3.564 \(\int (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=305 \[ -\frac {34 a b \left (53 a^2+38 b^2\right ) (e \cos (c+d x))^{9/2}}{6435 d e}-\frac {2 b \left (93 a^2+26 b^2\right ) (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{715 d e}+\frac {2 e^4 \left (55 a^4+60 a^2 b^2+4 b^4\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d \sqrt {e \cos (c+d x)}}+\frac {2 e^3 \left (55 a^4+60 a^2 b^2+4 b^4\right ) \sin (c+d x) \sqrt {e \cos (c+d x)}}{231 d}+\frac {2 e \left (55 a^4+60 a^2 b^2+4 b^4\right ) \sin (c+d x) (e \cos (c+d x))^{5/2}}{385 d}-\frac {2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^3}{15 d e}-\frac {14 a b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^2}{65 d e} \]

[Out]

-34/6435*a*b*(53*a^2+38*b^2)*(e*cos(d*x+c))^(9/2)/d/e+2/385*(55*a^4+60*a^2*b^2+4*b^4)*e*(e*cos(d*x+c))^(5/2)*s
in(d*x+c)/d-2/715*b*(93*a^2+26*b^2)*(e*cos(d*x+c))^(9/2)*(a+b*sin(d*x+c))/d/e-14/65*a*b*(e*cos(d*x+c))^(9/2)*(
a+b*sin(d*x+c))^2/d/e-2/15*b*(e*cos(d*x+c))^(9/2)*(a+b*sin(d*x+c))^3/d/e+2/231*(55*a^4+60*a^2*b^2+4*b^4)*e^4*(
cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/(e*cos
(d*x+c))^(1/2)+2/231*(55*a^4+60*a^2*b^2+4*b^4)*e^3*sin(d*x+c)*(e*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.55, antiderivative size = 305, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2692, 2862, 2669, 2635, 2642, 2641} \[ \frac {2 e^3 \left (60 a^2 b^2+55 a^4+4 b^4\right ) \sin (c+d x) \sqrt {e \cos (c+d x)}}{231 d}+\frac {2 e^4 \left (60 a^2 b^2+55 a^4+4 b^4\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d \sqrt {e \cos (c+d x)}}-\frac {34 a b \left (53 a^2+38 b^2\right ) (e \cos (c+d x))^{9/2}}{6435 d e}-\frac {2 b \left (93 a^2+26 b^2\right ) (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{715 d e}+\frac {2 e \left (60 a^2 b^2+55 a^4+4 b^4\right ) \sin (c+d x) (e \cos (c+d x))^{5/2}}{385 d}-\frac {2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^3}{15 d e}-\frac {14 a b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^2}{65 d e} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(7/2)*(a + b*Sin[c + d*x])^4,x]

[Out]

(-34*a*b*(53*a^2 + 38*b^2)*(e*Cos[c + d*x])^(9/2))/(6435*d*e) + (2*(55*a^4 + 60*a^2*b^2 + 4*b^4)*e^4*Sqrt[Cos[
c + d*x]]*EllipticF[(c + d*x)/2, 2])/(231*d*Sqrt[e*Cos[c + d*x]]) + (2*(55*a^4 + 60*a^2*b^2 + 4*b^4)*e^3*Sqrt[
e*Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (2*(55*a^4 + 60*a^2*b^2 + 4*b^4)*e*(e*Cos[c + d*x])^(5/2)*Sin[c + d*x]
)/(385*d) - (2*b*(93*a^2 + 26*b^2)*(e*Cos[c + d*x])^(9/2)*(a + b*Sin[c + d*x]))/(715*d*e) - (14*a*b*(e*Cos[c +
d*x])^(9/2)*(a + b*Sin[c + d*x])^2)/(65*d*e) - (2*b*(e*Cos[c + d*x])^(9/2)*(a + b*Sin[c + d*x])^3)/(15*d*e)

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2692

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[1/(m + p), Int[(g*Cos[e + f*x])^
p*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1)*Sin[e + f*x]), x], x] /; FreeQ[{
a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ[m
])

