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

Optimal. Leaf size=184 \[ \frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}+\frac {76 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{105 a^2 d}+\frac {2 \sin ^3(c+d x) \cos (c+d x)}{7 a d \sqrt {a \sin (c+d x)+a}}-\frac {16 \sin ^2(c+d x) \cos (c+d x)}{35 a d \sqrt {a \sin (c+d x)+a}}-\frac {344 \cos (c+d x)}{105 a d \sqrt {a \sin (c+d x)+a}} \]

[Out]

2*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-344/105*cos(d*x+c)/a/d/(a+a
*sin(d*x+c))^(1/2)-16/35*cos(d*x+c)*sin(d*x+c)^2/a/d/(a+a*sin(d*x+c))^(1/2)+2/7*cos(d*x+c)*sin(d*x+c)^3/a/d/(a
+a*sin(d*x+c))^(1/2)+76/105*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/a^2/d

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Rubi [A]  time = 0.59, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2879, 2983, 2968, 3023, 2751, 2649, 206} \[ \frac {76 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{105 a^2 d}+\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}+\frac {2 \sin ^3(c+d x) \cos (c+d x)}{7 a d \sqrt {a \sin (c+d x)+a}}-\frac {16 \sin ^2(c+d x) \cos (c+d x)}{35 a d \sqrt {a \sin (c+d x)+a}}-\frac {344 \cos (c+d x)}{105 a d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(a^(3/2)*d) - (344*Cos[c + d*x]
)/(105*a*d*Sqrt[a + a*Sin[c + d*x]]) - (16*Cos[c + d*x]*Sin[c + d*x]^2)/(35*a*d*Sqrt[a + a*Sin[c + d*x]]) + (2
*Cos[c + d*x]*Sin[c + d*x]^3)/(7*a*d*Sqrt[a + a*Sin[c + d*x]]) + (76*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(1
05*a^2*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 2879

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

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 \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac {\int \frac {\sin ^3(c+d x) (a-a \sin (c+d x))}{\sqrt {a+a \sin (c+d x)}} \, dx}{a^2}\\ &=\frac {2 \cos (c+d x) \sin ^3(c+d x)}{7 a d \sqrt {a+a \sin (c+d x)}}+\frac {2 \int \frac {\sin ^2(c+d x) \left (-3 a^2+4 a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{7 a^3}\\ &=-\frac {16 \cos (c+d x) \sin ^2(c+d x)}{35 a d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^3(c+d x)}{7 a d \sqrt {a+a \sin (c+d x)}}+\frac {4 \int \frac {\sin (c+d x) \left (8 a^3-\frac {19}{2} a^3 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{35 a^4}\\ &=-\frac {16 \cos (c+d x) \sin ^2(c+d x)}{35 a d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^3(c+d x)}{7 a d \sqrt {a+a \sin (c+d x)}}+\frac {4 \int \frac {8 a^3 \sin (c+d x)-\frac {19}{2} a^3 \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{35 a^4}\\ &=-\frac {16 \cos (c+d x) \sin ^2(c+d x)}{35 a d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^3(c+d x)}{7 a d \sqrt {a+a \sin (c+d x)}}+\frac {76 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{105 a^2 d}+\frac {8 \int \frac {-\frac {19 a^4}{4}+\frac {43}{2} a^4 \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{105 a^5}\\ &=-\frac {344 \cos (c+d x)}{105 a d \sqrt {a+a \sin (c+d x)}}-\frac {16 \cos (c+d x) \sin ^2(c+d x)}{35 a d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^3(c+d x)}{7 a d \sqrt {a+a \sin (c+d x)}}+\frac {76 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{105 a^2 d}-\frac {2 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{a}\\ &=-\frac {344 \cos (c+d x)}{105 a d \sqrt {a+a \sin (c+d x)}}-\frac {16 \cos (c+d x) \sin ^2(c+d x)}{35 a d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^3(c+d x)}{7 a d \sqrt {a+a \sin (c+d x)}}+\frac {76 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{105 a^2 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 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^{3/2} d}-\frac {344 \cos (c+d x)}{105 a d \sqrt {a+a \sin (c+d x)}}-\frac {16 \cos (c+d x) \sin ^2(c+d x)}{35 a d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sin ^3(c+d x)}{7 a d \sqrt {a+a \sin (c+d x)}}+\frac {76 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{105 a^2 d}\\ \end {align*}

