3.1066 \(\int \cot ^2(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=123 \[ \frac {\left (a^2+4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 d}+\frac {2 a b \cot (c+d x)}{3 d}-\frac {a b \cot (c+d x) \csc ^2(c+d x)}{6 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d} \]

[Out]

1/8*(a^2+4*b^2)*arctanh(cos(d*x+c))/d+2/3*a*b*cot(d*x+c)/d+1/8*(a^2-2*b^2)*cot(d*x+c)*csc(d*x+c)/d-1/6*a*b*cot
(d*x+c)*csc(d*x+c)^2/d-1/4*cot(d*x+c)*csc(d*x+c)^3*(a+b*sin(d*x+c))^2/d

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Rubi [A]  time = 0.36, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2889, 3048, 3031, 3021, 2748, 3767, 8, 3770} \[ \frac {\left (a^2+4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 d}+\frac {2 a b \cot (c+d x)}{3 d}-\frac {a b \cot (c+d x) \csc ^2(c+d x)}{6 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2 + 4*b^2)*ArcTanh[Cos[c + d*x]])/(8*d) + (2*a*b*Cot[c + d*x])/(3*d) + ((a^2 - 2*b^2)*Cot[c + d*x]*Csc[c +
d*x])/(8*d) - (a*b*Cot[c + d*x]*Csc[c + d*x]^2)/(6*d) - (Cot[c + d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^2)/
(4*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 3021

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

Rule 3031

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[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3048

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 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 \cot ^2(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx &=\int \csc ^5(c+d x) (a+b \sin (c+d x))^2 \left (1-\sin ^2(c+d x)\right ) \, dx\\ &=-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}+\frac {1}{4} \int \csc ^4(c+d x) (a+b \sin (c+d x)) \left (2 b-a \sin (c+d x)-3 b \sin ^2(c+d x)\right ) \, dx\\ &=-\frac {a b \cot (c+d x) \csc ^2(c+d x)}{6 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {1}{12} \int \csc ^3(c+d x) \left (3 \left (a^2-2 b^2\right )+8 a b \sin (c+d x)+9 b^2 \sin ^2(c+d x)\right ) \, dx\\ &=\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a b \cot (c+d x) \csc ^2(c+d x)}{6 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {1}{24} \int \csc ^2(c+d x) \left (16 a b+3 \left (a^2+4 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a b \cot (c+d x) \csc ^2(c+d x)}{6 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac {1}{3} (2 a b) \int \csc ^2(c+d x) \, dx-\frac {1}{8} \left (a^2+4 b^2\right ) \int \csc (c+d x) \, dx\\ &=\frac {\left (a^2+4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a b \cot (c+d x) \csc ^2(c+d x)}{6 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}+\frac {(2 a b) \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{3 d}\\ &=\frac {\left (a^2+4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{8 d}+\frac {2 a b \cot (c+d x)}{3 d}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{8 d}-\frac {a b \cot (c+d x) \csc ^2(c+d x)}{6 d}-\frac {\cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}\\ \end {align*}

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Mathematica [B]  time = 6.17, size = 579, normalized size = 4.71 \[ \frac {\left (a^2-4 b^2\right ) \sin ^2(c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^2}{32 d (a+b \sin (c+d x))^2}+\frac {\left (-a^2-4 b^2\right ) \sin ^2(c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \csc (c+d x)+b)^2}{8 d (a+b \sin (c+d x))^2}+\frac {\left (4 b^2-a^2\right ) \sin ^2(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^2}{32 d (a+b \sin (c+d x))^2}+\frac {\left (a^2+4 b^2\right ) \sin ^2(c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right ) (a \csc (c+d x)+b)^2}{8 d (a+b \sin (c+d x))^2}-\frac {a^2 \sin ^2(c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^2}{64 d (a+b \sin (c+d x))^2}+\frac {a^2 \sin ^2(c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^2}{64 d (a+b \sin (c+d x))^2}-\frac {a b \sin ^2(c+d x) \cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^2}{12 d (a+b \sin (c+d x))^2}+\frac {a b \sin ^2(c+d x) \cot \left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^2}{3 d (a+b \sin (c+d x))^2}-\frac {a b \sin ^2(c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^2}{3 d (a+b \sin (c+d x))^2}+\frac {a b \sin ^2(c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \csc (c+d x)+b)^2}{12 d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*Cot[(c + d*x)/2]*(b + a*Csc[c + d*x])^2*Sin[c + d*x]^2)/(3*d*(a + b*Sin[c + d*x])^2) + ((a^2 - 4*b^2)*Csc
[(c + d*x)/2]^2*(b + a*Csc[c + d*x])^2*Sin[c + d*x]^2)/(32*d*(a + b*Sin[c + d*x])^2) - (a*b*Cot[(c + d*x)/2]*C
sc[(c + d*x)/2]^2*(b + a*Csc[c + d*x])^2*Sin[c + d*x]^2)/(12*d*(a + b*Sin[c + d*x])^2) - (a^2*Csc[(c + d*x)/2]
^4*(b + a*Csc[c + d*x])^2*Sin[c + d*x]^2)/(64*d*(a + b*Sin[c + d*x])^2) + ((a^2 + 4*b^2)*(b + a*Csc[c + d*x])^
2*Log[Cos[(c + d*x)/2]]*Sin[c + d*x]^2)/(8*d*(a + b*Sin[c + d*x])^2) + ((-a^2 - 4*b^2)*(b + a*Csc[c + d*x])^2*
Log[Sin[(c + d*x)/2]]*Sin[c + d*x]^2)/(8*d*(a + b*Sin[c + d*x])^2) + ((-a^2 + 4*b^2)*(b + a*Csc[c + d*x])^2*Se
c[(c + d*x)/2]^2*Sin[c + d*x]^2)/(32*d*(a + b*Sin[c + d*x])^2) + (a^2*(b + a*Csc[c + d*x])^2*Sec[(c + d*x)/2]^
4*Sin[c + d*x]^2)/(64*d*(a + b*Sin[c + d*x])^2) - (a*b*(b + a*Csc[c + d*x])^2*Sin[c + d*x]^2*Tan[(c + d*x)/2])
/(3*d*(a + b*Sin[c + d*x])^2) + (a*b*(b + a*Csc[c + d*x])^2*Sec[(c + d*x)/2]^2*Sin[c + d*x]^2*Tan[(c + d*x)/2]
)/(12*d*(a + b*Sin[c + d*x])^2)

