3.1114 \(\int \cot ^4(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=236 \[ -\frac {\left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{16 d}+\frac {b \left (13 a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{60 a d}+\frac {\left (35 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{120 a^2 d}+\frac {b \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\left (15 a^4-80 a^2 b^2+12 b^4\right ) \cot (c+d x) \csc (c+d x)}{240 a^2 d}-\frac {2 a b \cot (c+d x)}{5 d}-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^3}{6 a d} \]

[Out]

-1/16*(a^2+6*b^2)*arctanh(cos(d*x+c))/d-2/5*a*b*cot(d*x+c)/d-1/240*(15*a^4-80*a^2*b^2+12*b^4)*cot(d*x+c)*csc(d
*x+c)/a^2/d+1/60*b*(13*a^2-2*b^2)*cot(d*x+c)*csc(d*x+c)^2/a/d+1/120*(35*a^2-6*b^2)*cot(d*x+c)*csc(d*x+c)^3*(a+
b*sin(d*x+c))^2/a^2/d+1/10*b*cot(d*x+c)*csc(d*x+c)^4*(a+b*sin(d*x+c))^3/a^2/d-1/6*cot(d*x+c)*csc(d*x+c)^5*(a+b
*sin(d*x+c))^3/a/d

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Rubi [A]  time = 0.60, antiderivative size = 236, 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 = {2893, 3047, 3031, 3021, 2748, 3767, 8, 3770} \[ -\frac {\left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{16 d}+\frac {b \left (13 a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{60 a d}-\frac {\left (-80 a^2 b^2+15 a^4+12 b^4\right ) \cot (c+d x) \csc (c+d x)}{240 a^2 d}+\frac {\left (35 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{120 a^2 d}+\frac {b \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {2 a b \cot (c+d x)}{5 d}-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^3}{6 a d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

-((a^2 + 6*b^2)*ArcTanh[Cos[c + d*x]])/(16*d) - (2*a*b*Cot[c + d*x])/(5*d) - ((15*a^4 - 80*a^2*b^2 + 12*b^4)*C
ot[c + d*x]*Csc[c + d*x])/(240*a^2*d) + (b*(13*a^2 - 2*b^2)*Cot[c + d*x]*Csc[c + d*x]^2)/(60*a*d) + ((35*a^2 -
6*b^2)*Cot[c + d*x]*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^2)/(120*a^2*d) + (b*Cot[c + d*x]*Csc[c + d*x]^4*(a +
b*Sin[c + d*x])^3)/(10*a^2*d) - (Cot[c + d*x]*Csc[c + d*x]^5*(a + b*Sin[c + d*x])^3)/(6*a*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 2893

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 1))/(a*d*f*(n + 1)), x] +
(-Dist[1/(a^2*d^2*(n + 1)*(n + 2)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 2)*Simp[a^2*n*(n + 2) -
b^2*(m + n + 2)*(m + n + 3) + a*b*m*Sin[e + f*x] - (a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 4))*Sin[e +
f*x]^2, x], x], x] - Simp[(b*(m + n + 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 2))/
(a^2*d^2*f*(n + 1)*(n + 2)), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || Intege
rsQ[2*m, 2*n]) &&  !m < -1 && LtQ[n, -1] && (LtQ[n, -2] || EqQ[m + n + 4, 0])

