3.1042 \(\int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx\)

Optimal. Leaf size=168 \[ \frac{g 2^{\frac{p+1}{2}} (1-\sin (e+f x))^{\frac{1-p}{2}} (a \sin (e+f x)+a)^{m+1} (g \cos (e+f x))^{p-1} (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (\frac{1}{2} (2 m+p+1);\frac{1-p}{2},-n;\frac{1}{2} (2 m+p+3);\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+p+1)} \]

[Out]

(2^((1 + p)/2)*g*AppellF1[(1 + 2*m + p)/2, (1 - p)/2, -n, (3 + 2*m + p)/2, (1 + Sin[e + f*x])/2, -((d*(1 + Sin
[e + f*x]))/(c - d))]*(g*Cos[e + f*x])^(-1 + p)*(1 - Sin[e + f*x])^((1 - p)/2)*(a + a*Sin[e + f*x])^(1 + m)*(c
+ d*Sin[e + f*x])^n)/(a*f*(1 + 2*m + p)*((c + d*Sin[e + f*x])/(c - d))^n)

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Rubi [A]  time = 0.28029, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {2921, 140, 139, 138} \[ \frac{g 2^{\frac{p+1}{2}} (1-\sin (e+f x))^{\frac{1-p}{2}} (a \sin (e+f x)+a)^{m+1} (g \cos (e+f x))^{p-1} (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (\frac{1}{2} (2 m+p+1);\frac{1-p}{2},-n;\frac{1}{2} (2 m+p+3);\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^((1 + p)/2)*g*AppellF1[(1 + 2*m + p)/2, (1 - p)/2, -n, (3 + 2*m + p)/2, (1 + Sin[e + f*x])/2, -((d*(1 + Sin
[e + f*x]))/(c - d))]*(g*Cos[e + f*x])^(-1 + p)*(1 - Sin[e + f*x])^((1 - p)/2)*(a + a*Sin[e + f*x])^(1 + m)*(c
+ d*Sin[e + f*x])^n)/(a*f*(1 + 2*m + p)*((c + d*Sin[e + f*x])/(c - d))^n)

Rule 2921

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

Rule 140

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

Rule 139

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

Rule 138

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

Rubi steps

\begin{align*} \int (g \cos (e+f x))^p (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx &=\frac{\left (g (g \cos (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}}\right ) \operatorname{Subst}\left (\int (a-a x)^{\frac{1}{2} (-1+p)} (a+a x)^{m+\frac{1}{2} (-1+p)} (c+d x)^n \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\left (2^{-\frac{1}{2}+\frac{p}{2}} g (g \cos (e+f x))^{-1+p} (a-a \sin (e+f x))^{-\frac{1}{2}+\frac{1-p}{2}+\frac{p}{2}} \left (\frac{a-a \sin (e+f x)}{a}\right )^{\frac{1}{2}-\frac{p}{2}} (a+a \sin (e+f x))^{\frac{1-p}{2}}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}-\frac{x}{2}\right )^{\frac{1}{2} (-1+p)} (a+a x)^{m+\frac{1}{2} (-1+p)} (c+d x)^n \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\left (2^{-\frac{1}{2}+\frac{p}{2}} g (g \cos (e+f x))^{-1+p} (a-a \sin (e+f x))^{-\frac{1}{2}+\frac{1-p}{2}+\frac{p}{2}} \left (\frac{a-a \sin (e+f x)}{a}\right )^{\frac{1}{2}-\frac{p}{2}} (a+a \sin (e+f x))^{\frac{1-p}{2}} (c+d \sin (e+f x))^n \left (\frac{a (c+d \sin (e+f x))}{a c-a d}\right )^{-n}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}-\frac{x}{2}\right )^{\frac{1}{2} (-1+p)} (a+a x)^{m+\frac{1}{2} (-1+p)} \left (\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}\right )^n \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{2^{\frac{1+p}{2}} g F_1\left (\frac{1}{2} (1+2 m+p);\frac{1-p}{2},-n;\frac{1}{2} (3+2 m+p);\frac{1}{2} (1+\sin (e+f x)),-\frac{d (1+\sin (e+f x))}{c-d}\right ) (g \cos (e+f x))^{-1+p} (1-\sin (e+f x))^{\frac{1-p}{2}} (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n}}{a f (1+2 m+p)}\\ \end{align*}

Mathematica [B]  time = 8.34151, size = 798, normalized size = 4.75 \[ -\frac{2 F_1\left (\frac{p+1}{2};m+n+p+1,-n;\frac{p+3}{2};-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right ),-\frac{(c-d) \tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right ) (g \cos (e+f x))^p \cos \left (\frac{1}{4} (2 e+2 f x+\pi )\right ) (a (\sin (e+f x)+1))^m (c+d \sin (e+f x))^n \sin \left (\frac{1}{4} (2 e+2 f x+\pi )\right )}{f \left (-\frac{d n F_1\left (\frac{p+1}{2};m+n+p+1,-n;\frac{p+3}{2};-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right ),-\frac{(c-d) \tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right ) \cos ^2(e+f x)}{c+d \sin (e+f x)}+\frac{2 (p+1) \left ((c-d) n F_1\left (\frac{p+3}{2};m+n+p+1,1-n;\frac{p+5}{2};-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right ),-\frac{(c-d) \tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )-(c+d) (m+n+p+1) F_1\left (\frac{p+3}{2};m+n+p+2,-n;\frac{p+5}{2};-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right ),-\frac{(c-d) \tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right ) \cot ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )}{(c+d) (p+3)}+2 (n+p) F_1\left (\frac{p+1}{2};m+n+p+1,-n;\frac{p+3}{2};-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right ),-\frac{(c-d) \tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right ) \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )-\frac{2 (c-d) n F_1\left (\frac{p+1}{2};m+n+p+1,-n;\frac{p+3}{2};-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right ),-\frac{(c-d) \tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right ) \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d \sin (e+f x)}+F_1\left (\frac{p+1}{2};m+n+p+1,-n;\frac{p+3}{2};-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right ),-\frac{(c-d) \tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )+p F_1\left (\frac{p+1}{2};m+n+p+1,-n;\frac{p+3}{2};-\tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right ),-\frac{(c-d) \tan ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right ) \sin (e+f x)\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 2.763, size = 0, normalized size = 0. \begin{align*} \int \left ( g\cos \left ( fx+e \right ) \right ) ^{p} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (g \cos \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 ) + c\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (g \cos \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 ) + c\right )}^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (g \cos \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 ) + c\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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