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Mathematica [C]  time = 1.88, size = 201, normalized size = 1.09 \[ \frac {\sqrt {a (\sin (c+d x)+1)} \left (1365 \sin \left (\frac {1}{2} (c+d x)\right )+245 \sin \left (\frac {3}{2} (c+d x)\right )-63 \sin \left (\frac {5}{2} (c+d x)\right )-15 \sin \left (\frac {7}{2} (c+d x)\right )-1365 \cos \left (\frac {1}{2} (c+d x)\right )+245 \cos \left (\frac {3}{2} (c+d x)\right )+63 \cos \left (\frac {5}{2} (c+d x)\right )-15 \cos \left (\frac {7}{2} (c+d x)\right )+(1680+1680 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \sec \left (\frac {d x}{4}\right ) \left (\cos \left (\frac {1}{4} (2 c+d x)\right )-\sin \left (\frac {1}{4} (2 c+d x)\right )\right )\right )\right )}{420 a^2 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*(1 + Sin[c + d*x])]*((1680 + 1680*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])] - 1365*Cos[(c + d*x)/2] + 245*Cos[(3*(c + d*x))/2] + 63*Cos[(5*(c + d*x))/2]
- 15*Cos[(7*(c + d*x))/2] + 1365*Sin[(c + d*x)/2] + 245*Sin[(3*(c + d*x))/2] - 63*Sin[(5*(c + d*x))/2] - 15*Si
n[(7*(c + d*x))/2]))/(420*a^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [A]  time = 0.51, size = 259, normalized size = 1.41 \[ \frac {\frac {105 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{\sqrt {a}} - 2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 24 \, \cos \left (d x + c\right )^{3} - 92 \, \cos \left (d x + c\right )^{2} + {\left (15 \, \cos \left (d x + c\right )^{3} + 39 \, \cos \left (d x + c\right )^{2} - 53 \, \cos \left (d x + c\right ) - 211\right )} \sin \left (d x + c\right ) + 158 \, \cos \left (d x + c\right ) + 211\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{105 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(105*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) - 2*(15*cos(d*x + c)^4 - 24*cos(d*x +
c)^3 - 92*cos(d*x + c)^2 + (15*cos(d*x + c)^3 + 39*cos(d*x + c)^2 - 53*cos(d*x + c) - 211)*sin(d*x + c) + 158*
cos(d*x + c) + 211)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) + a^2*d)

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giac [B]  time = 0.70, size = 375, normalized size = 2.04 \[ \frac {2 \, {\left (\frac {\sqrt {2} {\left (210 \, a \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) + 211 \, \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^{2}} - \frac {210 \, \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 \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {2 \, {\left ({\left ({\left ({\left ({\left ({\left ({\left (\frac {67 \, 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 {105 \, 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 {287 \, 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 {385 \, 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 {385 \, 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 {287 \, 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 {105 \, 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 {67 \, a^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {7}{2}}}\right )}}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(sqrt(2)*(210*a*arctan(sqrt(a)/sqrt(-a)) + 211*sqrt(-a)*sqrt(a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(sqrt(-a)
*a^2) - 210*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*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 2*(((((((67*a^2*tan(1/2*d*x + 1/2*c)/sgn(tan(1/
2*d*x + 1/2*c) + 1) - 105*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 287*a^2/sgn(tan(1/2*d*x +
1/2*c) + 1))*tan(1/2*d*x + 1/2*c) - 385*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 385*a^2/sgn(
tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) - 287*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c)
+ 105*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) - 67*a^2/sgn(tan(1/2*d*x + 1/2*c) + 1))/(a*tan(1
/2*d*x + 1/2*c)^2 + a)^(7/2))/d

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maple [A]  time = 1.30, size = 148, normalized size = 0.80 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (105 a^{\frac {7}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-15 \left (a -a \sin \left (d x +c \right )\right )^{\frac {7}{2}}+21 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} a -35 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} a^{2}-105 a^{3} \sqrt {a -a \sin \left (d x +c \right )}\right )}{105 a^{5} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(105*a^(7/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)
/a^(1/2))-15*(a-a*sin(d*x+c))^(7/2)+21*(a-a*sin(d*x+c))^(5/2)*a-35*(a-a*sin(d*x+c))^(3/2)*a^2-105*a^3*(a-a*sin
(d*x+c))^(1/2))/a^5/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 \[ \int \frac {\cos \left (d x + c\right )^{2} \sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^3}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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