3.482 \(\int \frac {\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=222 \[ \frac {4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{5/2} d}+\frac {472 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{315 a^3 d}-\frac {2 \sin ^4(c+d x) \cos (c+d x)}{9 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {38 \sin ^3(c+d x) \cos (c+d x)}{63 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {92 \sin ^2(c+d x) \cos (c+d x)}{105 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {2048 \cos (c+d x)}{315 a^2 d \sqrt {a \sin (c+d x)+a}} \]

[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)-2048/315*cos(d*x+c)/a^2/d/(
a+a*sin(d*x+c))^(1/2)-92/105*cos(d*x+c)*sin(d*x+c)^2/a^2/d/(a+a*sin(d*x+c))^(1/2)+38/63*cos(d*x+c)*sin(d*x+c)^
3/a^2/d/(a+a*sin(d*x+c))^(1/2)-2/9*cos(d*x+c)*sin(d*x+c)^4/a^2/d/(a+a*sin(d*x+c))^(1/2)+472/315*cos(d*x+c)*(a+
a*sin(d*x+c))^(1/2)/a^3/d

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Rubi [A]  time = 1.08, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 16, 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} \[ -\frac {2 \sin ^4(c+d x) \cos (c+d x)}{9 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {38 \sin ^3(c+d x) \cos (c+d x)}{63 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {92 \sin ^2(c+d x) \cos (c+d x)}{105 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {472 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{315 a^3 d}-\frac {2048 \cos (c+d x)}{315 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{5/2} d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x]^3)/(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) - (2048*Cos[c + d*x
])/(315*a^2*d*Sqrt[a + a*Sin[c + d*x]]) - (92*Cos[c + d*x]*Sin[c + d*x]^2)/(105*a^2*d*Sqrt[a + a*Sin[c + d*x]]
) + (38*Cos[c + d*x]*Sin[c + d*x]^3)/(63*a^2*d*Sqrt[a + a*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sin[c + d*x]^4)/(9*
a^2*d*Sqrt[a + a*Sin[c + d*x]]) + (472*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(315*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 ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=\frac {\int \frac {\sin ^3(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 ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{a^2}\\ &=\frac {4 \cos (c+d x) \sin ^3(c+d x)}{7 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^4(c+d x)}{9 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \int \frac {\sin ^3(c+d x) \left (\frac {17 a}{2}-\frac {1}{2} a \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{9 a^3}+\frac {2 \int \frac {\sin ^2(c+d x) (-6 a+a \sin (c+d x))}{\sqrt {a+a \sin (c+d x)}} \, dx}{7 a^3}\\ &=-\frac {4 \cos (c+d x) \sin ^2(c+d x)}{35 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {38 \cos (c+d x) \sin ^3(c+d x)}{63 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^4(c+d x)}{9 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \int \frac {\sin ^2(c+d x) \left (-\frac {3 a^2}{2}+30 a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{63 a^4}+\frac {4 \int \frac {\sin (c+d x) \left (2 a^2-\frac {31}{2} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{35 a^4}\\ &=-\frac {92 \cos (c+d x) \sin ^2(c+d x)}{105 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {38 \cos (c+d x) \sin ^3(c+d x)}{63 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^4(c+d x)}{9 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {8 \int \frac {\sin (c+d x) \left (60 a^3-\frac {75}{4} a^3 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{315 a^5}+\frac {4 \int \frac {2 a^2 \sin (c+d x)-\frac {31}{2} a^2 \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{35 a^4}\\ &=-\frac {92 \cos (c+d x) \sin ^2(c+d x)}{105 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {38 \cos (c+d x) \sin ^3(c+d x)}{63 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^4(c+d x)}{9 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {124 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{105 a^3 d}+\frac {8 \int \frac {60 a^3 \sin (c+d x)-\frac {75}{4} a^3 \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{315 a^5}+\frac {8 \int \frac {-\frac {31 a^3}{4}+\frac {37}{2} a^3 \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{105 a^5}\\ &=-\frac {296 \cos (c+d x)}{105 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {92 \cos (c+d x) \sin ^2(c+d x)}{105 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {38 \cos (c+d x) \sin ^3(c+d x)}{63 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^4(c+d x)}{9 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {472 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 a^3 d}+\frac {16 \int \frac {-\frac {75 a^4}{8}+\frac {435}{4} a^4 \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{945 a^6}-\frac {2 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{a^2}\\ &=-\frac {2048 \cos (c+d x)}{315 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {92 \cos (c+d x) \sin ^2(c+d x)}{105 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {38 \cos (c+d x) \sin ^3(c+d x)}{63 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^4(c+d x)}{9 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {472 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 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 {2048 \cos (c+d x)}{315 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {92 \cos (c+d x) \sin ^2(c+d x)}{105 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {38 \cos (c+d x) \sin ^3(c+d x)}{63 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^4(c+d x)}{9 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {472 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 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 {2048 \cos (c+d x)}{315 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {92 \cos (c+d x) \sin ^2(c+d x)}{105 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {38 \cos (c+d x) \sin ^3(c+d x)}{63 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sin ^4(c+d x)}{9 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {472 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{315 a^3 d}\\ \end {align*}

