3.481 \(\int \frac {\cos ^4(c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=260 \[ -\frac {4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{5/2} d}-\frac {1048 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{693 a^3 d}-\frac {2 \sin ^5(c+d x) \cos (c+d x)}{11 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {46 \sin ^4(c+d x) \cos (c+d x)}{99 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {424 \sin ^3(c+d x) \cos (c+d x)}{693 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {200 \sin ^2(c+d x) \cos (c+d x)}{231 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {4496 \cos (c+d x)}{693 a^2 d \sqrt {a \sin (c+d x)+a}} \]

[Out]

-4*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)+4496/693*cos(d*x+c)/a^2/d/
(a+a*sin(d*x+c))^(1/2)+200/231*cos(d*x+c)*sin(d*x+c)^2/a^2/d/(a+a*sin(d*x+c))^(1/2)-424/693*cos(d*x+c)*sin(d*x
+c)^3/a^2/d/(a+a*sin(d*x+c))^(1/2)+46/99*cos(d*x+c)*sin(d*x+c)^4/a^2/d/(a+a*sin(d*x+c))^(1/2)-2/11*cos(d*x+c)*
sin(d*x+c)^5/a^2/d/(a+a*sin(d*x+c))^(1/2)-1048/693*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/a^3/d

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Rubi [A]  time = 1.36, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 9, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {2880, 2778, 2983, 2968, 3023, 2751, 2649, 206, 3046} \[ -\frac {2 \sin ^5(c+d x) \cos (c+d x)}{11 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {46 \sin ^4(c+d x) \cos (c+d x)}{99 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {424 \sin ^3(c+d x) \cos (c+d x)}{693 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {200 \sin ^2(c+d x) \cos (c+d x)}{231 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {1048 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{693 a^3 d}+\frac {4496 \cos (c+d x)}{693 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{5/2} d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x]^4)/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-4*Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(a^(5/2)*d) + (4496*Cos[c + d*
x])/(693*a^2*d*Sqrt[a + a*Sin[c + d*x]]) + (200*Cos[c + d*x]*Sin[c + d*x]^2)/(231*a^2*d*Sqrt[a + a*Sin[c + d*x
]]) - (424*Cos[c + d*x]*Sin[c + d*x]^3)/(693*a^2*d*Sqrt[a + a*Sin[c + d*x]]) + (46*Cos[c + d*x]*Sin[c + d*x]^4
)/(99*a^2*d*Sqrt[a + a*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sin[c + d*x]^5)/(11*a^2*d*Sqrt[a + a*Sin[c + d*x]]) -
(1048*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(693*a^3*d)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 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 2778

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp
[(-2*d*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n - 1))/(f*(2*n - 1)*Sqrt[a + b*Sin[e + f*x]]), x] - Dist[1/(b*(2*n
- 1)), Int[((c + d*Sin[e + f*x])^(n - 2)*Simp[a*c*d - b*(2*d^2*(n - 1) + c^2*(2*n - 1)) + d*(a*d - b*c*(4*n -
3))*Sin[e + f*x], x])/Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 2880

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[-2/(a*b*d), Int[(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 2), x], x] + Dist[1/a^2
, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 2)*(1 + Sin[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}
, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -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]

