-p}{2};\cot ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )+(m+n-p+1) F_1\left (\frac{3-p}{2};-n,m+n-p+2;\frac{5-p}{2};\cot ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )\right )+(p-3) F_1\left (\frac{1-p}{2};-n,m+n-p+1;\frac{3-p}{2};\cot ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(g*(-3 + p)*AppellF1[(1 - p)/2, -n, 1 + m + n - p, (3 - p)/2, Cot[(2*e + Pi + 2*f*x)/4]^2, -Tan[(2*e - Pi + 2*
f*x)/4]^2]*(g*Sec[e + f*x])^(-1 + p)*(d*Sin[e + f*x])^n*(a*(1 + Sin[e + f*x]))^m)/(f*(-1 + p)*((-3 + p)*Appell
F1[(1 - p)/2, -n, 1 + m + n - p, (3 - p)/2, Cot[(2*e + Pi + 2*f*x)/4]^2, -Tan[(2*e - Pi + 2*f*x)/4]^2] + 2*(n*
AppellF1[(3 - p)/2, 1 - n, 1 + m + n - p, (5 - p)/2, Cot[(2*e + Pi + 2*f*x)/4]^2, -Tan[(2*e - Pi + 2*f*x)/4]^2
] + (1 + m + n - p)*AppellF1[(3 - p)/2, -n, 2 + m + n - p, (5 - p)/2, Cot[(2*e + Pi + 2*f*x)/4]^2, -Tan[(2*e -
Pi + 2*f*x)/4]^2])*Tan[(2*e - Pi + 2*f*x)/4]^2))

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Maple [F]  time = 7.592, size = 0, normalized size = 0. \begin{align*} \int \left ( g\sec \left ( fx+e \right ) \right ) ^{p} \left ( d\sin \left ( fx+e \right ) \right ) ^{n} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*sec(f*x+e))^p*(d*sin(f*x+e))^n*(a+a*sin(f*x+e))^m,x)

[Out]

int((g*sec(f*x+e))^p*(d*sin(f*x+e))^n*(a+a*sin(f*x+e))^m,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((g*sec(f*x + e))^p*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((g*sec(f*x + e))^p*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e))^n, 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((g*sec(f*x+e))**p*(d*sin(f*x+e))**n*(a+a*sin(f*x+e))**m,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((g*sec(f*x + e))^p*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e))^n, x)