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^(1/2+1/2*p)*g*AppellF1(1/2+m+1/2*p,-n,1/2-1/2*p,3/2+m+1/2*p,-d*(1+sin(f*x+e))/(c-d),1/2+1/2*sin(f*x+e))*(g*c
os(f*x+e))^(-1+p)*(1-sin(f*x+e))^(1/2-1/2*p)*(a+a*sin(f*x+e))^(1+m)*(c+d*sin(f*x+e))^n/a/f/(1+2*m+p)/(((c+d*si
n(f*x+e))/(c-d))^n)

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Rubi [A]  time = 0.28, antiderivative size = 168, normalized size of antiderivative = 1.00, 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 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])

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 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 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]

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*}

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Mathematica [B]  time = 10.40, 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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fricas [F]  time = 1.13, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (g \cos \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \]

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)

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maple [F]  time = 6.00, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x +e \right )\right )^{p} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{n}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x + e\right )\right )^{p} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (g\,\cos \left (e+f\,x\right )\right )}^p\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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