3.1286 \(\int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=191 \[ -\frac {a \left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}+\frac {\left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^3 d}-\frac {x \left (8 a^4-4 a^2 b^2-b^4\right )}{8 b^5}+\frac {2 a^3 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^5 d}-\frac {a \sin ^2(c+d x) \cos (c+d x)}{3 b^2 d}+\frac {\sin ^3(c+d x) \cos (c+d x)}{4 b d} \]

[Out]

-1/8*(8*a^4-4*a^2*b^2-b^4)*x/b^5-1/3*a*(3*a^2-b^2)*cos(d*x+c)/b^4/d+1/8*(4*a^2-b^2)*cos(d*x+c)*sin(d*x+c)/b^3/
d-1/3*a*cos(d*x+c)*sin(d*x+c)^2/b^2/d+1/4*cos(d*x+c)*sin(d*x+c)^3/b/d+2*a^3*arctan((b+a*tan(1/2*d*x+1/2*c))/(a
^2-b^2)^(1/2))*(a^2-b^2)^(1/2)/b^5/d

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Rubi [A]  time = 0.66, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2889, 3050, 3049, 3023, 2735, 2660, 618, 204} \[ -\frac {a \left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}+\frac {2 a^3 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^5 d}+\frac {\left (4 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 b^3 d}-\frac {x \left (-4 a^2 b^2+8 a^4-b^4\right )}{8 b^5}-\frac {a \sin ^2(c+d x) \cos (c+d x)}{3 b^2 d}+\frac {\sin ^3(c+d x) \cos (c+d x)}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*a^4 - 4*a^2*b^2 - b^4)*x)/(8*b^5) + (2*a^3*Sqrt[a^2 - b^2]*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2
]])/(b^5*d) - (a*(3*a^2 - b^2)*Cos[c + d*x])/(3*b^4*d) + ((4*a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x])/(8*b^3*d) -
(a*Cos[c + d*x]*Sin[c + d*x]^2)/(3*b^2*d) + (Cos[c + d*x]*Sin[c + d*x]^3)/(4*b*d)

Rule 204

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2889

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

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 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3050

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/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n
*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0
] && NeQ[c, 0])))

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\int \frac {\sin ^3(c+d x) \left (1-\sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx\\ &=\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}+\frac {\int \frac {\sin ^2(c+d x) \left (-3 a+b \sin (c+d x)+4 a \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{4 b}\\ &=-\frac {a \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}+\frac {\int \frac {\sin (c+d x) \left (8 a^2-a b \sin (c+d x)-3 \left (4 a^2-b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{12 b^2}\\ &=\frac {\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}+\frac {\int \frac {-3 a \left (4 a^2-b^2\right )+b \left (4 a^2+3 b^2\right ) \sin (c+d x)+8 a \left (3 a^2-b^2\right ) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{24 b^3}\\ &=-\frac {a \left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}+\frac {\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}+\frac {\int \frac {-3 a b \left (4 a^2-b^2\right )-3 \left (8 a^4-4 a^2 b^2-b^4\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{24 b^4}\\ &=-\frac {\left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^5}-\frac {a \left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}+\frac {\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}+\frac {\left (a^3 \left (a^2-b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^5}\\ &=-\frac {\left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^5}-\frac {a \left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}+\frac {\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}+\frac {\left (2 a^3 \left (a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^5 d}\\ &=-\frac {\left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^5}-\frac {a \left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}+\frac {\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}-\frac {\left (4 a^3 \left (a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^5 d}\\ &=-\frac {\left (8 a^4-4 a^2 b^2-b^4\right ) x}{8 b^5}+\frac {2 a^3 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^5 d}-\frac {a \left (3 a^2-b^2\right ) \cos (c+d x)}{3 b^4 d}+\frac {\left (4 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 b^3 d}-\frac {a \cos (c+d x) \sin ^2(c+d x)}{3 b^2 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d}\\ \end {align*}

