3.7.7 \(\int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^3 \, dx\) [607]
Optimal. Leaf size=320 \[ -\frac {d \left (d^2 (4+m)-c d \left (5-3 m-2 m^2\right )+2 c^2 \left (8+6 m+m^2\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {2^{\frac {1}{2}+m} \left (d^3 m \left (5+3 m+m^2\right )+3 c^2 d m \left (6+5 m+m^2\right )+3 c d^2 \left (3+4 m+4 m^2+m^3\right )+c^3 \left (6+11 m+6 m^2+m^3\right )\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {d^2 (d m+c (5+m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac {d \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)} \]
[Out]
-d*(d^2*(4+m)-c*d*(-2*m^2-3*m+5)+2*c^2*(m^2+6*m+8))*cos(f*x+e)*(a+a*sin(f*x+e))^m/f/(3+m)/(m^2+3*m+2)-2^(1/2+m
)*(d^3*m*(m^2+3*m+5)+3*c^2*d*m*(m^2+5*m+6)+3*c*d^2*(m^3+4*m^2+4*m+3)+c^3*(m^3+6*m^2+11*m+6))*cos(f*x+e)*hyperg
eom([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^2*
(d*m+c*(5+m))*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)/a/f/(2+m)/(3+m)-d*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e
))^2/f/(3+m)
________________________________________________________________________________________
Rubi [A]
time = 0.46, antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2862, 3047,
3102, 2830, 2731, 2730} \begin {gather*} -\frac {d \left (2 c^2 \left (m^2+6 m+8\right )-c d \left (-2 m^2-3 m+5\right )+d^2 (m+4)\right ) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (m+1) (m+2) (m+3)}-\frac {2^{m+\frac {1}{2}} \left (c^3 \left (m^3+6 m^2+11 m+6\right )+3 c^2 d m \left (m^2+5 m+6\right )+3 c d^2 \left (m^3+4 m^2+4 m+3\right )+d^3 m \left (m^2+3 m+5\right )\right ) \cos (e+f x) (\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 {d^2 (c (m+5)+d m) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (m+2) (m+3)}-\frac {d \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^2}{f (m+3)} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^3,x]
[Out]
-((d*(d^2*(4 + m) - c*d*(5 - 3*m - 2*m^2) + 2*c^2*(8 + 6*m + m^2))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1
+ m)*(2 + m)*(3 + m))) - (2^(1/2 + m)*(d^3*m*(5 + 3*m + m^2) + 3*c^2*d*m*(6 + 5*m + m^2) + 3*c*d^2*(3 + 4*m +
4*m^2 + m^3) + c^3*(6 + 11*m + 6*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^2*(d*m + c*(5 +
m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(2 + m)*(3 + m)) - (d*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*
(c + d*Sin[e + f*x])^2)/(f*(3 + m))
Rule 2730
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && GtQ[a, 0]
Rule 2731
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[a^IntPart[n]*((a + b*Sin[c + d*x])^FracPart
[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n]), Int[(1 + (b/a)*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && !GtQ[a, 0]
Rule 2830
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*S
in[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 2862
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(f*(m + n))), x] + Dist[1/(b*(m + n))
, Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 2)*Simp[d*(a*c*m + b*d*(n - 1)) + b*c^2*(m + n) + d*(a*
d*m + b*c*(m + 2*n - 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && E
qQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 1] && IntegerQ[n]
Rule 3047
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 3102
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 (c+d \sin (e+f x))^3 \, dx &=-\frac {d \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)) \left (a \left (c^2 (3+m)+d (2 d+c m)\right )+a d (d m+c (5+m)) \sin (e+f x)\right ) \, dx}{a (3+m)}\\ &=-\frac {d \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 \left (c^2 (3+m)+d (2 d+c m)\right )+\left (a c d (d m+c (5+m))+a d \left (c^2 (3+m)+d (2 d+c m)\right )\right ) \sin (e+f x)+a d^2 (d m+c (5+m)) \sin ^2(e+f x)\right ) \, dx}{a (3+m)}\\ &=-\frac {d^2 (d m+c (5+m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac {d \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 \left (c (2+m) \left (2 d^2+c d m+c^2 (3+m)\right )+d^2 (1+m) (d m+c (5+m))\right )+a^2 d \left (d^2 (4+m)-c d \left (5-3 m-2 m^2\right )+2 c^2 \left (8+6 m+m^2\right )\right ) \sin (e+f x)\right ) \, dx}{a^2 (2+m) (3+m)}\\ &=-\frac {d \left (d^2 (4+m)-c d \left (5-3 m-2 m^2\right )+2 c^2 \left (8+6 m+m^2\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {d^2 (d m+c (5+m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac {d \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac {\left (d^3 m \left (5+3 m+m^2\right )+3 c^2 d m \left (6+5 m+m^2\right )+3 c d^2 \left (3+4 m+4 m^2+m^3\right )+c^3 \left (6+11 m+6 m^2+m^3\right )\right ) \int (a+a \sin (e+f x))^m \, dx}{(1+m) (2+m) (3+m)}\\ &=-\frac {d \left (d^2 (4+m)-c d \left (5-3 m-2 m^2\right )+2 c^2 \left (8+6 m+m^2\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {d^2 (d m+c (5+m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac {d \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)}+\frac {\left (\left (d^3 m \left (5+3 m+m^2\right )+3 c^2 d m \left (6+5 m+m^2\right )+3 c d^2 \left (3+4 m+4 m^2+m^3\right )+c^3 \left (6+11 m+6 m^2+m^3\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 {d \left (d^2 (4+m)-c d \left (5-3 m-2 m^2\right )+2 c^2 \left (8+6 m+m^2\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 (d^3 m \left (5+3 m+m^2\right )+3 c^2 d m \left (6+5 m+m^2\right )+3 c d^2 \left (3+4 m+4 m^2+m^3\right )+c^3 \left (6+11 m+6 m^2+m^3\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^2 (d m+c (5+m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac {d \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 [B] Leaf count is larger than twice the leaf count of optimal. \(3599\) vs. \(2(320)=640\).
