3.819 \(\int (c (d \sin (e+f x))^p)^n (a+a \sin (e+f x))^m \, dx\)
Optimal. Leaf size=113 \[ -\frac{2^{m+\frac{1}{2}} \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \sin ^{-n p}(e+f x) F_1\left (\frac{1}{2};-n p,\frac{1}{2}-m;\frac{3}{2};1-\sin (e+f x),\frac{1}{2} (1-\sin (e+f x))\right ) \left (c (d \sin (e+f x))^p\right )^n}{f} \]
[Out]
-((2^(1/2 + m)*AppellF1[1/2, -(n*p), 1/2 - m, 3/2, 1 - Sin[e + f*x], (1 - Sin[e + f*x])/2]*Cos[e + f*x]*(c*(d*
Sin[e + f*x])^p)^n*(1 + Sin[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^m)/(f*Sin[e + f*x]^(n*p)))
________________________________________________________________________________________
Rubi [A] time = 0.232322, antiderivative size = 113, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used =
{2826, 2787, 2786, 2785, 133} \[ -\frac{2^{m+\frac{1}{2}} \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \sin ^{-n p}(e+f x) F_1\left (\frac{1}{2};-n p,\frac{1}{2}-m;\frac{3}{2};1-\sin (e+f x),\frac{1}{2} (1-\sin (e+f x))\right ) \left (c (d \sin (e+f x))^p\right )^n}{f} \]
Antiderivative was successfully verified.
[In]
Int[(c*(d*Sin[e + f*x])^p)^n*(a + a*Sin[e + f*x])^m,x]
[Out]
-((2^(1/2 + m)*AppellF1[1/2, -(n*p), 1/2 - m, 3/2, 1 - Sin[e + f*x], (1 - Sin[e + f*x])/2]*Cos[e + f*x]*(c*(d*
Sin[e + f*x])^p)^n*(1 + Sin[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^m)/(f*Sin[e + f*x]^(n*p)))
Rule 2826
Int[((c_.)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(p_))^(n_)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol]
:> Dist[(c^IntPart[n]*(c*(d*Sin[e + f*x])^p)^FracPart[n])/(d*Sin[e + f*x])^(p*FracPart[n]), Int[(a + b*Sin[e
+ f*x])^m*(d*Sin[e + f*x])^(n*p), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[n]
Rule 2787
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Sin[e + f*x])^FracPart[m])/(1 + (b*Sin[e + f*x])/a)^FracPart[m], Int[(1 + (b*Sin[e + f*x])/a)^
m*(d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m] && !GtQ
[a, 0]
Rule 2786
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[((d/b
)^IntPart[n]*(d*Sin[e + f*x])^FracPart[n])/(b*Sin[e + f*x])^FracPart[n], Int[(a + b*Sin[e + f*x])^m*(b*Sin[e +
f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m] && GtQ[a, 0] && !Gt
Q[d/b, 0]
Rule 2785
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Dist[(b*(d
/b)^n*Cos[e + f*x])/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]]), Subst[Int[((a - x)^n*(2*a - x)^(m -
1/2))/Sqrt[x], x], x, a - b*Sin[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !In
tegerQ[m] && GtQ[a, 0] && GtQ[d/b, 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 \left (c (d \sin (e+f x))^p\right )^n (a+a \sin (e+f x))^m \, dx &=\left ((d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int (d \sin (e+f x))^{n p} (a+a \sin (e+f x))^m \, dx\\ &=\left ((d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n (1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (d \sin (e+f x))^{n p} (1+\sin (e+f x))^m \, dx\\ &=\left (\sin ^{-n p}(e+f x) \left (c (d \sin (e+f x))^p\right )^n (1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int \sin ^{n p}(e+f x) (1+\sin (e+f x))^m \, dx\\ &=-\frac{\left (\cos (e+f x) \sin ^{-n p}(e+f x) \left (c (d \sin (e+f x))^p\right )^n (1+\sin (e+f x))^{-\frac{1}{2}-m} (a+a \sin (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{(1-x)^{n p} (2-x)^{-\frac{1}{2}+m}}{\sqrt{x}} \, dx,x,1-\sin (e+f x)\right )}{f \sqrt{1-\sin (e+f x)}}\\ &=-\frac{2^{\frac{1}{2}+m} F_1\left (\frac{1}{2};-n p,\frac{1}{2}-m;\frac{3}{2};1-\sin (e+f x),\frac{1}{2} (1-\sin (e+f x))\right ) \cos (e+f x) \sin ^{-n p}(e+f x) \left (c (d \sin (e+f x))^p\right )^n (1+\sin (e+f x))^{-\frac{1}{2}-m} (a+a \sin (e+f x))^m}{f}\\ \end{align*}