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Mathematica [A]  time = 1.11, size = 146, normalized size = 0.76 \[ \frac {24 a^2 b^2 \sin (2 (c+d x))+24 a b \left (b^2-4 a^2\right ) \cos (c+d x)-12 \left (8 a^4-4 a^2 b^2-b^4\right ) (c+d x)+192 a^3 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )+8 a b^3 \cos (3 (c+d x))-3 b^4 \sin (4 (c+d x))}{96 b^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*(8*a^4 - 4*a^2*b^2 - b^4)*(c + d*x) + 192*a^3*Sqrt[a^2 - b^2]*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 -
b^2]] + 24*a*b*(-4*a^2 + b^2)*Cos[c + d*x] + 8*a*b^3*Cos[3*(c + d*x)] + 24*a^2*b^2*Sin[2*(c + d*x)] - 3*b^4*Si
n[4*(c + d*x)])/(96*b^5*d)

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fricas [A]  time = 0.86, size = 380, normalized size = 1.99 \[ \left [\frac {8 \, a b^{3} \cos \left (d x + c\right )^{3} - 24 \, a^{3} b \cos \left (d x + c\right ) + 12 \, \sqrt {-a^{2} + b^{2}} a^{3} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 3 \, {\left (8 \, a^{4} - 4 \, a^{2} b^{2} - b^{4}\right )} d x - 3 \, {\left (2 \, b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, b^{5} d}, \frac {8 \, a b^{3} \cos \left (d x + c\right )^{3} - 24 \, a^{3} b \cos \left (d x + c\right ) - 24 \, \sqrt {a^{2} - b^{2}} a^{3} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 3 \, {\left (8 \, a^{4} - 4 \, a^{2} b^{2} - b^{4}\right )} d x - 3 \, {\left (2 \, b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, b^{5} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(8*a*b^3*cos(d*x + c)^3 - 24*a^3*b*cos(d*x + c) + 12*sqrt(-a^2 + b^2)*a^3*log(-((2*a^2 - b^2)*cos(d*x +
c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^
2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 3*(8*a^4 - 4*a^2*b^2 - b^4)*d*x - 3*(2*b^4*cos(d*x + c)^
3 - (4*a^2*b^2 + b^4)*cos(d*x + c))*sin(d*x + c))/(b^5*d), 1/24*(8*a*b^3*cos(d*x + c)^3 - 24*a^3*b*cos(d*x + c
) - 24*sqrt(a^2 - b^2)*a^3*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - 3*(8*a^4 - 4*a^2*b^2
- b^4)*d*x - 3*(2*b^4*cos(d*x + c)^3 - (4*a^2*b^2 + b^4)*cos(d*x + c))*sin(d*x + c))/(b^5*d)]

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giac [B]  time = 0.18, size = 366, normalized size = 1.92 \[ -\frac {\frac {3 \, {\left (8 \, a^{4} - 4 \, a^{2} b^{2} - b^{4}\right )} {\left (d x + c\right )}}{b^{5}} - \frac {48 \, {\left (a^{5} - a^{3} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{5}} + \frac {2 \, {\left (12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a^{3} - 8 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} b^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*(8*a^4 - 4*a^2*b^2 - b^4)*(d*x + c)/b^5 - 48*(a^5 - a^3*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a)
+ arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*b^5) + 2*(12*a^2*b*tan(1/2*d*x + 1/2
*c)^7 - 3*b^3*tan(1/2*d*x + 1/2*c)^7 + 24*a^3*tan(1/2*d*x + 1/2*c)^6 - 24*a*b^2*tan(1/2*d*x + 1/2*c)^6 + 12*a^
2*b*tan(1/2*d*x + 1/2*c)^5 + 21*b^3*tan(1/2*d*x + 1/2*c)^5 + 72*a^3*tan(1/2*d*x + 1/2*c)^4 - 24*a*b^2*tan(1/2*
d*x + 1/2*c)^4 - 12*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 21*b^3*tan(1/2*d*x + 1/2*c)^3 + 72*a^3*tan(1/2*d*x + 1/2*c)
^2 - 8*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 12*a^2*b*tan(1/2*d*x + 1/2*c) + 3*b^3*tan(1/2*d*x + 1/2*c) + 24*a^3 - 8*
a*b^2)/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*b^4))/d