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Mathematica [C]  time = 3.28, size = 225, normalized size = 1.01 \[ \frac {\sqrt {a (\sin (c+d x)+1)} \left (16380 \sin \left (\frac {1}{2} (c+d x)\right )+3150 \sin \left (\frac {3}{2} (c+d x)\right )-882 \sin \left (\frac {5}{2} (c+d x)\right )-225 \sin \left (\frac {7}{2} (c+d x)\right )+35 \sin \left (\frac {9}{2} (c+d x)\right )-16380 \cos \left (\frac {1}{2} (c+d x)\right )+3150 \cos \left (\frac {3}{2} (c+d x)\right )+882 \cos \left (\frac {5}{2} (c+d x)\right )-225 \cos \left (\frac {7}{2} (c+d x)\right )-35 \cos \left (\frac {9}{2} (c+d x)\right )+(20160+20160 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \sec \left (\frac {d x}{4}\right ) \left (\cos \left (\frac {1}{4} (2 c+d x)\right )-\sin \left (\frac {1}{4} (2 c+d x)\right )\right )\right )\right )}{2520 a^3 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^3)/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(Sqrt[a*(1 + Sin[c + d*x])]*((20160 + 20160*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*Sec[(d*x)/4]*(Cos[(2*
c + d*x)/4] - Sin[(2*c + d*x)/4])] - 16380*Cos[(c + d*x)/2] + 3150*Cos[(3*(c + d*x))/2] + 882*Cos[(5*(c + d*x)
)/2] - 225*Cos[(7*(c + d*x))/2] - 35*Cos[(9*(c + d*x))/2] + 16380*Sin[(c + d*x)/2] + 3150*Sin[(3*(c + d*x))/2]
- 882*Sin[(5*(c + d*x))/2] - 225*Sin[(7*(c + d*x))/2] + 35*Sin[(9*(c + d*x))/2]))/(2520*a^3*d*(Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]))

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fricas [A]  time = 0.52, size = 280, normalized size = 1.26 \[ \frac {2 \, {\left (\frac {315 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} - {\left (35 \, \cos \left (d x + c\right )^{5} + 130 \, \cos \left (d x + c\right )^{4} - 208 \, \cos \left (d x + c\right )^{3} - 634 \, \cos \left (d x + c\right )^{2} - {\left (35 \, \cos \left (d x + c\right )^{4} - 95 \, \cos \left (d x + c\right )^{3} - 303 \, \cos \left (d x + c\right )^{2} + 331 \, \cos \left (d x + c\right ) + 1292\right )} \sin \left (d x + c\right ) + 961 \, \cos \left (d x + c\right ) + 1292\right )} \sqrt {a \sin \left (d x + c\right ) + a}\right )}}{315 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

2/315*(315*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) - (35*cos(d*x + c)^5 + 130*cos(d*x + c
)^4 - 208*cos(d*x + c)^3 - 634*cos(d*x + c)^2 - (35*cos(d*x + c)^4 - 95*cos(d*x + c)^3 - 303*cos(d*x + c)^2 +
331*cos(d*x + c) + 1292)*sin(d*x + c) + 961*cos(d*x + c) + 1292)*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 = 0.91, size = 434, normalized size = 1.95 \[ \frac {8 \, {\left (\frac {\sqrt {2} {\left (315 \, a \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) + 323 \, \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 {315 \, \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 {{\left ({\left ({\left ({\left ({\left ({\left ({\left ({\left (\frac {197 \, a^{2} \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 {315 \, a^{2}}{\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 ) + \frac {1044 \, a^{2}}{\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 ) - \frac {1470 \, a^{2}}{\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 ) + \frac {2142 \, a^{2}}{\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 ) - \frac {2142 \, a^{2}}{\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 ) + \frac {1470 \, a^{2}}{\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 ) - \frac {1044 \, a^{2}}{\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 ) + \frac {315 \, a^{2}}{\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 ) - \frac {197 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {9}{2}}}\right )}}{315 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

8/315*(sqrt(2)*(315*a*arctan(sqrt(a)/sqrt(-a)) + 323*sqrt(-a)*sqrt(a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(sqrt(-a)
*a^3) - 315*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) + s
qrt(a))/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) + (((((((((197*a^2*tan(1/2*d*x + 1/2*c)/sgn(tan
(1/2*d*x + 1/2*c) + 1) - 315*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 1044*a^2/sgn(tan(1/2*d*
x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) - 1470*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 2142*a^
2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) - 2142*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x +
1/2*c) + 1470*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) - 1044*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1)
)*tan(1/2*d*x + 1/2*c) + 315*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) - 197*a^2/sgn(tan(1/2*d*x
+ 1/2*c) + 1))/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(9/2))/d

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maple [A]  time = 1.28, size = 166, normalized size = 0.75 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (630 a^{\frac {9}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-35 \left (a -a \sin \left (d x +c \right )\right )^{\frac {9}{2}}+45 a \left (a -a \sin \left (d x +c \right )\right )^{\frac {7}{2}}-63 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} a^{2}-105 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} a^{3}-630 a^{4} \sqrt {a -a \sin \left (d x +c \right )}\right )}{315 a^{7} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x)

[Out]

2/315*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(630*a^(9/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)
/a^(1/2))-35*(a-a*sin(d*x+c))^(9/2)+45*a*(a-a*sin(d*x+c))^(7/2)-63*(a-a*sin(d*x+c))^(5/2)*a^2-105*(a-a*sin(d*x
+c))^(3/2)*a^3-630*a^4*(a-a*sin(d*x+c))^(1/2))/a^7/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 \[ \int \frac {\cos \left (d x + c\right )^{4} \sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^4*sin(d*x + c)^3/(a*sin(d*x + c) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^3}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*sin(c + d*x)^3)/(a + a*sin(c + d*x))^(5/2),x)

[Out]

int((cos(c + d*x)^4*sin(c + d*x)^3)/(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)**3/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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