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 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + 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*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(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, B, 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 ^4(c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac {b \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^3}{6 a d}-\frac {\int \csc ^5(c+d x) (a+b \sin (c+d x))^2 \left (35 a^2-6 b^2+2 a b \sin (c+d x)-3 \left (10 a^2-b^2\right ) \sin ^2(c+d x)\right ) \, dx}{30 a^2}\\ &=\frac {\left (35 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{120 a^2 d}+\frac {b \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^3}{6 a d}-\frac {\int \csc ^4(c+d x) (a+b \sin (c+d x)) \left (6 b \left (13 a^2-2 b^2\right )-a \left (15 a^2-2 b^2\right ) \sin (c+d x)-b \left (85 a^2-6 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{120 a^2}\\ &=\frac {b \left (13 a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{60 a d}+\frac {\left (35 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{120 a^2 d}+\frac {b \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^3}{6 a d}+\frac {\int \csc ^3(c+d x) \left (3 \left (15 a^4-80 a^2 b^2+12 b^4\right )+144 a^3 b \sin (c+d x)+3 b^2 \left (85 a^2-6 b^2\right ) \sin ^2(c+d x)\right ) \, dx}{360 a^2}\\ &=-\frac {\left (15 a^4-80 a^2 b^2+12 b^4\right ) \cot (c+d x) \csc (c+d x)}{240 a^2 d}+\frac {b \left (13 a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{60 a d}+\frac {\left (35 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{120 a^2 d}+\frac {b \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^3}{6 a d}+\frac {\int \csc ^2(c+d x) \left (288 a^3 b+45 a^2 \left (a^2+6 b^2\right ) \sin (c+d x)\right ) \, dx}{720 a^2}\\ &=-\frac {\left (15 a^4-80 a^2 b^2+12 b^4\right ) \cot (c+d x) \csc (c+d x)}{240 a^2 d}+\frac {b \left (13 a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{60 a d}+\frac {\left (35 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{120 a^2 d}+\frac {b \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^3}{6 a d}+\frac {1}{5} (2 a b) \int \csc ^2(c+d x) \, dx+\frac {1}{16} \left (a^2+6 b^2\right ) \int \csc (c+d x) \, dx\\ &=-\frac {\left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac {\left (15 a^4-80 a^2 b^2+12 b^4\right ) \cot (c+d x) \csc (c+d x)}{240 a^2 d}+\frac {b \left (13 a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{60 a d}+\frac {\left (35 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{120 a^2 d}+\frac {b \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^3}{6 a d}-\frac {(2 a b) \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{5 d}\\ &=-\frac {\left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{16 d}-\frac {2 a b \cot (c+d x)}{5 d}-\frac {\left (15 a^4-80 a^2 b^2+12 b^4\right ) \cot (c+d x) \csc (c+d x)}{240 a^2 d}+\frac {b \left (13 a^2-2 b^2\right ) \cot (c+d x) \csc ^2(c+d x)}{60 a d}+\frac {\left (35 a^2-6 b^2\right ) \cot (c+d x) \csc ^3(c+d x) (a+b \sin (c+d x))^2}{120 a^2 d}+\frac {b \cot (c+d x) \csc ^4(c+d x) (a+b \sin (c+d x))^3}{10 a^2 d}-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^3}{6 a d}\\ \end {align*}

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Mathematica [A]  time = 0.89, size = 319, normalized size = 1.35 \[ \frac {-30 \left (a^2-10 b^2\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )+6 \csc ^4\left (\frac {1}{2} (c+d x)\right ) \left (5 a^2+14 a b \sin (c+d x)-5 b^2\right )+5 a^2 \sec ^6\left (\frac {1}{2} (c+d x)\right )-30 a^2 \sec ^4\left (\frac {1}{2} (c+d x)\right )+30 a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )+120 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-120 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+384 a b \tan \left (\frac {1}{2} (c+d x)\right )-384 a b \cot \left (\frac {1}{2} (c+d x)\right )-a \csc ^6\left (\frac {1}{2} (c+d x)\right ) (5 a+12 b \sin (c+d x))+768 a b \sin ^6\left (\frac {1}{2} (c+d x)\right ) \csc ^5(c+d x)-1344 a b \sin ^4\left (\frac {1}{2} (c+d x)\right ) \csc ^3(c+d x)+30 b^2 \sec ^4\left (\frac {1}{2} (c+d x)\right )-300 b^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )+720 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-720 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{1920 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*Csc[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]

[Out]