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maple [B]  time = 0.28, size = 657, normalized size = 3.44 \[ -\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {6 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {2 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{3 d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {2 a^{3}}{d \,b^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {2 a}{3 d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}{d \,b^{5}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d b}+\frac {2 a^{3} \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*a^2+1/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/
2*c)^7-2/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^6*a^3+2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*
d*x+1/2*c)^6*a-1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*a^2-7/4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^4*
tan(1/2*d*x+1/2*c)^5-6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^4*a^3+2/d/b^2/(1+tan(1/2*d*x+1/2*c)
^2)^4*tan(1/2*d*x+1/2*c)^4*a+1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*a^2+7/4/d/b/(1+tan(1/2*d*
x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3-6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^2*a^3+2/3/d/b^2/(1+ta
n(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^2*a+1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*a^2-1/4/d/b
/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)-2/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^4*a^3+2/3/d/b^2/(1+tan(1/2*d*x
+1/2*c)^2)^4*a-2/d/b^5*arctan(tan(1/2*d*x+1/2*c))*a^4+1/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a^2+1/4/d/b*arctan(ta
n(1/2*d*x+1/2*c))+2/d*a^3*(a^2-b^2)^(1/2)/b^5*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 13.35, size = 2616, normalized size = 13.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7*tan(c/2 + (d*x)/2)^3)/(4*d*(b + 4*b*tan(c/2 + (d*x)/2)^2 + 6*b*tan(c/2 + (d*x)/2)^4 + 4*b*tan(c/2 + (d*x)/2
)^6 + b*tan(c/2 + (d*x)/2)^8)) - (7*tan(c/2 + (d*x)/2)^5)/(4*d*(b + 4*b*tan(c/2 + (d*x)/2)^2 + 6*b*tan(c/2 + (
d*x)/2)^4 + 4*b*tan(c/2 + (d*x)/2)^6 + b*tan(c/2 + (d*x)/2)^8)) + tan(c/2 + (d*x)/2)^7/(4*d*(b + 4*b*tan(c/2 +
(d*x)/2)^2 + 6*b*tan(c/2 + (d*x)/2)^4 + 4*b*tan(c/2 + (d*x)/2)^6 + b*tan(c/2 + (d*x)/2)^8)) + (2*a)/(3*d*(4*b
