3.336 \(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=351 \[ -\frac {2^{m+\frac {1}{2}} \cos (e+f x) \left (A (m+3) \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right )+B \left (c^2 m \left (m^2+5 m+6\right )+2 c d \left (m^3+4 m^2+4 m+3\right )+d^2 m \left (m^2+3 m+5\right )\right )\right ) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2) (m+3)}+\frac {\cos (e+f x) \left (d (A d (m+3)+B (2 c+d m))-2 (m+2) \left (A c d (m+3)+B \left (c^2+c d m+d^2\right )\right )\right ) (a \sin (e+f x)+a)^m}{f (m+1) (m+2) (m+3)}-\frac {d \cos (e+f x) (A d (m+3)+B (2 c+d m)) (a \sin (e+f x)+a)^{m+1}}{a f (m+2) (m+3)}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^2}{f (m+3)} \]

[Out]

(d*(A*d*(3+m)+B*(d*m+2*c))-2*(2+m)*(A*c*d*(3+m)+B*(c*d*m+c^2+d^2)))*cos(f*x+e)*(a+a*sin(f*x+e))^m/f/(1+m)/(2+m
)/(3+m)-2^(1/2+m)*(A*(3+m)*(2*c*d*m*(2+m)+d^2*(m^2+m+1)+c^2*(m^2+3*m+2))+B*(d^2*m*(m^2+3*m+5)+c^2*m*(m^2+5*m+6
)+2*c*d*(m^3+4*m^2+4*m+3)))*cos(f*x+e)*hypergeom([1/2, 1/2-m],[3/2],1/2-1/2*sin(f*x+e))*(1+sin(f*x+e))^(-1/2-m
)*(a+a*sin(f*x+e))^m/f/(3+m)/(m^2+3*m+2)-d*(A*d*(3+m)+B*(d*m+2*c))*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)/a/f/(2+m)
/(3+m)-B*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^2/f/(3+m)

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Rubi [A]  time = 0.99, antiderivative size = 351, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2983, 2968, 3023, 2751, 2652, 2651} \[ -\frac {2^{m+\frac {1}{2}} \cos (e+f x) \left (A (m+3) \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right )+B \left (c^2 m \left (m^2+5 m+6\right )+2 c d \left (m^3+4 m^2+4 m+3\right )+d^2 m \left (m^2+3 m+5\right )\right )\right ) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{f (m+1) (m+2) (m+3)}+\frac {\cos (e+f x) \left (d (A d (m+3)+B (2 c+d m))-2 (m+2) \left (A c d (m+3)+B \left (c^2+c d m+d^2\right )\right )\right ) (a \sin (e+f x)+a)^m}{f (m+1) (m+2) (m+3)}-\frac {d \cos (e+f x) (A d (m+3)+B (2 c+d m)) (a \sin (e+f x)+a)^{m+1}}{a f (m+2) (m+3)}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^2}{f (m+3)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2,x]

[Out]

((d*(A*d*(3 + m) + B*(2*c + d*m)) - 2*(2 + m)*(A*c*d*(3 + m) + B*(c^2 + d^2 + c*d*m)))*Cos[e + f*x]*(a + a*Sin
[e + f*x])^m)/(f*(1 + m)*(2 + m)*(3 + m)) - (2^(1/2 + m)*(A*(3 + m)*(2*c*d*m*(2 + m) + d^2*(1 + m + m^2) + c^2
*(2 + 3*m + m^2)) + B*(d^2*m*(5 + 3*m + m^2) + c^2*m*(6 + 5*m + m^2) + 2*c*d*(3 + 4*m + 4*m^2 + m^3)))*Cos[e +
f*x]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-1/2 - m)*(a + a*Sin[e +
f*x])^m)/(f*(1 + m)*(2 + m)*(3 + m)) - (d*(A*d*(3 + m) + B*(2*c + d*m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 +
m))/(a*f*(2 + m)*(3 + m)) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^2)/(f*(3 + m))

Rule 2651

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(2^(n + 1/2)*a^(n - 1/2)*b*Cos[c + d*x]*Hy
pergeometric2F1[1/2, 1/2 - n, 3/2, (1*(1 - (b*Sin[c + d*x])/a))/2])/(d*Sqrt[a + b*Sin[c + d*x]]), x] /; FreeQ[
{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2652

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Sin[c + d*x])^FracPart
[n])/(1 + (b*Sin[c + d*x])/a)^FracPart[n], Int[(1 + (b*Sin[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 2983

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[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c*(
m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (I
ntegerQ[n] || EqQ[m + 1/2, 0])