(-384*a*b*Cot[(c + d*x)/2] - 30*(a^2 - 10*b^2)*Csc[(c + d*x)/2]^2 - 120*a^2*Log[Cos[(c + d*x)/2]] - 720*b^2*Lo
g[Cos[(c + d*x)/2]] + 120*a^2*Log[Sin[(c + d*x)/2]] + 720*b^2*Log[Sin[(c + d*x)/2]] + 30*a^2*Sec[(c + d*x)/2]^
2 - 300*b^2*Sec[(c + d*x)/2]^2 - 30*a^2*Sec[(c + d*x)/2]^4 + 30*b^2*Sec[(c + d*x)/2]^4 + 5*a^2*Sec[(c + d*x)/2
]^6 - 1344*a*b*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 768*a*b*Csc[c + d*x]^5*Sin[(c + d*x)/2]^6 - a*Csc[(c + d*x)
/2]^6*(5*a + 12*b*Sin[c + d*x]) + 6*Csc[(c + d*x)/2]^4*(5*a^2 - 5*b^2 + 14*a*b*Sin[c + d*x]) + 384*a*b*Tan[(c
+ d*x)/2])/(1920*d)

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fricas [A]  time = 0.80, size = 274, normalized size = 1.16 \[ \frac {192 \, a b \cos \left (d x + c\right )^{5} \sin \left (d x + c\right ) + 30 \, {\left (a^{2} - 10 \, b^{2}\right )} \cos \left (d x + c\right )^{5} + 80 \, {\left (a^{2} + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{3} - 30 \, {\left (a^{2} + 6 \, b^{2}\right )} \cos \left (d x + c\right ) - 15 \, {\left ({\left (a^{2} + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{2} + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{2} + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 6 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left ({\left (a^{2} + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{2} + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{2} + 6 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 6 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{480 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^7*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/480*(192*a*b*cos(d*x + c)^5*sin(d*x + c) + 30*(a^2 - 10*b^2)*cos(d*x + c)^5 + 80*(a^2 + 6*b^2)*cos(d*x + c)^
3 - 30*(a^2 + 6*b^2)*cos(d*x + c) - 15*((a^2 + 6*b^2)*cos(d*x + c)^6 - 3*(a^2 + 6*b^2)*cos(d*x + c)^4 + 3*(a^2
+ 6*b^2)*cos(d*x + c)^2 - a^2 - 6*b^2)*log(1/2*cos(d*x + c) + 1/2) + 15*((a^2 + 6*b^2)*cos(d*x + c)^6 - 3*(a^
2 + 6*b^2)*cos(d*x + c)^4 + 3*(a^2 + 6*b^2)*cos(d*x + c)^2 - a^2 - 6*b^2)*log(-1/2*cos(d*x + c) + 1/2))/(d*cos
(d*x + c)^6 - 3*d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^2 - d)

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giac [A]  time = 0.27, size = 309, normalized size = 1.31 \[ \frac {5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 24 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 30 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 240 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 240 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, {\left (a^{2} + 6 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {294 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 1764 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 240 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 240 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 30 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6}}}{1920 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^7*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/1920*(5*a^2*tan(1/2*d*x + 1/2*c)^6 + 24*a*b*tan(1/2*d*x + 1/2*c)^5 - 15*a^2*tan(1/2*d*x + 1/2*c)^4 + 30*b^2*
tan(1/2*d*x + 1/2*c)^4 - 120*a*b*tan(1/2*d*x + 1/2*c)^3 - 15*a^2*tan(1/2*d*x + 1/2*c)^2 - 240*b^2*tan(1/2*d*x
+ 1/2*c)^2 + 240*a*b*tan(1/2*d*x + 1/2*c) + 120*(a^2 + 6*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) - (294*a^2*tan(1/
2*d*x + 1/2*c)^6 + 1764*b^2*tan(1/2*d*x + 1/2*c)^6 + 240*a*b*tan(1/2*d*x + 1/2*c)^5 - 15*a^2*tan(1/2*d*x + 1/2
*c)^4 - 240*b^2*tan(1/2*d*x + 1/2*c)^4 - 120*a*b*tan(1/2*d*x + 1/2*c)^3 - 15*a^2*tan(1/2*d*x + 1/2*c)^2 + 30*b
^2*tan(1/2*d*x + 1/2*c)^2 + 24*a*b*tan(1/2*d*x + 1/2*c) + 5*a^2)/tan(1/2*d*x + 1/2*c)^6)/d