^2*tan(c/2 + (d*x)/2)^2 + 6*b^2*tan(c/2 + (d*x)/2)^4 + 4*b^2*tan(c/2 + (d*x)/2)^6 + b^2*tan(c/2 + (d*x)/2)^8 +
b^2)) - (2*a^3)/(d*(4*b^4*tan(c/2 + (d*x)/2)^2 + 6*b^4*tan(c/2 + (d*x)/2)^4 + 4*b^4*tan(c/2 + (d*x)/2)^6 + b^
4*tan(c/2 + (d*x)/2)^8 + b^4)) + atan((11*a^3*tan(c/2 + (d*x)/2))/(8*((a*b^2)/8 + (11*a^3)/8 + (3*a^5)/(2*b^2)
- (11*a^7)/b^4 + (8*a^9)/b^6)) + (3*a^5*tan(c/2 + (d*x)/2))/(2*((a*b^4)/8 + (3*a^5)/2 + (11*a^3*b^2)/8 - (11*
a^7)/b^2 + (8*a^9)/b^4)) - (11*a^7*tan(c/2 + (d*x)/2))/((a*b^6)/8 - 11*a^7 + (11*a^3*b^4)/8 + (3*a^5*b^2)/2 +
(8*a^9)/b^2) + (8*a^9*tan(c/2 + (d*x)/2))/((a*b^8)/8 + 8*a^9 + (11*a^3*b^6)/8 + (3*a^5*b^4)/2 - 11*a^7*b^2) +
(a*b*tan(c/2 + (d*x)/2))/(8*((a*b)/8 + (11*a^3)/(8*b) + (3*a^5)/(2*b^3) - (11*a^7)/b^5 + (8*a^9)/b^7)))/(4*b*d
) - tan(c/2 + (d*x)/2)/(4*d*(b + 4*b*tan(c/2 + (d*x)/2)^2 + 6*b*tan(c/2 + (d*x)/2)^4 + 4*b*tan(c/2 + (d*x)/2)^
6 + b*tan(c/2 + (d*x)/2)^8)) + (2*a*tan(c/2 + (d*x)/2)^2)/(3*d*(4*b^2*tan(c/2 + (d*x)/2)^2 + 6*b^2*tan(c/2 + (
d*x)/2)^4 + 4*b^2*tan(c/2 + (d*x)/2)^6 + b^2*tan(c/2 + (d*x)/2)^8 + b^2)) + (2*a*tan(c/2 + (d*x)/2)^4)/(d*(4*b
^2*tan(c/2 + (d*x)/2)^2 + 6*b^2*tan(c/2 + (d*x)/2)^4 + 4*b^2*tan(c/2 + (d*x)/2)^6 + b^2*tan(c/2 + (d*x)/2)^8 +
b^2)) + (2*a*tan(c/2 + (d*x)/2)^6)/(d*(4*b^2*tan(c/2 + (d*x)/2)^2 + 6*b^2*tan(c/2 + (d*x)/2)^4 + 4*b^2*tan(c/
2 + (d*x)/2)^6 + b^2*tan(c/2 + (d*x)/2)^8 + b^2)) + (a^2*tan(c/2 + (d*x)/2))/(d*(4*b^3*tan(c/2 + (d*x)/2)^2 +
6*b^3*tan(c/2 + (d*x)/2)^4 + 4*b^3*tan(c/2 + (d*x)/2)^6 + b^3*tan(c/2 + (d*x)/2)^8 + b^3)) + (a^2*atan((11*a^3
*tan(c/2 + (d*x)/2))/(8*((a*b^2)/8 + (11*a^3)/8 + (3*a^5)/(2*b^2) - (11*a^7)/b^4 + (8*a^9)/b^6)) + (3*a^5*tan(
c/2 + (d*x)/2))/(2*((a*b^4)/8 + (3*a^5)/2 + (11*a^3*b^2)/8 - (11*a^7)/b^2 + (8*a^9)/b^4)) - (11*a^7*tan(c/2 +
(d*x)/2))/((a*b^6)/8 - 11*a^7 + (11*a^3*b^4)/8 + (3*a^5*b^2)/2 + (8*a^9)/b^2) + (8*a^9*tan(c/2 + (d*x)/2))/((a
*b^8)/8 + 8*a^9 + (11*a^3*b^6)/8 + (3*a^5*b^4)/2 - 11*a^7*b^2) + (a*b*tan(c/2 + (d*x)/2))/(8*((a*b)/8 + (11*a^
3)/(8*b) + (3*a^5)/(2*b^3) - (11*a^7)/b^5 + (8*a^9)/b^7))))/(b^3*d) - (2*a^4*atan((11*a^3*tan(c/2 + (d*x)/2))/
(8*((a*b^2)/8 + (11*a^3)/8 + (3*a^5)/(2*b^2) - (11*a^7)/b^4 + (8*a^9)/b^6)) + (3*a^5*tan(c/2 + (d*x)/2))/(2*((
a*b^4)/8 + (3*a^5)/2 + (11*a^3*b^2)/8 - (11*a^7)/b^2 + (8*a^9)/b^4)) - (11*a^7*tan(c/2 + (d*x)/2))/((a*b^6)/8
