3.390 \(\int \cot ^4(c+d x) \csc ^6(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=168 \[ -\frac {a^2 \cot ^9(c+d x)}{9 d}-\frac {3 a^2 \cot ^7(c+d x)}{7 d}-\frac {2 a^2 \cot ^5(c+d x)}{5 d}-\frac {3 a^2 \tanh ^{-1}(\cos (c+d x))}{64 d}-\frac {a^2 \cot ^3(c+d x) \csc ^5(c+d x)}{4 d}+\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{8 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{32 d}-\frac {3 a^2 \cot (c+d x) \csc (c+d x)}{64 d} \]

[Out]

-3/64*a^2*arctanh(cos(d*x+c))/d-2/5*a^2*cot(d*x+c)^5/d-3/7*a^2*cot(d*x+c)^7/d-1/9*a^2*cot(d*x+c)^9/d-3/64*a^2*
cot(d*x+c)*csc(d*x+c)/d-1/32*a^2*cot(d*x+c)*csc(d*x+c)^3/d+1/8*a^2*cot(d*x+c)*csc(d*x+c)^5/d-1/4*a^2*cot(d*x+c
)^3*csc(d*x+c)^5/d

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Rubi [A]  time = 0.27, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2873, 2607, 14, 2611, 3768, 3770, 270} \[ -\frac {a^2 \cot ^9(c+d x)}{9 d}-\frac {3 a^2 \cot ^7(c+d x)}{7 d}-\frac {2 a^2 \cot ^5(c+d x)}{5 d}-\frac {3 a^2 \tanh ^{-1}(\cos (c+d x))}{64 d}-\frac {a^2 \cot ^3(c+d x) \csc ^5(c+d x)}{4 d}+\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{8 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{32 d}-\frac {3 a^2 \cot (c+d x) \csc (c+d x)}{64 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*Csc[c + d*x]^6*(a + a*Sin[c + d*x])^2,x]

[Out]

(-3*a^2*ArcTanh[Cos[c + d*x]])/(64*d) - (2*a^2*Cot[c + d*x]^5)/(5*d) - (3*a^2*Cot[c + d*x]^7)/(7*d) - (a^2*Cot
[c + d*x]^9)/(9*d) - (3*a^2*Cot[c + d*x]*Csc[c + d*x])/(64*d) - (a^2*Cot[c + d*x]*Csc[c + d*x]^3)/(32*d) + (a^
2*Cot[c + d*x]*Csc[c + d*x]^5)/(8*d) - (a^2*Cot[c + d*x]^3*Csc[c + d*x]^5)/(4*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2873

