3.1071 \(\int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=136 \[ -\frac {a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a \left (a^2-2 b^2\right ) \cos (c+d x)}{2 d}+\frac {b \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} b x \left (12 a^2+b^2\right )+\frac {a \cos (c+d x) (a+b \sin (c+d x))^2}{4 d}+\frac {\cos (c+d x) (a+b \sin (c+d x))^3}{4 d} \]

[Out]

1/8*b*(12*a^2+b^2)*x-a^3*arctanh(cos(d*x+c))/d+1/2*a*(a^2-2*b^2)*cos(d*x+c)/d+1/8*b*(2*a^2-b^2)*cos(d*x+c)*sin
(d*x+c)/d+1/4*a*cos(d*x+c)*(a+b*sin(d*x+c))^2/d+1/4*cos(d*x+c)*(a+b*sin(d*x+c))^3/d

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Rubi [A]  time = 0.41, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2889, 3050, 3049, 3033, 3023, 2735, 3770} \[ \frac {a \left (a^2-2 b^2\right ) \cos (c+d x)}{2 d}+\frac {b \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} b x \left (12 a^2+b^2\right )-\frac {a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^2}{4 d}+\frac {\cos (c+d x) (a+b \sin (c+d x))^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(12*a^2 + b^2)*x)/8 - (a^3*ArcTanh[Cos[c + d*x]])/d + (a*(a^2 - 2*b^2)*Cos[c + d*x])/(2*d) + (b*(2*a^2 - b^
2)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*Cos[c + d*x]*(a + b*Sin[c + d*x])^2)/(4*d) + (Cos[c + d*x]*(a + b*Sin
[c + d*x])^3)/(4*d)

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 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!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])))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \cos (c+d x) \cot (c+d x) (a+b \sin (c+d x))^3 \, dx &=\int \csc (c+d x) (a+b \sin (c+d x))^3 \left (1-\sin ^2(c+d x)\right ) \, dx\\ &=\frac {\cos (c+d x) (a+b \sin (c+d x))^3}{4 d}+\frac {1}{4} \int \csc (c+d x) (a+b \sin (c+d x))^2 \left (4 a+b \sin (c+d x)-3 a \sin ^2(c+d x)\right ) \, dx\\ &=\frac {a \cos (c+d x) (a+b \sin (c+d x))^2}{4 d}+\frac {\cos (c+d x) (a+b \sin (c+d x))^3}{4 d}+\frac {1}{12} \int \csc (c+d x) (a+b \sin (c+d x)) \left (12 a^2+9 a b \sin (c+d x)-3 \left (2 a^2-b^2\right ) \sin ^2(c+d x)\right ) \, dx\\ &=\frac {b \left (2 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^2}{4 d}+\frac {\cos (c+d x) (a+b \sin (c+d x))^3}{4 d}+\frac {1}{24} \int \csc (c+d x) \left (24 a^3+3 b \left (12 a^2+b^2\right ) \sin (c+d x)-12 a \left (a^2-2 b^2\right ) \sin ^2(c+d x)\right ) \, dx\\ &=\frac {a \left (a^2-2 b^2\right ) \cos (c+d x)}{2 d}+\frac {b \left (2 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^2}{4 d}+\frac {\cos (c+d x) (a+b \sin (c+d x))^3}{4 d}+\frac {1}{24} \int \csc (c+d x) \left (24 a^3+3 b \left (12 a^2+b^2\right ) \sin (c+d x)\right ) \, dx\\ &=\frac {1}{8} b \left (12 a^2+b^2\right ) x+\frac {a \left (a^2-2 b^2\right ) \cos (c+d x)}{2 d}+\frac {b \left (2 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^2}{4 d}+\frac {\cos (c+d x) (a+b \sin (c+d x))^3}{4 d}+a^3 \int \csc (c+d x) \, dx\\ &=\frac {1}{8} b \left (12 a^2+b^2\right ) x-\frac {a^3 \tanh ^{-1}(\cos (c+d x))}{d}+\frac {a \left (a^2-2 b^2\right ) \cos (c+d x)}{2 d}+\frac {b \left (2 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a \cos (c+d x) (a+b \sin (c+d x))^2}{4 d}+\frac {\cos (c+d x) (a+b \sin (c+d x))^3}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 129, normalized size = 0.95 \[ \frac {32 a^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-32 a^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+8 a \left (4 a^2-3 b^2\right ) \cos (c+d x)+24 a^2 b \sin (2 (c+d x))+48 a^2 b c+48 a^2 b d x-8 a b^2 \cos (3 (c+d x))-b^3 \sin (4 (c+d x))+4 b^3 c+4 b^3 d x}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(48*a^2*b*c + 4*b^3*c + 48*a^2*b*d*x + 4*b^3*d*x + 8*a*(4*a^2 - 3*b^2)*Cos[c + d*x] - 8*a*b^2*Cos[3*(c + d*x)]
- 32*a^3*Log[Cos[(c + d*x)/2]] + 32*a^3*Log[Sin[(c + d*x)/2]] + 24*a^2*b*Sin[2*(c + d*x)] - b^3*Sin[4*(c + d*
x)])/(32*d)