- 11*a^7 + (11*a^3*b^4)/8 + (3*a^5*b^2)/2 + (8*a^9)/b^2) + (8*a^9*tan(c/2 + (d*x)/2))/((a*b^8)/8 + 8*a^9 + (11
*a^3*b^6)/8 + (3*a^5*b^4)/2 - 11*a^7*b^2) + (a*b*tan(c/2 + (d*x)/2))/(8*((a*b)/8 + (11*a^3)/(8*b) + (3*a^5)/(2
*b^3) - (11*a^7)/b^5 + (8*a^9)/b^7))))/(b^5*d) + (a^2*tan(c/2 + (d*x)/2)^3)/(d*(4*b^3*tan(c/2 + (d*x)/2)^2 + 6
*b^3*tan(c/2 + (d*x)/2)^4 + 4*b^3*tan(c/2 + (d*x)/2)^6 + b^3*tan(c/2 + (d*x)/2)^8 + b^3)) - (a^2*tan(c/2 + (d*
x)/2)^5)/(d*(4*b^3*tan(c/2 + (d*x)/2)^2 + 6*b^3*tan(c/2 + (d*x)/2)^4 + 4*b^3*tan(c/2 + (d*x)/2)^6 + b^3*tan(c/
2 + (d*x)/2)^8 + b^3)) - (a^2*tan(c/2 + (d*x)/2)^7)/(d*(4*b^3*tan(c/2 + (d*x)/2)^2 + 6*b^3*tan(c/2 + (d*x)/2)^
4 + 4*b^3*tan(c/2 + (d*x)/2)^6 + b^3*tan(c/2 + (d*x)/2)^8 + b^3)) - (6*a^3*tan(c/2 + (d*x)/2)^2)/(d*(4*b^4*tan
(c/2 + (d*x)/2)^2 + 6*b^4*tan(c/2 + (d*x)/2)^4 + 4*b^4*tan(c/2 + (d*x)/2)^6 + b^4*tan(c/2 + (d*x)/2)^8 + b^4))
- (6*a^3*tan(c/2 + (d*x)/2)^4)/(d*(4*b^4*tan(c/2 + (d*x)/2)^2 + 6*b^4*tan(c/2 + (d*x)/2)^4 + 4*b^4*tan(c/2 +
(d*x)/2)^6 + b^4*tan(c/2 + (d*x)/2)^8 + b^4)) - (2*a^3*tan(c/2 + (d*x)/2)^6)/(d*(4*b^4*tan(c/2 + (d*x)/2)^2 +
6*b^4*tan(c/2 + (d*x)/2)^4 + 4*b^4*tan(c/2 + (d*x)/2)^6 + b^4*tan(c/2 + (d*x)/2)^8 + b^4)) - (2*a^3*atanh((a^5
*(b^2 - a^2)^(1/2))/(a^5*b + (7*a^7)/b - (8*a^9)/b^3 + 14*a^6*tan(c/2 + (d*x)/2) + 2*a^4*b^2*tan(c/2 + (d*x)/2
) - (16*a^8*tan(c/2 + (d*x)/2))/b^2) + (8*a^7*(b^2 - a^2)^(1/2))/(7*a^7*b + a^5*b^3 - (8*a^9)/b - 16*a^8*tan(c
/2 + (d*x)/2) + 2*a^4*b^4*tan(c/2 + (d*x)/2) + 14*a^6*b^2*tan(c/2 + (d*x)/2)) + (2*a^4*tan(c/2 + (d*x)/2)*(b^2
- a^2)^(1/2))/(a^5 + (7*a^7)/b^2 - (8*a^9)/b^4 + 2*a^4*b*tan(c/2 + (d*x)/2) + (14*a^6*tan(c/2 + (d*x)/2))/b -
(16*a^8*tan(c/2 + (d*x)/2))/b^3) + (15*a^6*tan(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(7*a^7 + a^5*b^2 - (8*a^9)/b
^2 + 14*a^6*b*tan(c/2 + (d*x)/2) + 2*a^4*b^3*tan(c/2 + (d*x)/2) - (16*a^8*tan(c/2 + (d*x)/2))/b) - (8*a^8*tan(
c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(a^5*b^4 - 8*a^9 + 7*a^7*b^2 - 16*a^8*b*tan(c/2 + (d*x)/2) + 2*a^4*b^5*tan(c
/2 + (d*x)/2) + 14*a^6*b^3*tan(c/2 + (d*x)/2)))*(b^2 - a^2)^(1/2))/(b^5*d)

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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+b*sin(d*x+c)),x)

[Out]

Timed out

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