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

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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 ^4(c+d x) \csc ^6(c+d x) (a+a \sin (c+d x))^2 \, dx &=\int \left (a^2 \cot ^4(c+d x) \csc ^4(c+d x)+2 a^2 \cot ^4(c+d x) \csc ^5(c+d x)+a^2 \cot ^4(c+d x) \csc ^6(c+d x)\right ) \, dx\\ &=a^2 \int \cot ^4(c+d x) \csc ^4(c+d x) \, dx+a^2 \int \cot ^4(c+d x) \csc ^6(c+d x) \, dx+\left (2 a^2\right ) \int \cot ^4(c+d x) \csc ^5(c+d x) \, dx\\ &=-\frac {a^2 \cot ^3(c+d x) \csc ^5(c+d x)}{4 d}-\frac {1}{4} \left (3 a^2\right ) \int \cot ^2(c+d x) \csc ^5(c+d x) \, dx+\frac {a^2 \operatorname {Subst}\left (\int x^4 \left (1+x^2\right ) \, dx,x,-\cot (c+d x)\right )}{d}+\frac {a^2 \operatorname {Subst}\left (\int x^4 \left (1+x^2\right )^2 \, dx,x,-\cot (c+d x)\right )}{d}\\ &=\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{8 d}-\frac {a^2 \cot ^3(c+d x) \csc ^5(c+d x)}{4 d}+\frac {1}{8} a^2 \int \csc ^5(c+d x) \, dx+\frac {a^2 \operatorname {Subst}\left (\int \left (x^4+x^6\right ) \, dx,x,-\cot (c+d x)\right )}{d}+\frac {a^2 \operatorname {Subst}\left (\int \left (x^4+2 x^6+x^8\right ) \, dx,x,-\cot (c+d x)\right )}{d}\\ &=-\frac {2 a^2 \cot ^5(c+d x)}{5 d}-\frac {3 a^2 \cot ^7(c+d x)}{7 d}-\frac {a^2 \cot ^9(c+d x)}{9 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{32 d}+\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{8 d}-\frac {a^2 \cot ^3(c+d x) \csc ^5(c+d x)}{4 d}+\frac {1}{32} \left (3 a^2\right ) \int \csc ^3(c+d x) \, dx\\ &=-\frac {2 a^2 \cot ^5(c+d x)}{5 d}-\frac {3 a^2 \cot ^7(c+d x)}{7 d}-\frac {a^2 \cot ^9(c+d x)}{9 d}-\frac {3 a^2 \cot (c+d x) \csc (c+d x)}{64 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{32 d}+\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{8 d}-\frac {a^2 \cot ^3(c+d x) \csc ^5(c+d x)}{4 d}+\frac {1}{64} \left (3 a^2\right ) \int \csc (c+d x) \, dx\\ &=-\frac {3 a^2 \tanh ^{-1}(\cos (c+d x))}{64 d}-\frac {2 a^2 \cot ^5(c+d x)}{5 d}-\frac {3 a^2 \cot ^7(c+d x)}{7 d}-\frac {a^2 \cot ^9(c+d x)}{9 d}-\frac {3 a^2 \cot (c+d x) \csc (c+d x)}{64 d}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x)}{32 d}+\frac {a^2 \cot (c+d x) \csc ^5(c+d x)}{8 d}-\frac {a^2 \cot ^3(c+d x) \csc ^5(c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 1.37, size = 313, normalized size = 1.86 \[ -\frac {a^2 \csc ^9(c+d x) \left (212940 \sin (2 (c+d x))+195300 \sin (4 (c+d x))+16380 \sin (6 (c+d x))-1890 \sin (8 (c+d x))+451584 \cos (c+d x)+155904 \cos (3 (c+d x))-20736 \cos (5 (c+d x))-14976 \cos (7 (c+d x))+1664 \cos (9 (c+d x))-119070 \sin (c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+79380 \sin (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-34020 \sin (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+8505 \sin (7 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-945 \sin (9 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+119070 \sin (c+d x) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-79380 \sin (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+34020 \sin (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-8505 \sin (7 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+945 \sin (9 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{5160960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*Csc[c + d*x]^6*(a + a*Sin[c + d*x])^2,x]

[Out]

-1/5160960*(a^2*Csc[c + d*x]^9*(451584*Cos[c + d*x] + 155904*Cos[3*(c + d*x)] - 20736*Cos[5*(c + d*x)] - 14976
*Cos[7*(c + d*x)] + 1664*Cos[9*(c + d*x)] + 119070*Log[Cos[(c + d*x)/2]]*Sin[c + d*x] - 119070*Log[Sin[(c + d*
x)/2]]*Sin[c + d*x] + 212940*Sin[2*(c + d*x)] - 79380*Log[Cos[(c + d*x)/2]]*Sin[3*(c + d*x)] + 79380*Log[Sin[(
c + d*x)/2]]*Sin[3*(c + d*x)] + 195300*Sin[4*(c + d*x)] + 34020*Log[Cos[(c + d*x)/2]]*Sin[5*(c + d*x)] - 34020
*Log[Sin[(c + d*x)/2]]*Sin[5*(c + d*x)] + 16380*Sin[6*(c + d*x)] - 8505*Log[Cos[(c + d*x)/2]]*Sin[7*(c + d*x)]
+ 8505*Log[Sin[(c + d*x)/2]]*Sin[7*(c + d*x)] - 1890*Sin[8*(c + d*x)] + 945*Log[Cos[(c + d*x)/2]]*Sin[9*(c +
d*x)] - 945*Log[Sin[(c + d*x)/2]]*Sin[9*(c + d*x)]))/d