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fricas [A]  time = 0.79, size = 200, normalized size = 1.63 \[ -\frac {32 \, a b \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + 6 \, {\left (a^{2} - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right ) - 3 \, {\left ({\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 4 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left ({\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 4 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(32*a*b*cos(d*x + c)^3*sin(d*x + c) + 6*(a^2 - 4*b^2)*cos(d*x + c)^3 + 6*(a^2 + 4*b^2)*cos(d*x + c) - 3*
((a^2 + 4*b^2)*cos(d*x + c)^4 - 2*(a^2 + 4*b^2)*cos(d*x + c)^2 + a^2 + 4*b^2)*log(1/2*cos(d*x + c) + 1/2) + 3*
((a^2 + 4*b^2)*cos(d*x + c)^4 - 2*(a^2 + 4*b^2)*cos(d*x + c)^2 + a^2 + 4*b^2)*log(-1/2*cos(d*x + c) + 1/2))/(d
*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

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giac [A]  time = 0.22, size = 182, normalized size = 1.48 \[ \frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 16 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, {\left (a^{2} + 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {50 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 200 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 16 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*a^2*tan(1/2*d*x + 1/2*c)^4 + 16*a*b*tan(1/2*d*x + 1/2*c)^3 + 24*b^2*tan(1/2*d*x + 1/2*c)^2 - 48*a*b*t
an(1/2*d*x + 1/2*c) - 24*(a^2 + 4*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) + (50*a^2*tan(1/2*d*x + 1/2*c)^4 + 200*b
^2*tan(1/2*d*x + 1/2*c)^4 + 48*a*b*tan(1/2*d*x + 1/2*c)^3 - 24*b^2*tan(1/2*d*x + 1/2*c)^2 - 16*a*b*tan(1/2*d*x
+ 1/2*c) - 3*a^2)/tan(1/2*d*x + 1/2*c)^4)/d

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maple [A]  time = 0.46, size = 173, normalized size = 1.41 \[ -\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{4}}-\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}-\frac {a^{2} \cos \left (d x +c \right )}{8 d}-\frac {a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d}-\frac {2 a b \left (\cos ^{3}\left (d x +c \right )\right )}{3 d \sin \left (d x +c \right )^{3}}-\frac {b^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{2 d \sin \left (d x +c \right )^{2}}-\frac {b^{2} \cos \left (d x +c \right )}{2 d}-\frac {b^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*csc(d*x+c)^5*(a+b*sin(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.41, size = 129, normalized size = 1.05 \[ -\frac {3 \, a^{2} {\left (\frac {2 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 12 \, b^{2} {\left (\frac {2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + \log \left (\cos \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {32 \, a b}{\tan \left (d x + c\right )^{3}}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(3*a^2*(2*(cos(d*x + c)^3 + cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - log(cos(d*x + c) + 1
) + log(cos(d*x + c) - 1)) - 12*b^2*(2*cos(d*x + c)/(cos(d*x + c)^2 - 1) + log(cos(d*x + c) + 1) - log(cos(d*x
+ c) - 1)) + 32*a*b/tan(d*x + c)^3)/d

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mupad [B]  time = 9.33, size = 165, normalized size = 1.34 \[ \frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {a^2}{8}+\frac {b^2}{2}\right )}{d}+\frac {b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^2}{4}-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+2\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{16\,d}+\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12\,d}-\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*tan(c/2 + (d*x)/2)^4)/(64*d) - (log(tan(c/2 + (d*x)/2))*(a^2/8 + b^2/2))/d + (b^2*tan(c/2 + (d*x)/2)^2)/(
8*d) - (cot(c/2 + (d*x)/2)^4*(2*b^2*tan(c/2 + (d*x)/2)^2 + a^2/4 - 4*a*b*tan(c/2 + (d*x)/2)^3 + (4*a*b*tan(c/2
+ (d*x)/2))/3))/(16*d) + (a*b*tan(c/2 + (d*x)/2)^3)/(12*d) - (a*b*tan(c/2 + (d*x)/2))/(4*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*csc(d*x+c)**5*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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