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maple [A]  time = 0.49, size = 253, normalized size = 1.07 \[ -\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{6 d \sin \left (d x +c \right )^{6}}-\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{24 d \sin \left (d x +c \right )^{4}}+\frac {a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{48 d \sin \left (d x +c \right )^{2}}+\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{48 d}+\frac {a^{2} \cos \left (d x +c \right )}{16 d}+\frac {a^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{16 d}-\frac {2 a b \left (\cos ^{5}\left (d x +c \right )\right )}{5 d \sin \left (d x +c \right )^{5}}-\frac {b^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{4 d \sin \left (d x +c \right )^{4}}+\frac {b^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{8 d \sin \left (d x +c \right )^{2}}+\frac {b^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{8 d}+\frac {3 b^{2} \cos \left (d x +c \right )}{8 d}+\frac {3 b^{2} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*csc(d*x+c)^7*(a+b*sin(d*x+c))^2,x)

[Out]

-1/6/d*a^2/sin(d*x+c)^6*cos(d*x+c)^5-1/24/d*a^2/sin(d*x+c)^4*cos(d*x+c)^5+1/48/d*a^2/sin(d*x+c)^2*cos(d*x+c)^5
+1/48*a^2*cos(d*x+c)^3/d+1/16*a^2*cos(d*x+c)/d+1/16/d*a^2*ln(csc(d*x+c)-cot(d*x+c))-2/5/d*a*b/sin(d*x+c)^5*cos
(d*x+c)^5-1/4/d*b^2/sin(d*x+c)^4*cos(d*x+c)^5+1/8/d*b^2/sin(d*x+c)^2*cos(d*x+c)^5+1/8*b^2*cos(d*x+c)^3/d+3/8*b
^2*cos(d*x+c)/d+3/8/d*b^2*ln(csc(d*x+c)-cot(d*x+c))

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maxima [A]  time = 0.41, size = 180, normalized size = 0.76 \[ \frac {5 \, a^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{5} + 8 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 30 \, b^{2} {\left (\frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - \frac {192 \, a b}{\tan \left (d x + c\right )^{5}}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^7*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/480*(5*a^2*(2*(3*cos(d*x + c)^5 + 8*cos(d*x + c)^3 - 3*cos(d*x + c))/(cos(d*x + c)^6 - 3*cos(d*x + c)^4 + 3*
cos(d*x + c)^2 - 1) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)) - 30*b^2*(2*(5*cos(d*x + c)^3 - 3*cos
(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) + 3*log(cos(d*x + c) + 1) - 3*log(cos(d*x + c) - 1)) - 192*
a*b/tan(d*x + c)^5)/d

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mupad [B]  time = 9.67, size = 262, normalized size = 1.11 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {a^2}{16}+\frac {3\,b^2}{8}\right )}{d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{384\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2}{2}-b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^2}{2}+8\,b^2\right )-\frac {a^2}{6}+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-8\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}\right )}{64\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2}{128}+\frac {b^2}{8}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^2}{128}-\frac {b^2}{64}\right )}{d}-\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{16\,d}+\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{80\,d}+\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*(a + b*sin(c + d*x))^2)/sin(c + d*x)^7,x)

[Out]

(log(tan(c/2 + (d*x)/2))*(a^2/16 + (3*b^2)/8))/d + (a^2*tan(c/2 + (d*x)/2)^6)/(384*d) + (cot(c/2 + (d*x)/2)^6*
(tan(c/2 + (d*x)/2)^2*(a^2/2 - b^2) + tan(c/2 + (d*x)/2)^4*(a^2/2 + 8*b^2) - a^2/6 + 4*a*b*tan(c/2 + (d*x)/2)^
3 - 8*a*b*tan(c/2 + (d*x)/2)^5 - (4*a*b*tan(c/2 + (d*x)/2))/5))/(64*d) - (tan(c/2 + (d*x)/2)^2*(a^2/128 + b^2/
8))/d - (tan(c/2 + (d*x)/2)^4*(a^2/128 - b^2/64))/d - (a*b*tan(c/2 + (d*x)/2)^3)/(16*d) + (a*b*tan(c/2 + (d*x)
/2)^5)/(80*d) + (a*b*tan(c/2 + (d*x)/2))/(8*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)**4*csc(d*x+c)**7*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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