3.11.49 \(\int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx\) [1049]

Optimal. Leaf size=175 \[ \frac {2^{\frac {1}{2}-\frac {p}{2}} F_1\left (\frac {1}{2} (1+2 m-p);\frac {1+p}{2},-n;\frac {1}{2} (3+2 m-p);\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \sec (e+f x) (g \sec (e+f x))^p (1-\sin (e+f x))^{\frac {1+p}{2}} (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}}{a f (1+2 m-p)} \]

[Out]

2^(1/2-1/2*p)*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))*sec(f*
x+e)*(g*sec(f*x+e))^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*sin(f*x+e))/(c-d))^n)

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Rubi [A]
time = 0.27, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3005, 3000, 145, 144, 143} \begin {gather*} \frac {2^{\frac {1}{2}-\frac {p}{2}} \sec (e+f x) (1-\sin (e+f x))^{\frac {p+1}{2}} (a \sin (e+f x)+a)^{m+1} (g \sec (e+f x))^p (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 {p+1}{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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(1/2 - p/2)*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))]*Sec[e + f*x]*(g*Sec[e + f*x])^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 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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 144

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 145

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 3000

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*S
in[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 3005

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]

Rubi steps

\begin {align*} \int (g \sec (e+f x))^p (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx &=\left ((g \cos (e+f x))^p (g \sec (e+f x))^p\right ) \int (g \cos (e+f x))^{-p} (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, 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 ) \text {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}} \sec (e+f x) (g \sec (e+f x))^p (a-a \sin (e+f x))^{-\frac {1}{2}-\frac {p}{2}+\frac {1+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 ) \text {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}} \sec (e+f x) (g \sec (e+f x))^p (a-a \sin (e+f x))^{-\frac {1}{2}-\frac {p}{2}+\frac {1+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 ) \text {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}{2}-\frac {p}{2}} 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 ) \sec (e+f x) (g \sec (e+f x))^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] Leaf count is larger than twice the leaf count of optimal. \(852\) vs. \(2(175)=350\).
time = 8.82, size = 852, normalized size = 4.87 \begin {gather*} \frac {2 F_1\left (\frac {1-p}{2};1+m+n-p,-n;\frac {3-p}{2};-\tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right ),-\frac {(c-d) \tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right ) \cos \left (\frac {1}{4} (2 e+\pi +2 f x)\right ) (g \sec (e+f x))^p (a (1+\sin (e+f x)))^m (c+d \sin (e+f x))^n \sin \left (\frac {1}{4} (2 e+\pi +2 f x)\right )}{f \left (-F_1\left (\frac {1-p}{2};1+m+n-p,-n;\frac {3-p}{2};-\tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right ),-\frac {(c-d) \tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )+\frac {2 (1-p) \left (-\frac {(c-d) n F_1\left (\frac {3-p}{2};1+m+n-p,1-n;\frac {5-p}{2};-\tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right ),-\frac {(c-d) \tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )}{c+d}+(1+m+n-p) F_1\left (\frac {3-p}{2};2+m+n-p,-n;\frac {5-p}{2};-\tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right ),-\frac {(c-d) \tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )\right ) \cot ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )}{3-p}+p F_1\left (\frac {1-p}{2};1+m+n-p,-n;\frac {3-p}{2};-\tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right ),-\frac {(c-d) \tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right ) \sin (e+f x)+\frac {d n F_1\left (\frac {1-p}{2};1+m+n-p,-n;\frac {3-p}{2};-\tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right ),-\frac {(c-d) \tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right ) \cos ^2(e+f x)}{c+d \sin (e+f x)}-2 (n-p) F_1\left (\frac {1-p}{2};1+m+n-p,-n;\frac {3-p}{2};-\tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right ),-\frac {(c-d) \tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right ) \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )+\frac {2 (c-d) n F_1\left (\frac {1-p}{2};1+m+n-p,-n;\frac {3-p}{2};-\tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right ),-\frac {(c-d) \tan ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right ) \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d \sin (e+f x)}\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(g*Sec[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))]*Cos[(2*e + Pi + 2*f*x)/4]*(g*Sec[e + f*x])^p*(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)/(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))]*Si
n[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)*T
an[(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 = 0.23, size = 0, normalized size = 0.00 \[\int \left (g \sec \left (f x +e \right )\right )^{p} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((g*sec(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((g*sec(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((g*sec(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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3006 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((g*sec(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 \begin {gather*} \int {\left (\frac {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 \end {gather*}

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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