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx &=-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac {\int (a+a \sin (e+f x))^m (c+d \sin (e+f x)) (a (A c (3+m)+B (2 d+c m))+a (A d (3+m)+B (2 c+d m)) \sin (e+f x)) \, dx}{a (3+m)}\\ &=-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac {\int (a+a \sin (e+f x))^m \left (a c (A c (3+m)+B (2 d+c m))+(a d (A c (3+m)+B (2 d+c m))+a c (A d (3+m)+B (2 c+d m))) \sin (e+f x)+a d (A d (3+m)+B (2 c+d m)) \sin ^2(e+f x)\right ) \, dx}{a (3+m)}\\ &=-\frac {d (A d (3+m)+B (2 c+d m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac {\int (a+a \sin (e+f x))^m \left (a^2 (c (2+m) (A c (3+m)+B (2 d+c m))+d (1+m) (A d (3+m)+B (2 c+d m)))-a^2 \left (d (A d (3+m)+B (2 c+d m))-2 (2+m) \left (A c d (3+m)+B \left (c^2+d^2+c d m\right )\right )\right ) \sin (e+f x)\right ) \, dx}{a^2 (2+m) (3+m)}\\ &=\frac {\left (d (A d (3+m)+B (2 c+d m))-2 (2+m) \left (A c d (3+m)+B \left (c^2+d^2+c d m\right )\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {d (A d (3+m)+B (2 c+d m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac {\left (A (3+m) \left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right )+B \left (d^2 m \left (5+3 m+m^2\right )+c^2 m \left (6+5 m+m^2\right )+2 c d \left (3+4 m+4 m^2+m^3\right )\right )\right ) \int (a+a \sin (e+f x))^m \, dx}{(1+m) (2+m) (3+m)}\\ &=\frac {\left (d (A d (3+m)+B (2 c+d m))-2 (2+m) \left (A c d (3+m)+B \left (c^2+d^2+c d m\right )\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {d (A d (3+m)+B (2 c+d m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac {\left (\left (A (3+m) \left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right )+B \left (d^2 m \left (5+3 m+m^2\right )+c^2 m \left (6+5 m+m^2\right )+2 c d \left (3+4 m+4 m^2+m^3\right )\right )\right ) (1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (1+\sin (e+f x))^m \, dx}{(1+m) (2+m) (3+m)}\\ &=\frac {\left (d (A d (3+m)+B (2 c+d m))-2 (2+m) \left (A c d (3+m)+B \left (c^2+d^2+c d m\right )\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {2^{\frac {1}{2}+m} \left (A (3+m) \left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right )+B \left (d^2 m \left (5+3 m+m^2\right )+c^2 m \left (6+5 m+m^2\right )+2 c d \left (3+4 m+4 m^2+m^3\right )\right )\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {d (A d (3+m)+B (2 c+d m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}\\ \end {align*}

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Mathematica [A]  time = 7.73, size = 300, normalized size = 0.85 \[ -\frac {\csc ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right )^m (a (\sin (e+f x)+1))^m \left (-2 (A+B) (c+d)^2 \tan \left (\frac {1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac {1}{2},m+4;\frac {3}{2};-\tan ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )\right )-\frac {2}{7} (A-B) (c-d)^2 \tan ^7\left (\frac {1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac {7}{2},m+4;\frac {9}{2};-\tan ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )\right )-\frac {2}{5} (c-d) (A (3 c+d)-B (c+3 d)) \tan ^5\left (\frac {1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac {5}{2},m+4;\frac {7}{2};-\tan ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )\right )-\frac {2}{3} (c+d) (3 A c-A d+B c-3 B d) \tan ^3\left (\frac {1}{4} (2 e+2 f x-\pi )\right ) \, _2F_1\left (\frac {3}{2},m+4;\frac {5}{2};-\tan ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )\right )\right )}{f} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2,x]

[Out]

-(((Csc[(2*e + Pi + 2*f*x)/4]^2)^m*(a*(1 + Sin[e + f*x]))^m*(-2*(A + B)*(c + d)^2*Hypergeometric2F1[1/2, 4 + m
, 3/2, -Tan[(2*e - Pi + 2*f*x)/4]^2]*Tan[(2*e - Pi + 2*f*x)/4] - (2*(c + d)*(3*A*c + B*c - A*d - 3*B*d)*Hyperg
eometric2F1[3/2, 4 + m, 5/2, -Tan[(2*e - Pi + 2*f*x)/4]^2]*Tan[(2*e - Pi + 2*f*x)/4]^3)/3 - (2*(c - d)*(A*(3*c
+ d) - B*(c + 3*d))*Hypergeometric2F1[5/2, 4 + m, 7/2, -Tan[(2*e - Pi + 2*f*x)/4]^2]*Tan[(2*e - Pi + 2*f*x)/4
]^5)/5 - (2*(A - B)*(c - d)^2*Hypergeometric2F1[7/2, 4 + m, 9/2, -Tan[(2*e - Pi + 2*f*x)/4]^2]*Tan[(2*e - Pi +
2*f*x)/4]^7)/7))/f)

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (A c^{2} + 2 \, B c d + A d^{2} - {\left (2 \, B c d + A d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (B d^{2} \cos \left (f x + e\right )^{2} - B c^{2} - 2 \, A c d - B d^{2}\right )} \sin \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

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+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

integral((A*c^2 + 2*B*c*d + A*d^2 - (2*B*c*d + A*d^2)*cos(f*x + e)^2 - (B*d^2*cos(f*x + e)^2 - B*c^2 - 2*A*c*d
- B*d^2)*sin(f*x + e))*(a*sin(f*x + e) + a)^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

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+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^2*(a*sin(f*x + e) + a)^m, x)

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maple [F]  time = 9.53, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{2}\, dx \]

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+d*sin(f*x+e))^2,x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

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+d*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^2*(a*sin(f*x + e) + a)^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2 \,d x \]

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 + d*sin(e + f*x))^2,x)

[Out]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )^{2}\, dx \]

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+d*sin(f*x+e))**2,x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(A + B*sin(e + f*x))*(c + d*sin(e + f*x))**2, x)

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