Mathematica [B] time = 15.1318, size = 2967, normalized size = 26.26 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(c*(d*Sin[e + f*x])^p)^n*(a + a*Sin[e + f*x])^m,x]
[Out]
(-3*AppellF1[1/2, -(n*p), 1 + m + n*p, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Cos[e + f
*x]*Sin[e + f*x]^(n*p)*(c*(d*Sin[e + f*x])^p)^n*(a + a*Sin[e + f*x])^m)/(f*(Sec[(-e + Pi/2 - f*x)/2]^2)^m*(3*A
ppellF1[1/2, -(n*p), 1 + m + n*p, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] - 2*((1 + m +
n*p)*AppellF1[3/2, -(n*p), 2 + m + n*p, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] + n*p*Ap
pellF1[3/2, 1 - n*p, 1 + m + n*p, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2])*Tan[(-e + Pi/
2 - f*x)/2]^2)*((-3*n*p*AppellF1[1/2, -(n*p), 1 + m + n*p, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 -
f*x)/2]^2]*Cos[e + f*x]^2*Sin[e + f*x]^(-1 + n*p))/((Sec[(-e + Pi/2 - f*x)/2]^2)^m*(3*AppellF1[1/2, -(n*p), 1
+ m + n*p, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] - 2*((1 + m + n*p)*AppellF1[3/2, -(n*
p), 2 + m + n*p, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] + n*p*AppellF1[3/2, 1 - n*p, 1
+ m + n*p, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2])*Tan[(-e + Pi/2 - f*x)/2]^2)) + (3*Ap
pellF1[1/2, -(n*p), 1 + m + n*p, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Sin[e + f*x]^(1
+ n*p))/((Sec[(-e + Pi/2 - f*x)/2]^2)^m*(3*AppellF1[1/2, -(n*p), 1 + m + n*p, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2
, -Tan[(-e + Pi/2 - f*x)/2]^2] - 2*((1 + m + n*p)*AppellF1[3/2, -(n*p), 2 + m + n*p, 5/2, Tan[(-e + Pi/2 - f*x
)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] + n*p*AppellF1[3/2, 1 - n*p, 1 + m + n*p, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2
, -Tan[(-e + Pi/2 - f*x)/2]^2])*Tan[(-e + Pi/2 - f*x)/2]^2)) - (3*m*AppellF1[1/2, -(n*p), 1 + m + n*p, 3/2, Ta
n[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Cos[e + f*x]*Sin[e + f*x]^(n*p)*Tan[(-e + Pi/2 - f*x)/2
])/((Sec[(-e + Pi/2 - f*x)/2]^2)^m*(3*AppellF1[1/2, -(n*p), 1 + m + n*p, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan
[(-e + Pi/2 - f*x)/2]^2] - 2*((1 + m + n*p)*AppellF1[3/2, -(n*p), 2 + m + n*p, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2
, -Tan[(-e + Pi/2 - f*x)/2]^2] + n*p*AppellF1[3/2, 1 - n*p, 1 + m + n*p, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan
[(-e + Pi/2 - f*x)/2]^2])*Tan[(-e + Pi/2 - f*x)/2]^2)) + (3*Cos[e + f*x]*Sin[e + f*x]^(n*p)*(-((1 + m + n*p)*A
ppellF1[3/2, -(n*p), 2 + m + n*p, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Sec[(-e + Pi/2
- f*x)/2]^2*Tan[(-e + Pi/2 - f*x)/2])/3 - (n*p*AppellF1[3/2, 1 - n*p, 1 + m + n*p, 5/2, Tan[(-e + Pi/2 - f*x)
/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Sec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2 - f*x)/2])/3))/((Sec[(-e + Pi/2
- f*x)/2]^2)^m*(3*AppellF1[1/2, -(n*p), 1 + m + n*p, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2
]^2] - 2*((1 + m + n*p)*AppellF1[3/2, -(n*p), 2 + m + n*p, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 -