Rule 3046

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

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x) \sin ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=\frac {\int \frac {\sin ^4(c+d x) \left (1+\sin ^2(c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{a^2}-\frac {2 \int \frac {\sin ^5(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{a^2}\\ &=\frac {4 \cos (c+d x) \sin ^4(c+d x)}{9 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^5(c+d x)}{11 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \int \frac {\sin ^4(c+d x) \left (\frac {21 a}{2}-\frac {1}{2} a \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{11 a^3}+\frac {2 \int \frac {\sin ^3(c+d x) (-8 a+a \sin (c+d x))}{\sqrt {a+a \sin (c+d x)}} \, dx}{9 a^3}\\ &=-\frac {4 \cos (c+d x) \sin ^3(c+d x)}{63 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {46 \cos (c+d x) \sin ^4(c+d x)}{99 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^5(c+d x)}{11 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \int \frac {\sin ^3(c+d x) \left (-2 a^2+\frac {95}{2} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{99 a^4}+\frac {4 \int \frac {\sin ^2(c+d x) \left (3 a^2-\frac {57}{2} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{63 a^4}\\ &=\frac {76 \cos (c+d x) \sin ^2(c+d x)}{105 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {424 \cos (c+d x) \sin ^3(c+d x)}{693 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {46 \cos (c+d x) \sin ^4(c+d x)}{99 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^5(c+d x)}{11 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {8 \int \frac {\sin ^2(c+d x) \left (\frac {285 a^3}{2}-\frac {123}{4} a^3 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{693 a^5}+\frac {8 \int \frac {\sin (c+d x) \left (-57 a^3+\frac {87}{4} a^3 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{315 a^5}\\ &=\frac {200 \cos (c+d x) \sin ^2(c+d x)}{231 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {424 \cos (c+d x) \sin ^3(c+d x)}{693 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {46 \cos (c+d x) \sin ^4(c+d x)}{99 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^5(c+d x)}{11 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {16 \int \frac {\sin (c+d x) \left (-\frac {123 a^4}{2}+\frac {2973}{8} a^4 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{3465 a^6}+\frac {8 \int \frac {-57 a^3 \sin (c+d x)+\frac {87}{4} a^3 \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{315 a^5}\\ &=\frac {200 \cos (c+d x) \sin ^2(c+d x)}{231 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {424 \cos (c+d x) \sin ^3(c+d x)}{693 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {46 \cos (c+d x) \sin ^4(c+d x)}{99 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^5(c+d x)}{11 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {116 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 a^3 d}+\frac {16 \int \frac {-\frac {123}{2} a^4 \sin (c+d x)+\frac {2973}{8} a^4 \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{3465 a^6}+\frac {16 \int \frac {\frac {87 a^4}{8}-\frac {429}{4} a^4 \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{945 a^6}\\ &=\frac {1144 \cos (c+d x)}{315 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {200 \cos (c+d x) \sin ^2(c+d x)}{231 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {424 \cos (c+d x) \sin ^3(c+d x)}{693 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {46 \cos (c+d x) \sin ^4(c+d x)}{99 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^5(c+d x)}{11 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {1048 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{693 a^3 d}+\frac {32 \int \frac {\frac {2973 a^5}{16}-\frac {3711}{8} a^5 \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{10395 a^7}+\frac {2 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{a^2}\\ &=\frac {4496 \cos (c+d x)}{693 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {200 \cos (c+d x) \sin ^2(c+d x)}{231 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {424 \cos (c+d x) \sin ^3(c+d x)}{693 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {46 \cos (c+d x) \sin ^4(c+d x)}{99 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^5(c+d x)}{11 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {1048 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{693 a^3 d}+\frac {2 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{a^2}-\frac {4 \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^2 d}\\ &=-\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{a^{5/2} d}+\frac {4496 \cos (c+d x)}{693 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {200 \cos (c+d x) \sin ^2(c+d x)}{231 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {424 \cos (c+d x) \sin ^3(c+d x)}{693 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {46 \cos (c+d x) \sin ^4(c+d x)}{99 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^5(c+d x)}{11 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {1048 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{693 a^3 d}-\frac {4 \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^2 d}\\ &=-\frac {4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{a^{5/2} d}+\frac {4496 \cos (c+d x)}{693 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {200 \cos (c+d x) \sin ^2(c+d x)}{231 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {424 \cos (c+d x) \sin ^3(c+d x)}{693 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {46 \cos (c+d x) \sin ^4(c+d x)}{99 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^5(c+d x)}{11 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {1048 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{693 a^3 d}\\ \end {align*}

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Mathematica [C]  time = 1.08, size = 224, normalized size = 0.86 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5 \left (-73458 \sin \left (\frac {1}{2} (c+d x)\right )-15246 \sin \left (\frac {3}{2} (c+d x)\right )+4851 \sin \left (\frac {5}{2} (c+d x)\right )+1485 \sin \left (\frac {7}{2} (c+d x)\right )-385 \sin \left (\frac {9}{2} (c+d x)\right )-63 \sin \left (\frac {11}{2} (c+d x)\right )+73458 \cos \left (\frac {1}{2} (c+d x)\right )-15246 \cos \left (\frac {3}{2} (c+d x)\right )-4851 \cos \left (\frac {5}{2} (c+d x)\right )+1485 \cos \left (\frac {7}{2} (c+d x)\right )+385 \cos \left (\frac {9}{2} (c+d x)\right )-63 \cos \left (\frac {11}{2} (c+d x)\right )+(88704+88704 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (c+d x)\right )-1\right )\right )\right )}{11088 d (a (\sin (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^4)/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5*((88704 + 88704*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Ta
n[(c + d*x)/4])] + 73458*Cos[(c + d*x)/2] - 15246*Cos[(3*(c + d*x))/2] - 4851*Cos[(5*(c + d*x))/2] + 1485*Cos[
(7*(c + d*x))/2] + 385*Cos[(9*(c + d*x))/2] - 63*Cos[(11*(c + d*x))/2] - 73458*Sin[(c + d*x)/2] - 15246*Sin[(3
*(c + d*x))/2] + 4851*Sin[(5*(c + d*x))/2] + 1485*Sin[(7*(c + d*x))/2] - 385*Sin[(9*(c + d*x))/2] - 63*Sin[(11
*(c + d*x))/2]))/(11088*d*(a*(1 + Sin[c + d*x]))^(5/2))