time = 63.06, size = 3599, normalized size = 11.25 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^3,x]
[Out]
(-22*(Cos[(-e + Pi/2 - f*x)/2]^2)^(1/2 - m)*(1 - Sin[(-e + Pi/2 - f*x)/2]^2)^(-1/2 + m)*(c + d - 2*d*Sin[(-e +
Pi/2 - f*x)/2]^2)^3*(945*(c + d)^3*Gamma[1/2 - m]*Hypergeometric2F1[1/2, 1/2 - m, 9/2, Sin[(-e + Pi/2 - f*x)/
2]^2] - 1890*d*(c + d)^2*Gamma[1/2 - m]*Hypergeometric2F1[1/2, 1/2 - m, 9/2, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(
-e + Pi/2 - f*x)/2]^2 + 142*(c + d)^3*Gamma[3/2 - m]*Hypergeometric2F1[3/2, 3/2 - m, 11/2, Sin[(-e + Pi/2 - f*
x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^2 + 60*(c + d)^3*Gamma[3/2 - m]*HypergeometricPFQ[{3/2, 2, 3/2 - m}, {1, 11/
2}, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^2 + 8*(c + d)^3*Gamma[3/2 - m]*HypergeometricPFQ[{3/2
, 2, 2, 3/2 - m}, {1, 1, 11/2}, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^2 + 2268*d^2*(c + d)*Gamm
a[1/2 - m]*Hypergeometric2F1[1/2, 1/2 - m, 9/2, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^4 - 564*d
*(c + d)^2*Gamma[3/2 - m]*Hypergeometric2F1[3/2, 3/2 - m, 11/2, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f
*x)/2]^4 - 312*d*(c + d)^2*Gamma[3/2 - m]*HypergeometricPFQ[{3/2, 2, 3/2 - m}, {1, 11/2}, Sin[(-e + Pi/2 - f*x
)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^4 - 48*d*(c + d)^2*Gamma[3/2 - m]*HypergeometricPFQ[{3/2, 2, 2, 3/2 - m}, {1,
1, 11/2}, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^4 - 1080*d^3*Gamma[1/2 - m]*Hypergeometric2F1[
1/2, 1/2 - m, 9/2, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^6 + 744*d^2*(c + d)*Gamma[3/2 - m]*Hyp
ergeometric2F1[3/2, 3/2 - m, 11/2, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^6 + 528*d^2*(c + d)*Ga
mma[3/2 - m]*HypergeometricPFQ[{3/2, 2, 3/2 - m}, {1, 11/2}, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)
/2]^6 + 96*d^2*(c + d)*Gamma[3/2 - m]*HypergeometricPFQ[{3/2, 2, 2, 3/2 - m}, {1, 1, 11/2}, Sin[(-e + Pi/2 - f
*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^6 - 368*d^3*Gamma[3/2 - m]*Hypergeometric2F1[3/2, 3/2 - m, 11/2, Sin[(-e +
Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^8 - 288*d^3*Gamma[3/2 - m]*HypergeometricPFQ[{3/2, 2, 3/2 - m}, {1,
11/2}, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^8 - 64*d^3*Gamma[3/2 - m]*HypergeometricPFQ[{3/2,
2, 2, 3/2 - m}, {1, 1, 11/2}, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^8)*(a + a*Sin[e + f*x])^m*
Tan[(-e + Pi/2 - f*x)/2])/(3*f*(3465*(c + d)^3*Gamma[1/2 - m]*Hypergeometric2F1[1/2, 1/2 - m, 9/2, Sin[(-e + P
i/2 - f*x)/2]^2] - 20790*d*(c + d)^2*Gamma[1/2 - m]*Hypergeometric2F1[1/2, 1/2 - m, 9/2, Sin[(-e + Pi/2 - f*x)
/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^2 - 385*(c + d)^3*(-1 + 2*m)*Gamma[1/2 - m]*Hypergeometric2F1[3/2, 3/2 - m, 11
/2, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^2 + 1562*(c + d)^3*Gamma[3/2 - m]*Hypergeometric2F1[3
/2, 3/2 - m, 11/2, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^2 + 660*(c + d)^3*Gamma[3/2 - m]*Hyper
geometricPFQ[{3/2, 2, 3/2 - m}, {1, 11/2}, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^2 + 88*(c + d)
^3*Gamma[3/2 - m]*HypergeometricPFQ[{3/2, 2, 2, 3/2 - m}, {1, 1, 11/2}, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e +
Pi/2 - f*x)/2]^2 + 41580*d^2*(c + d)*Gamma[1/2 - m]*Hypergeometric2F1[1/2, 1/2 - m, 9/2, Sin[(-e + Pi/2 - f*x)