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fricas [A]  time = 0.75, size = 116, normalized size = 0.85 \[ -\frac {8 \, a b^{2} \cos \left (d x + c\right )^{3} - 8 \, a^{3} \cos \left (d x + c\right ) + 4 \, a^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 4 \, a^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (12 \, a^{2} b + b^{3}\right )} d x + {\left (2 \, b^{3} \cos \left (d x + c\right )^{3} - {\left (12 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*a*b^2*cos(d*x + c)^3 - 8*a^3*cos(d*x + c) + 4*a^3*log(1/2*cos(d*x + c) + 1/2) - 4*a^3*log(-1/2*cos(d*x
+ c) + 1/2) - (12*a^2*b + b^3)*d*x + (2*b^3*cos(d*x + c)^3 - (12*a^2*b + b^3)*cos(d*x + c))*sin(d*x + c))/d

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giac [B]  time = 0.23, size = 293, normalized size = 2.15 \[ \frac {8 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + {\left (12 \, a^{2} b + b^{3}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 7 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, a^{3} + 8 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(8*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + (12*a^2*b + b^3)*(d*x + c) - 2*(12*a^2*b*tan(1/2*d*x + 1/2*c)^7 -
b^3*tan(1/2*d*x + 1/2*c)^7 - 8*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 + 7*b^3*tan(1/2*d*x + 1/2*c)^5 - 24*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 - 7*b^3*tan(1/2*d*x + 1/2*c)^3 - 24*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) + b^3*tan(1/2*d*x + 1/2*c) - 8*a^3 + 8*a*b^2)/(tan(1/2*
d*x + 1/2*c)^2 + 1)^4)/d

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maple [A]  time = 0.43, size = 150, normalized size = 1.10 \[ \frac {a^{3} \cos \left (d x +c \right )}{d}+\frac {a^{3} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{d}+\frac {3 a^{2} b \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {3 a^{2} b x}{2}+\frac {3 a^{2} b c}{2 d}-\frac {a \,b^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{d}-\frac {b^{3} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4 d}+\frac {b^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{8 d}+\frac {b^{3} x}{8}+\frac {b^{3} c}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*cos(d*x+c)/d+1/d*a^3*ln(csc(d*x+c)-cot(d*x+c))+3/2/d*a^2*b*cos(d*x+c)*sin(d*x+c)+3/2*a^2*b*x+3/2/d*a^2*b*c
-a*b^2*cos(d*x+c)^3/d-1/4/d*b^3*cos(d*x+c)^3*sin(d*x+c)+1/8*b^3*cos(d*x+c)*sin(d*x+c)/d+1/8*b^3*x+1/8/d*b^3*c

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maxima [A]  time = 0.40, size = 101, normalized size = 0.74 \[ -\frac {32 \, a b^{2} \cos \left (d x + c\right )^{3} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} b - {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} b^{3} - 16 \, a^{3} {\left (2 \, \cos \left (d x + c\right ) - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(32*a*b^2*cos(d*x + c)^3 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2*b - (4*d*x + 4*c - sin(4*d*x + 4*c))*
b^3 - 16*a^3*(2*cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)))/d