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fricas [A]  time = 0.50, size = 304, normalized size = 1.81 \[ -\frac {3328 \, a^{2} \cos \left (d x + c\right )^{9} - 14976 \, a^{2} \cos \left (d x + c\right )^{7} + 16128 \, a^{2} \cos \left (d x + c\right )^{5} + 945 \, {\left (a^{2} \cos \left (d x + c\right )^{8} - 4 \, a^{2} \cos \left (d x + c\right )^{6} + 6 \, a^{2} \cos \left (d x + c\right )^{4} - 4 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 945 \, {\left (a^{2} \cos \left (d x + c\right )^{8} - 4 \, a^{2} \cos \left (d x + c\right )^{6} + 6 \, a^{2} \cos \left (d x + c\right )^{4} - 4 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 630 \, {\left (3 \, a^{2} \cos \left (d x + c\right )^{7} - 11 \, a^{2} \cos \left (d x + c\right )^{5} - 11 \, a^{2} \cos \left (d x + c\right )^{3} + 3 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{40320 \, {\left (d \cos \left (d x + c\right )^{8} - 4 \, d \cos \left (d x + c\right )^{6} + 6 \, d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^10*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/40320*(3328*a^2*cos(d*x + c)^9 - 14976*a^2*cos(d*x + c)^7 + 16128*a^2*cos(d*x + c)^5 + 945*(a^2*cos(d*x + c
)^8 - 4*a^2*cos(d*x + c)^6 + 6*a^2*cos(d*x + c)^4 - 4*a^2*cos(d*x + c)^2 + a^2)*log(1/2*cos(d*x + c) + 1/2)*si
n(d*x + c) - 945*(a^2*cos(d*x + c)^8 - 4*a^2*cos(d*x + c)^6 + 6*a^2*cos(d*x + c)^4 - 4*a^2*cos(d*x + c)^2 + a^
2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 630*(3*a^2*cos(d*x + c)^7 - 11*a^2*cos(d*x + c)^5 - 11*a^2*cos(
d*x + c)^3 + 3*a^2*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^8 - 4*d*cos(d*x + c)^6 + 6*d*cos(d*x + c)^4 -
4*d*cos(d*x + c)^2 + d)*sin(d*x + c))

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giac [A]  time = 0.30, size = 261, normalized size = 1.55 \[ \frac {70 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 315 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 450 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1008 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2520 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3360 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15120 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 11340 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {42774 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 11340 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 3360 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 2520 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1008 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 450 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 315 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 70 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9}}}{322560 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^10*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/322560*(70*a^2*tan(1/2*d*x + 1/2*c)^9 + 315*a^2*tan(1/2*d*x + 1/2*c)^8 + 450*a^2*tan(1/2*d*x + 1/2*c)^7 - 10
08*a^2*tan(1/2*d*x + 1/2*c)^5 - 2520*a^2*tan(1/2*d*x + 1/2*c)^4 - 3360*a^2*tan(1/2*d*x + 1/2*c)^3 + 15120*a^2*
log(abs(tan(1/2*d*x + 1/2*c))) + 11340*a^2*tan(1/2*d*x + 1/2*c) - (42774*a^2*tan(1/2*d*x + 1/2*c)^9 + 11340*a^
2*tan(1/2*d*x + 1/2*c)^8 - 3360*a^2*tan(1/2*d*x + 1/2*c)^6 - 2520*a^2*tan(1/2*d*x + 1/2*c)^5 - 1008*a^2*tan(1/
2*d*x + 1/2*c)^4 + 450*a^2*tan(1/2*d*x + 1/2*c)^2 + 315*a^2*tan(1/2*d*x + 1/2*c) + 70*a^2)/tan(1/2*d*x + 1/2*c
)^9)/d