f*x)/2]^2] + n*p*AppellF1[3/2, 1 - n*p, 1 + m + n*p, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2
]^2])*Tan[(-e + Pi/2 - f*x)/2]^2)) - (3*AppellF1[1/2, -(n*p), 1 + m + n*p, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -T
an[(-e + Pi/2 - f*x)/2]^2]*Cos[e + f*x]*Sin[e + f*x]^(n*p)*(-2*((1 + m + n*p)*AppellF1[3/2, -(n*p), 2 + m + n*
p, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] + n*p*AppellF1[3/2, 1 - n*p, 1 + m + n*p, 5/2
, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2])*Sec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2 - f*x)/2
] + 3*(-((1 + m + n*p)*AppellF1[3/2, -(n*p), 2 + m + n*p, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f
*x)/2]^2]*Sec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2 - f*x)/2])/3 - (n*p*AppellF1[3/2, 1 - n*p, 1 + m + n*p, 5/
2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Sec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2 - f*x)/2
])/3) - 2*Tan[(-e + Pi/2 - f*x)/2]^2*((1 + m + n*p)*((-3*(2 + m + n*p)*AppellF1[5/2, -(n*p), 3 + m + n*p, 7/2,
Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Sec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2 - f*x)/2])
/5 - (3*n*p*AppellF1[5/2, 1 - n*p, 2 + m + n*p, 7/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*
Sec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2 - f*x)/2])/5) + n*p*((-3*(1 + m + n*p)*AppellF1[5/2, 1 - n*p, 2 + m
+ n*p, 7/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2]*Sec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2
- f*x)/2])/5 + (3*(1 - n*p)*AppellF1[5/2, 2 - n*p, 1 + m + n*p, 7/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + P
i/2 - f*x)/2]^2]*Sec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2 - f*x)/2])/5))))/((Sec[(-e + Pi/2 - f*x)/2]^2)^m*(3
*AppellF1[1/2, -(n*p), 1 + m + n*p, 3/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] - 2*((1 + m
+ n*p)*AppellF1[3/2, -(n*p), 2 + m + n*p, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2] + n*p*
AppellF1[3/2, 1 - n*p, 1 + m + n*p, 5/2, Tan[(-e + Pi/2 - f*x)/2]^2, -Tan[(-e + Pi/2 - f*x)/2]^2])*Tan[(-e + P
i/2 - f*x)/2]^2)^2)))
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Maple [F] time = 0.41, size = 0, normalized size = 0. \begin{align*} \int \left ( c \left ( d\sin \left ( fx+e \right ) \right ) ^{p} \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((c*(d*sin(f*x+e))^p)^n*(a+a*sin(f*x+e))^m,x)
[Out]
int((c*(d*sin(f*x+e))^p)^n*(a+a*sin(f*x+e))^m,x)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*(d*sin(f*x+e))^p)^n*(a+a*sin(f*x+e))^m,x, algorithm="maxima")
[Out]
integrate(((d*sin(f*x + e))^p*c)^n*(a*sin(f*x + e) + a)^m, x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*(d*sin(f*x+e))^p)^n*(a+a*sin(f*x+e))^m,x, algorithm="fricas")
[Out]
integral(((d*sin(f*x + e))^p*c)^n*(a*sin(f*x + e) + a)^m, x)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sin{\left (e + f x \right )} + 1\right )\right )^{m} \left (c \left (d \sin{\left (e + f x \right )}\right )^{p}\right )^{n}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*(d*sin(f*x+e))**p)**n*(a+a*sin(f*x+e))**m,x)
[Out]
Integral((a*(sin(e + f*x) + 1))**m*(c*(d*sin(e + f*x))**p)**n, x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*(d*sin(f*x+e))^p)^n*(a+a*sin(f*x+e))^m,x, algorithm="giac")
[Out]
integrate(((d*sin(f*x + e))^p*c)^n*(a*sin(f*x + e) + a)^m, x)