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fricas [A]  time = 0.51, size = 299, normalized size = 1.15 \[ \frac {2 \, {\left (\frac {693 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} - {\left (63 \, \cos \left (d x + c\right )^{6} - 161 \, \cos \left (d x + c\right )^{5} - 562 \, \cos \left (d x + c\right )^{4} + 622 \, \cos \left (d x + c\right )^{3} + 1759 \, \cos \left (d x + c\right )^{2} + {\left (63 \, \cos \left (d x + c\right )^{5} + 224 \, \cos \left (d x + c\right )^{4} - 338 \, \cos \left (d x + c\right )^{3} - 960 \, \cos \left (d x + c\right )^{2} + 799 \, \cos \left (d x + c\right ) + 2984\right )} \sin \left (d x + c\right ) - 2185 \, \cos \left (d x + c\right ) - 2984\right )} \sqrt {a \sin \left (d x + c\right ) + a}\right )}}{693 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

2/693*(693*sqrt(2)*(a*cos(d*x + c) + a*sin(d*x + c) + a)*log(-(cos(d*x + c)^2 - (cos(d*x + c) - 2)*sin(d*x + c
) - 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*(cos(d*x + c) - sin(d*x + c) + 1)/sqrt(a) + 3*cos(d*x + c) + 2)/(cos(d*
x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2))/sqrt(a) - (63*cos(d*x + c)^6 - 161*cos(d*x + c
)^5 - 562*cos(d*x + c)^4 + 622*cos(d*x + c)^3 + 1759*cos(d*x + c)^2 + (63*cos(d*x + c)^5 + 224*cos(d*x + c)^4
- 338*cos(d*x + c)^3 - 960*cos(d*x + c)^2 + 799*cos(d*x + c) + 2984)*sin(d*x + c) - 2185*cos(d*x + c) - 2984)*
sqrt(a*sin(d*x + c) + a))/(a^3*d*cos(d*x + c) + a^3*d*sin(d*x + c) + a^3*d)

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giac [B]  time = 0.96, size = 504, normalized size = 1.94 \[ -\frac {8 \, {\left (\frac {\sqrt {2} {\left (693 \, a \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) + 746 \, \sqrt {-a} \sqrt {a}\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {-a} a^{3}} - \frac {693 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} + \sqrt {a}\right )}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {\frac {431 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {693 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {2717 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {3927 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {7326 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {8778 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {8778 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {7326 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {3927 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - {\left (\frac {2717 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + {\left (\frac {431 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {693 \, a^{3}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {11}{2}}}\right )}}{693 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-8/693*(sqrt(2)*(693*a*arctan(sqrt(a)/sqrt(-a)) + 746*sqrt(-a)*sqrt(a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(sqrt(-a
)*a^3) - 693*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) +
sqrt(a))/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) - (431*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (69
3*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (2717*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (3927*a^3/sgn(tan(1/2*d*x + 1/
2*c) + 1) - (7326*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (8778*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (8778*a^3/sgn(
tan(1/2*d*x + 1/2*c) + 1) - (7326*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) - (3927*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1)
- (2717*a^3/sgn(tan(1/2*d*x + 1/2*c) + 1) + (431*a^3*tan(1/2*d*x + 1/2*c)/sgn(tan(1/2*d*x + 1/2*c) + 1) - 693*
a^3/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d
*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c))*tan(1/2*d
*x + 1/2*c))*tan(1/2*d*x + 1/2*c))/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(11/2))/d

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maple [A]  time = 1.23, size = 166, normalized size = 0.64 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (-1386 a^{\frac {11}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )+63 \left (a -a \sin \left (d x +c \right )\right )^{\frac {11}{2}}-154 a \left (a -a \sin \left (d x +c \right )\right )^{\frac {9}{2}}+198 \left (a -a \sin \left (d x +c \right )\right )^{\frac {7}{2}} a^{2}+231 a^{4} \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}+1386 a^{5} \sqrt {a -a \sin \left (d x +c \right )}\right )}{693 a^{8} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x)

[Out]

2/693/a^8*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(-1386*a^(11/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*
2^(1/2)/a^(1/2))+63*(a-a*sin(d*x+c))^(11/2)-154*a*(a-a*sin(d*x+c))^(9/2)+198*(a-a*sin(d*x+c))^(7/2)*a^2+231*a^
4*(a-a*sin(d*x+c))^(3/2)+1386*a^5*(a-a*sin(d*x+c))^(1/2))/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^4*sin(d*x + c)^4/(a*sin(d*x + c) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^4}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*sin(c + d*x)^4)/(a + a*sin(c + d*x))^(5/2),x)

[Out]

int((cos(c + d*x)^4*sin(c + d*x)^4)/(a + a*sin(c + d*x))^(5/2), 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(cos(d*x+c)**4*sin(d*x+c)**4/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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