Rule 2862

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

Rubi steps

\begin {align*} \int (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^4 \, dx &=-\frac {2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^3}{15 d e}+\frac {2}{15} \int (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2 \left (\frac {15 a^2}{2}+3 b^2+\frac {21}{2} a b \sin (c+d x)\right ) \, dx\\ &=-\frac {14 a b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^2}{65 d e}-\frac {2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^3}{15 d e}+\frac {4}{195} \int (e \cos (c+d x))^{7/2} (a+b \sin (c+d x)) \left (\frac {3}{4} a \left (65 a^2+54 b^2\right )+\frac {3}{4} b \left (93 a^2+26 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac {2 b \left (93 a^2+26 b^2\right ) (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{715 d e}-\frac {14 a b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^2}{65 d e}-\frac {2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^3}{15 d e}+\frac {8 \int (e \cos (c+d x))^{7/2} \left (\frac {39}{8} \left (55 a^4+60 a^2 b^2+4 b^4\right )+\frac {51}{8} a b \left (53 a^2+38 b^2\right ) \sin (c+d x)\right ) \, dx}{2145}\\ &=-\frac {34 a b \left (53 a^2+38 b^2\right ) (e \cos (c+d x))^{9/2}}{6435 d e}-\frac {2 b \left (93 a^2+26 b^2\right ) (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{715 d e}-\frac {14 a b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^2}{65 d e}-\frac {2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^3}{15 d e}+\frac {1}{55} \left (55 a^4+60 a^2 b^2+4 b^4\right ) \int (e \cos (c+d x))^{7/2} \, dx\\ &=-\frac {34 a b \left (53 a^2+38 b^2\right ) (e \cos (c+d x))^{9/2}}{6435 d e}+\frac {2 \left (55 a^4+60 a^2 b^2+4 b^4\right ) e (e \cos (c+d x))^{5/2} \sin (c+d x)}{385 d}-\frac {2 b \left (93 a^2+26 b^2\right ) (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{715 d e}-\frac {14 a b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^2}{65 d e}-\frac {2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^3}{15 d e}+\frac {1}{77} \left (\left (55 a^4+60 a^2 b^2+4 b^4\right ) e^2\right ) \int (e \cos (c+d x))^{3/2} \, dx\\ &=-\frac {34 a b \left (53 a^2+38 b^2\right ) (e \cos (c+d x))^{9/2}}{6435 d e}+\frac {2 \left (55 a^4+60 a^2 b^2+4 b^4\right ) e^3 \sqrt {e \cos (c+d x)} \sin (c+d x)}{231 d}+\frac {2 \left (55 a^4+60 a^2 b^2+4 b^4\right ) e (e \cos (c+d x))^{5/2} \sin (c+d x)}{385 d}-\frac {2 b \left (93 a^2+26 b^2\right ) (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{715 d e}-\frac {14 a b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^2}{65 d e}-\frac {2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^3}{15 d e}+\frac {1}{231} \left (\left (55 a^4+60 a^2 b^2+4 b^4\right ) e^4\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx\\ &=-\frac {34 a b \left (53 a^2+38 b^2\right ) (e \cos (c+d x))^{9/2}}{6435 d e}+\frac {2 \left (55 a^4+60 a^2 b^2+4 b^4\right ) e^3 \sqrt {e \cos (c+d x)} \sin (c+d x)}{231 d}+\frac {2 \left (55 a^4+60 a^2 b^2+4 b^4\right ) e (e \cos (c+d x))^{5/2} \sin (c+d x)}{385 d}-\frac {2 b \left (93 a^2+26 b^2\right ) (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{715 d e}-\frac {14 a b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^2}{65 d e}-\frac {2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^3}{15 d e}+\frac {\left (\left (55 a^4+60 a^2 b^2+4 b^4\right ) e^4 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{231 \sqrt {e \cos (c+d x)}}\\ &=-\frac {34 a b \left (53 a^2+38 b^2\right ) (e \cos (c+d x))^{9/2}}{6435 d e}+\frac {2 \left (55 a^4+60 a^2 b^2+4 b^4\right ) e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d \sqrt {e \cos (c+d x)}}+\frac {2 \left (55 a^4+60 a^2 b^2+4 b^4\right ) e^3 \sqrt {e \cos (c+d x)} \sin (c+d x)}{231 d}+\frac {2 \left (55 a^4+60 a^2 b^2+4 b^4\right ) e (e \cos (c+d x))^{5/2} \sin (c+d x)}{385 d}-\frac {2 b \left (93 a^2+26 b^2\right ) (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))}{715 d e}-\frac {14 a b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^2}{65 d e}-\frac {2 b (e \cos (c+d x))^{9/2} (a+b \sin (c+d x))^3}{15 d e}\\ \end {align*}

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Mathematica [A]  time = 4.81, size = 251, normalized size = 0.82 \[ \frac {(e \cos (c+d x))^{7/2} \left (-154 a b \left (26 a^2+11 b^2\right ) \sqrt {\cos (c+d x)}+104 \left (55 a^4+60 a^2 b^2+4 b^4\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {1}{120} \sqrt {\cos (c+d x)} \left (-28 b \cos (4 (c+d x)) \left (39 b \left (180 a^2+b^2\right ) \sin (c+d x)+220 a \left (26 a^2-b^2\right )\right )+156 \left (5720 a^4+2460 a^2 b^2+87 b^4\right ) \sin (c+d x)+\cos (2 (c+d x)) \left (78 \left (2640 a^4-7200 a^2 b^2-557 b^4\right ) \sin (c+d x)-3080 \left (208 a^3 b+73 a b^3\right )\right )+462 b^3 \cos (6 (c+d x)) (60 a+13 b \sin (c+d x))\right )\right )}{12012 d \cos ^{\frac {7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(7/2)*(a + b*Sin[c + d*x])^4,x]