/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^4 + 770*d*(c + d)^2*(-1 + 2*m)*Gamma[1/2 - m]*Hypergeometric2F1[3/2, 3/2 - m,
11/2, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^4 - 10340*d*(c + d)^2*Gamma[3/2 - m]*Hypergeometric
2F1[3/2, 3/2 - m, 11/2, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^4 - 142*(c + d)^3*(-3 + 2*m)*Gamm
a[3/2 - m]*Hypergeometric2F1[5/2, 5/2 - m, 13/2, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^4 - 5720
*d*(c + d)^2*Gamma[3/2 - m]*HypergeometricPFQ[{3/2, 2, 3/2 - m}, {1, 11/2}, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-
e + Pi/2 - f*x)/2]^4 - 120*(c + d)^3*(-3 + 2*m)*Gamma[3/2 - m]*HypergeometricPFQ[{5/2, 3, 5/2 - m}, {2, 13/2},
Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^4 - 880*d*(c + d)^2*Gamma[3/2 - m]*HypergeometricPFQ[{3/
2, 2, 2, 3/2 - m}, {1, 1, 11/2}, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^4 - 32*(c + d)^3*(-3 + 2
*m)*Gamma[3/2 - m]*HypergeometricPFQ[{5/2, 3, 3, 5/2 - m}, {2, 2, 13/2}, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e +
Pi/2 - f*x)/2]^4 - 27720*d^3*Gamma[1/2 - m]*Hypergeometric2F1[1/2, 1/2 - m, 9/2, Sin[(-e + Pi/2 - f*x)/2]^2]*
Sin[(-e + Pi/2 - f*x)/2]^6 - 924*d^2*(c + d)*(-1 + 2*m)*Gamma[1/2 - m]*Hypergeometric2F1[3/2, 3/2 - m, 11/2, S
in[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^6 + 19096*d^2*(c + d)*Gamma[3/2 - m]*Hypergeometric2F1[3/2
, 3/2 - m, 11/2, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^6 + 564*d*(c + d)^2*(-3 + 2*m)*Gamma[3/2
- m]*Hypergeometric2F1[5/2, 5/2 - m, 13/2, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^6 + 13552*d^2
*(c + d)*Gamma[3/2 - m]*HypergeometricPFQ[{3/2, 2, 3/2 - m}, {1, 11/2}, Sin[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e +
Pi/2 - f*x)/2]^6 + 624*d*(c + d)^2*(-3 + 2*m)*Gamma[3/2 - m]*HypergeometricPFQ[{5/2, 3, 5/2 - m}, {2, 13/2}, S
in[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2]^6 + 2464*d^2*(c + d)*Gamma[3/2 - m]*HypergeometricPFQ[{3/2
, 2, 2, 3/2 - m}, {1, 1, 11/2}, Sin[(-e + Pi/2 ...
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Maple [F]
time = 0.43, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{3}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^3,x)
[Out]
int((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^3,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*(c+d*sin(f*x+e))^3,x, algorithm="maxima")
[Out]
integrate((d*sin(f*x + e) + c)^3*(a*sin(f*x + e) + a)^m, 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*(c+d*sin(f*x+e))^3,x, algorithm="fricas")
[Out]
integral(-(3*c*d^2*cos(f*x + e)^2 - c^3 - 3*c*d^2 + (d^3*cos(f*x + e)^2 - 3*c^2*d - d^3)*sin(f*x + e))*(a*sin(
f*x + e) + a)^m, 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 + d \sin {\left (e + f x \right )}\right )^{3}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))**m*(c+d*sin(f*x+e))**3,x)
[Out]
Integral((a*(sin(e + f*x) + 1))**m*(c + d*sin(e + f*x))**3, 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*(c+d*sin(f*x+e))^3,x, algorithm="giac")
[Out]
integrate((d*sin(f*x + e) + c)^3*(a*sin(f*x + e) + a)^m, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^3,x)
[Out]
int((a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^3, x)
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