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maple [A]  time = 0.41, size = 224, normalized size = 1.33 \[ -\frac {13 a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{63 d \sin \left (d x +c \right )^{7}}-\frac {26 a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{315 d \sin \left (d x +c \right )^{5}}-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{8}}-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{6}}-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{32 d \sin \left (d x +c \right )^{4}}+\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{64 d \sin \left (d x +c \right )^{2}}+\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{64 d}+\frac {3 a^{2} \cos \left (d x +c \right )}{64 d}+\frac {3 a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{64 d}-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{9 d \sin \left (d x +c \right )^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^10*(a+a*sin(d*x+c))^2,x)

[Out]

-13/63/d*a^2/sin(d*x+c)^7*cos(d*x+c)^5-26/315/d*a^2/sin(d*x+c)^5*cos(d*x+c)^5-1/4/d*a^2/sin(d*x+c)^8*cos(d*x+c
)^5-1/8/d*a^2/sin(d*x+c)^6*cos(d*x+c)^5-1/32/d*a^2/sin(d*x+c)^4*cos(d*x+c)^5+1/64/d*a^2/sin(d*x+c)^2*cos(d*x+c
)^5+1/64*a^2*cos(d*x+c)^3/d+3/64*a^2*cos(d*x+c)/d+3/64/d*a^2*ln(csc(d*x+c)-cot(d*x+c))-1/9/d*a^2/sin(d*x+c)^9*
cos(d*x+c)^5

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maxima [A]  time = 0.34, size = 177, normalized size = 1.05 \[ \frac {315 \, a^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{7} - 11 \, \cos \left (d x + c\right )^{5} - 11 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{8} - 4 \, \cos \left (d x + c\right )^{6} + 6 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - \frac {1152 \, {\left (7 \, \tan \left (d x + c\right )^{2} + 5\right )} a^{2}}{\tan \left (d x + c\right )^{7}} - \frac {128 \, {\left (63 \, \tan \left (d x + c\right )^{4} + 90 \, \tan \left (d x + c\right )^{2} + 35\right )} a^{2}}{\tan \left (d x + c\right )^{9}}}{40320 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^10*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/40320*(315*a^2*(2*(3*cos(d*x + c)^7 - 11*cos(d*x + c)^5 - 11*cos(d*x + c)^3 + 3*cos(d*x + c))/(cos(d*x + c)^
8 - 4*cos(d*x + c)^6 + 6*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + 1) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c)
- 1)) - 1152*(7*tan(d*x + c)^2 + 5)*a^2/tan(d*x + c)^7 - 128*(63*tan(d*x + c)^4 + 90*tan(d*x + c)^2 + 35)*a^2
/tan(d*x + c)^9)/d

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mupad [B]  time = 12.07, size = 387, normalized size = 2.30 \[ \frac {a^2\,\left (70\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}-70\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}+315\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}-315\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+450\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-1008\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-2520\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-3360\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+11340\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-11340\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+3360\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2520\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+1008\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-450\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+15120\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\right )}{322560\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*(a + a*sin(c + d*x))^2)/sin(c + d*x)^10,x)

[Out]

(a^2*(70*sin(c/2 + (d*x)/2)^18 - 70*cos(c/2 + (d*x)/2)^18 + 315*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^17 - 315
*cos(c/2 + (d*x)/2)^17*sin(c/2 + (d*x)/2) + 450*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^16 - 1008*cos(c/2 + (d
*x)/2)^4*sin(c/2 + (d*x)/2)^14 - 2520*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^13 - 3360*cos(c/2 + (d*x)/2)^6*s
in(c/2 + (d*x)/2)^12 + 11340*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^10 - 11340*cos(c/2 + (d*x)/2)^10*sin(c/2
+ (d*x)/2)^8 + 3360*cos(c/2 + (d*x)/2)^12*sin(c/2 + (d*x)/2)^6 + 2520*cos(c/2 + (d*x)/2)^13*sin(c/2 + (d*x)/2)
^5 + 1008*cos(c/2 + (d*x)/2)^14*sin(c/2 + (d*x)/2)^4 - 450*cos(c/2 + (d*x)/2)^16*sin(c/2 + (d*x)/2)^2 + 15120*
log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2)^9))/(322560*d*cos(c/2 + (d*
x)/2)^9*sin(c/2 + (d*x)/2)^9)

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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*csc(d*x+c)**10*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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