[Out]

((e*Cos[c + d*x])^(7/2)*(-154*a*b*(26*a^2 + 11*b^2)*Sqrt[Cos[c + d*x]] + 104*(55*a^4 + 60*a^2*b^2 + 4*b^4)*Ell
ipticF[(c + d*x)/2, 2] + (Sqrt[Cos[c + d*x]]*(156*(5720*a^4 + 2460*a^2*b^2 + 87*b^4)*Sin[c + d*x] + 462*b^3*Co
s[6*(c + d*x)]*(60*a + 13*b*Sin[c + d*x]) - 28*b*Cos[4*(c + d*x)]*(220*a*(26*a^2 - b^2) + 39*b*(180*a^2 + b^2)
*Sin[c + d*x]) + Cos[2*(c + d*x)]*(-3080*(208*a^3*b + 73*a*b^3) + 78*(2640*a^4 - 7200*a^2*b^2 - 557*b^4)*Sin[c
+ d*x])))/120))/(12012*d*Cos[c + d*x]^(7/2))

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{4} e^{3} \cos \left (d x + c\right )^{7} - 2 \, {\left (3 \, a^{2} b^{2} + b^{4}\right )} e^{3} \cos \left (d x + c\right )^{5} + {\left (a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} e^{3} \cos \left (d x + c\right )^{3} - 4 \, {\left (a b^{3} e^{3} \cos \left (d x + c\right )^{5} - {\left (a^{3} b + a b^{3}\right )} e^{3} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(7/2)*(a+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

integral((b^4*e^3*cos(d*x + c)^7 - 2*(3*a^2*b^2 + b^4)*e^3*cos(d*x + c)^5 + (a^4 + 6*a^2*b^2 + b^4)*e^3*cos(d*
x + c)^3 - 4*(a*b^3*e^3*cos(d*x + c)^5 - (a^3*b + a*b^3)*e^3*cos(d*x + c)^3)*sin(d*x + c))*sqrt(e*cos(d*x + c)
), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}} {\left (b \sin \left (d x + c\right ) + a\right )}^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(7/2)*(a+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(7/2)*(b*sin(d*x + c) + a)^4, x)

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maple [B]  time = 3.41, size = 863, normalized size = 2.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(7/2)*(a+b*sin(d*x+c))^4,x)

[Out]

-2/45045/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^4*(3739008*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*
x+1/2*c)^12-2620800*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10+102960*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^8+946608*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-154440*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-14
4456*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+120120*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-34320*a^4*
cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+780*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-2690688*b^4*cos(1/2*d*
x+1/2*c)*sin(1/2*d*x+1/2*c)^14+768768*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^16+8673280*a*b^3*sin(1/2*d*x+1
/2*c)^11-6209280*a*b^3*sin(1/2*d*x+1/2*c)^13+1774080*a*b^3*sin(1/2*d*x+1/2*c)^15-640640*a^3*b*sin(1/2*d*x+1/2*
c)^11+1601600*a^3*b*sin(1/2*d*x+1/2*c)^9-6160000*a*b^3*sin(1/2*d*x+1/2*c)^9-1601600*a^3*b*sin(1/2*d*x+1/2*c)^7
+2279200*a*b^3*sin(1/2*d*x+1/2*c)^7+800800*a^3*b*sin(1/2*d*x+1/2*c)^5-363440*a*b^3*sin(1/2*d*x+1/2*c)^5-200200
*a^3*b*sin(1/2*d*x+1/2*c)^3-6160*a*b^3*sin(1/2*d*x+1/2*c)^3+20020*a^3*b*sin(1/2*d*x+1/2*c)+6160*a*b^3*sin(1/2*
d*x+1/2*c)+11700*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^
(1/2))*a^2*b^2-1572480*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^12+3931200*a^2*b^2*cos(1/2*d*x+1/2*c)*sin
(1/2*d*x+1/2*c)^10-3818880*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+1797120*a^2*b^2*cos(1/2*d*x+1/2*c)*
sin(1/2*d*x+1/2*c)^6-360360*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+11700*a^2*b^2*cos(1/2*d*x+1/2*c)*s
in(1/2*d*x+1/2*c)^2+10725*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+
1/2*c),2^(1/2))*a^4+780*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/
2*c),2^(1/2))*b^4)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}} {\left (b \sin \left (d x + c\right ) + a\right )}^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(7/2)*(a+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(7/2)*(b*sin(d*x + c) + a)^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^4 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^(7/2)*(a + b*sin(c + d*x))^4,x)

[Out]

int((e*cos(c + d*x))^(7/2)*(a + b*sin(c + d*x))^4, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(7/2)*(a+b*sin(d*x+c))**4,x)

[Out]

Timed out

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