∫ (a + a sin(e + fx))m(A + B sin(e + fx))(c − c sin(e + fx))n dx [195]

3.2.95 \(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^n \, dx\) [195]

Optimal. Leaf size=174 \[ \frac {2^{\frac {1}{2}+n} c (B (m-n)+A (1+m+n)) \cos (e+f x) \, _2F_1\left (\frac {1}{2} (1+2 m),\frac {1}{2} (1-2 n);\frac {1}{2} (3+2 m);\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n}}{f (1+2 m) (1+m+n)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n)} \]

[Out]

2^(1/2+n)*c*(B*(m-n)+A*(1+m+n))*cos(f*x+e)*hypergeom([1/2+m, 1/2-n],[3/2+m],1/2+1/2*sin(f*x+e))*(1-sin(f*x+e))

^(1/2-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+n)/f/(1+2*m)/(1+m+n)-B*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin

(f*x+e))^n/f/(1+m+n)

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Rubi [A]
time = 0.22, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3052, 2824, 2768, 72, 71} \begin {gather*} \frac {c 2^{n+\frac {1}{2}} (A (m+n+1)+B (m-n)) \cos (e+f x) (1-\sin (e+f x))^{\frac {1}{2}-n} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-1} \, _2F_1\left (\frac {1}{2} (2 m+1),\frac {1}{2} (1-2 n);\frac {1}{2} (2 m+3);\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1) (m+n+1)}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n}{f (m+n+1)} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^n,x]

[Out]

(2^(1/2 + n)*c*(B*(m - n) + A*(1 + m + n))*Cos[e + f*x]*Hypergeometric2F1[(1 + 2*m)/2, (1 - 2*n)/2, (3 + 2*m)/

2, (1 + Sin[e + f*x])/2]*(1 - Sin[e + f*x])^(1/2 - n)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 + n))/(f

*(1 + 2*m)*(1 + m + n)) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n)/(f*(1 + m + n))

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c

- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}

, x] && NeQ[b*c - a*d, 0] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(Ra

tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), 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*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -

a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[m] && !IntegerQ[n] &&

(RationalQ[m] || !SimplerQ[n + 1, m + 1])

Rule 2768

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[a^2*(

(g*Cos[e + f*x])^(p + 1)/(f*g*(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), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&

EqQ[a^2 - b^2, 0] && !IntegerQ[m]

Rule 2824

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist

[a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*((c + d*Sin[e + f*x])^FracPart[m]/Cos[e + f*x]^(2*

FracPart[m])), Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, m, n},

x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (FractionQ[m] || !FractionQ[n])

Rule 3052

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_

.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*

(m + n + 1))), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Si

n[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &&

!LtQ[m, -2^(-1)] && NeQ[m + n + 1, 0]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^n \, dx &=-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n)}+\left (A+\frac {B (m-n)}{1+m+n}\right ) \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx\\ &=-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n)}+\left (\left (A+\frac {B (m-n)}{1+m+n}\right ) \cos ^{-2 m}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \cos ^{2 m}(e+f x) (c-c \sin (e+f x))^{-m+n} \, dx\\ &=-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n)}+\frac {\left (c^2 \left (A+\frac {B (m-n)}{1+m+n}\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{\frac {1}{2} (-1-2 m)+m} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 m)}\right ) \text {Subst}\left (\int (c-c x)^{-m+\frac {1}{2} (-1+2 m)+n} (c+c x)^{\frac {1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n)}+\frac {\left (2^{-\frac {1}{2}+n} c^2 \left (A+\frac {B (m-n)}{1+m+n}\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-\frac {1}{2}+\frac {1}{2} (-1-2 m)+m+n} \left (\frac {c-c \sin (e+f x)}{c}\right )^{\frac {1}{2}-n} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 m)}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{-m+\frac {1}{2} (-1+2 m)+n} (c+c x)^{\frac {1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {2^{\frac {1}{2}+n} c \left (A+\frac {B (m-n)}{1+m+n}\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2} (1+2 m),\frac {1}{2} (1-2 n);\frac {1}{2} (3+2 m);\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n}}{f (1+2 m)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f (1+m+n)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 12.41, size = 1931, normalized size = 11.10 \begin {gather*} \text {Too large to display} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^n,x]