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mupad [B]  time = 11.12, size = 567, normalized size = 4.17 \[ \frac {a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {2\,a\,b^2-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2\,b-\frac {b^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a\,b^2-6\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (6\,a\,b^2-2\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (6\,a\,b^2-6\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (3\,a^2\,b-\frac {b^3}{4}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^2\,b+\frac {7\,b^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (3\,a^2\,b+\frac {7\,b^3}{4}\right )-2\,a^3}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {b\,\mathrm {atan}\left (\frac {\frac {b\,\left (12\,a^2+b^2\right )\,\left (3\,a^2\,b+\frac {b^3}{4}+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^2+b^2\right )\,3{}\mathrm {i}}{4}\right )}{8}+\frac {b\,\left (12\,a^2+b^2\right )\,\left (3\,a^2\,b+\frac {b^3}{4}+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^2+b^2\right )\,3{}\mathrm {i}}{4}\right )}{8}}{2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,a^4\,b^2+\frac {3\,a^2\,b^4}{2}+\frac {b^6}{16}\right )+6\,a^5\,b+\frac {a^3\,b^3}{2}-\frac {b\,\left (12\,a^2+b^2\right )\,\left (3\,a^2\,b+\frac {b^3}{4}+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^2+b^2\right )\,3{}\mathrm {i}}{4}\right )\,1{}\mathrm {i}}{8}+\frac {b\,\left (12\,a^2+b^2\right )\,\left (3\,a^2\,b+\frac {b^3}{4}+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^2+b^2\right )\,3{}\mathrm {i}}{4}\right )\,1{}\mathrm {i}}{8}}\right )\,\left (12\,a^2+b^2\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*log(tan(c/2 + (d*x)/2)))/d - (2*a*b^2 - tan(c/2 + (d*x)/2)*(3*a^2*b - b^3/4) + tan(c/2 + (d*x)/2)^2*(2*a*
b^2 - 6*a^3) + tan(c/2 + (d*x)/2)^6*(6*a*b^2 - 2*a^3) + tan(c/2 + (d*x)/2)^4*(6*a*b^2 - 6*a^3) + tan(c/2 + (d*
x)/2)^7*(3*a^2*b - b^3/4) - tan(c/2 + (d*x)/2)^3*(3*a^2*b + (7*b^3)/4) + tan(c/2 + (d*x)/2)^5*(3*a^2*b + (7*b^
3)/4) - 2*a^3)/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/
2)^8 + 1)) + (b*atan(((b*(12*a^2 + b^2)*(3*a^2*b + b^3/4 + 2*a^3*tan(c/2 + (d*x)/2) - (b*tan(c/2 + (d*x)/2)*(1
2*a^2 + b^2)*3i)/4))/8 + (b*(12*a^2 + b^2)*(3*a^2*b + b^3/4 + 2*a^3*tan(c/2 + (d*x)/2) + (b*tan(c/2 + (d*x)/2)
*(12*a^2 + b^2)*3i)/4))/8)/(2*tan(c/2 + (d*x)/2)*(b^6/16 + (3*a^2*b^4)/2 + 9*a^4*b^2) + 6*a^5*b + (a^3*b^3)/2
- (b*(12*a^2 + b^2)*(3*a^2*b + b^3/4 + 2*a^3*tan(c/2 + (d*x)/2) - (b*tan(c/2 + (d*x)/2)*(12*a^2 + b^2)*3i)/4)*
1i)/8 + (b*(12*a^2 + b^2)*(3*a^2*b + b^3/4 + 2*a^3*tan(c/2 + (d*x)/2) + (b*tan(c/2 + (d*x)/2)*(12*a^2 + b^2)*3
i)/4)*1i)/8))*(12*a^2 + b^2))/(4*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*csc(d*x+c)*(a+b*sin(d*x+c))**3,x)

[Out]

Integral((a + b*sin(c + d*x))**3*cos(c + d*x)**2*csc(c + d*x), x)

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