[Out]

(-4*(8*B*AppellF1[1/2 + n, -2*m, 2*(1 + m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]

^2] - (A + B)*AppellF1[1/2 + n, -2*m, 1 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*

x)/4]^2] - 8*B*AppellF1[1/2 + n, -2*m, 3 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f

*x)/4]^2])*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^n*Tan[(-e + Pi/2 - f*x)/4])/(f*((8

*B*AppellF1[1/2 + n, -2*m, 2*(1 + m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] -

(A + B)*AppellF1[1/2 + n, -2*m, 1 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^

2] - 8*B*AppellF1[1/2 + n, -2*m, 3 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]

^2])*Sec[(-e + Pi/2 - f*x)/4]^2 + 4*n*(8*B*AppellF1[1/2 + n, -2*m, 2*(1 + m + n), 3/2 + n, Tan[(-e + Pi/2 - f*

x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - (A + B)*AppellF1[1/2 + n, -2*m, 1 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2

- f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 8*B*AppellF1[1/2 + n, -2*m, 3 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/

2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Cot[(-e + Pi/2 - f*x)/2]*Tan[(-e + Pi/2 - f*x)/4] + 4*(m + n)*(8*

B*AppellF1[1/2 + n, -2*m, 2*(1 + m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - (

A + B)*AppellF1[1/2 + n, -2*m, 1 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2

] - 8*B*AppellF1[1/2 + n, -2*m, 3 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^

2])*Tan[(-e + Pi/2 - f*x)/4]^2 + (2*(1/2 + n)*((A + B)*(2*m*AppellF1[3/2 + n, 1 - 2*m, 1 + 2*(m + n), 5/2 + n,

Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + (1 + 2*(m + n))*AppellF1[3/2 + n, -2*m, 2*(1 + m +

n), 5/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]) + 8*B*(2*m*AppellF1[3/2 + n, 1 - 2*m,

3 + 2*(m + n), 5/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + (3 + 2*(m + n))*AppellF1[3/

2 + n, -2*m, 2*(2 + m + n), 5/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]) - 16*B*(m*Appel

lF1[3/2 + n, 1 - 2*m, 2*(1 + m + n), 5/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + (1 +

m + n)*AppellF1[3/2 + n, -2*m, 3 + 2*m + 2*n, 5/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2

]))*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]^2)/(3/2 + n) - (4*m*(8*B*AppellF1[1/2 + n, -2*m, 2*(1

+ m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - (A + B)*AppellF1[1/2 + n, -2*m,

1 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 8*B*AppellF1[1/2 + n, -2*m,

3 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Sec[(-e + Pi/2 - f*x)/4]^2*

Tan[(-e + Pi/2 - f*x)/4]^2)/(-1 + Tan[(-e + Pi/2 - f*x)/4]^2) - 4*m*(8*B*AppellF1[1/2 + n, -2*m, 2*(1 + m + n)

, 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - (A + B)*AppellF1[1/2 + n, -2*m, 1 + 2*(m

+ n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 8*B*AppellF1[1/2 + n, -2*m, 3 + 2*(

m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Tan[(-e + Pi/2 - f*x)/4]*Tan[(-e +

Pi/2 - f*x)/2]))

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Maple [F]
time = 0.32, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{n}\, dx\]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^n,x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*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((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(-c*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((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^n,x, algorithm="fricas")

[Out]

integral((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^n, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{n} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**n,x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(-c*(sin(e + f*x) - 1))**n*(A + B*sin(e + f*x)), x)

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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((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(-c*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 (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^n \,d x \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^n,x)

[Out]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^n, x)

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