3.1234 \(\int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=226 \[ \frac {b \csc ^4(c+d x)}{2 a^3 d}-\frac {\csc ^5(c+d x)}{5 a^2 d}-\frac {b \left (a^2-b^2\right )^2}{a^6 d (a+b \sin (c+d x))}-\frac {2 b \left (a^2-b^2\right ) \csc ^2(c+d x)}{a^5 d}+\frac {\left (2 a^2-3 b^2\right ) \csc ^3(c+d x)}{3 a^4 d}-\frac {2 b \left (a^4-4 a^2 b^2+3 b^4\right ) \log (\sin (c+d x))}{a^7 d}+\frac {2 b \left (a^4-4 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))}{a^7 d}-\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \csc (c+d x)}{a^6 d} \]

[Out]

-(a^4-6*a^2*b^2+5*b^4)*csc(d*x+c)/a^6/d-2*b*(a^2-b^2)*csc(d*x+c)^2/a^5/d+1/3*(2*a^2-3*b^2)*csc(d*x+c)^3/a^4/d+
1/2*b*csc(d*x+c)^4/a^3/d-1/5*csc(d*x+c)^5/a^2/d-2*b*(a^4-4*a^2*b^2+3*b^4)*ln(sin(d*x+c))/a^7/d+2*b*(a^4-4*a^2*
b^2+3*b^4)*ln(a+b*sin(d*x+c))/a^7/d-b*(a^2-b^2)^2/a^6/d/(a+b*sin(d*x+c))

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Rubi [A]  time = 0.26, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2837, 12, 894} \[ -\frac {b \left (a^2-b^2\right )^2}{a^6 d (a+b \sin (c+d x))}+\frac {\left (2 a^2-3 b^2\right ) \csc ^3(c+d x)}{3 a^4 d}-\frac {2 b \left (a^2-b^2\right ) \csc ^2(c+d x)}{a^5 d}-\frac {\left (-6 a^2 b^2+a^4+5 b^4\right ) \csc (c+d x)}{a^6 d}-\frac {2 b \left (-4 a^2 b^2+a^4+3 b^4\right ) \log (\sin (c+d x))}{a^7 d}+\frac {2 b \left (-4 a^2 b^2+a^4+3 b^4\right ) \log (a+b \sin (c+d x))}{a^7 d}+\frac {b \csc ^4(c+d x)}{2 a^3 d}-\frac {\csc ^5(c+d x)}{5 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^5*Csc[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

-(((a^4 - 6*a^2*b^2 + 5*b^4)*Csc[c + d*x])/(a^6*d)) - (2*b*(a^2 - b^2)*Csc[c + d*x]^2)/(a^5*d) + ((2*a^2 - 3*b
^2)*Csc[c + d*x]^3)/(3*a^4*d) + (b*Csc[c + d*x]^4)/(2*a^3*d) - Csc[c + d*x]^5/(5*a^2*d) - (2*b*(a^4 - 4*a^2*b^
2 + 3*b^4)*Log[Sin[c + d*x]])/(a^7*d) + (2*b*(a^4 - 4*a^2*b^2 + 3*b^4)*Log[a + b*Sin[c + d*x]])/(a^7*d) - (b*(
a^2 - b^2)^2)/(a^6*d*(a + b*Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2837

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

Rubi steps

\begin {align*} \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {b^6 \left (b^2-x^2\right )^2}{x^6 (a+x)^2} \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac {b \operatorname {Subst}\left (\int \frac {\left (b^2-x^2\right )^2}{x^6 (a+x)^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b \operatorname {Subst}\left (\int \left (\frac {b^4}{a^2 x^6}-\frac {2 b^4}{a^3 x^5}+\frac {-2 a^2 b^2+3 b^4}{a^4 x^4}+\frac {4 b^2 \left (a^2-b^2\right )}{a^5 x^3}+\frac {a^4-6 a^2 b^2+5 b^4}{a^6 x^2}-\frac {2 \left (a^4-4 a^2 b^2+3 b^4\right )}{a^7 x}+\frac {\left (a^2-b^2\right )^2}{a^6 (a+x)^2}+\frac {2 \left (a^4-4 a^2 b^2+3 b^4\right )}{a^7 (a+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \csc (c+d x)}{a^6 d}-\frac {2 b \left (a^2-b^2\right ) \csc ^2(c+d x)}{a^5 d}+\frac {\left (2 a^2-3 b^2\right ) \csc ^3(c+d x)}{3 a^4 d}+\frac {b \csc ^4(c+d x)}{2 a^3 d}-\frac {\csc ^5(c+d x)}{5 a^2 d}-\frac {2 b \left (a^4-4 a^2 b^2+3 b^4\right ) \log (\sin (c+d x))}{a^7 d}+\frac {2 b \left (a^4-4 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))}{a^7 d}-\frac {b \left (a^2-b^2\right )^2}{a^6 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 3.17, size = 220, normalized size = 0.97 \[ \frac {-6 a^6 \csc ^6(c+d x)+9 a^5 b \csc ^5(c+d x)+\left (30 a^3 b^3-40 a^5 b\right ) \csc ^3(c+d x)+5 a^4 \left (4 a^2-3 b^2\right ) \csc ^4(c+d x)-30 a^2 \left (a^4-4 a^2 b^2+3 b^4\right ) \csc ^2(c+d x)-60 b^2 \left (a^4-4 a^2 b^2+3 b^4\right ) (\log (\sin (c+d x))-\log (a+b \sin (c+d x)))-60 a b \left (a^4-4 a^2 b^2+3 b^4\right ) \csc (c+d x) (-\log (a+b \sin (c+d x))+\log (\sin (c+d x))+1)}{30 a^7 d (a \csc (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^5*Csc[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

(-30*a^2*(a^4 - 4*a^2*b^2 + 3*b^4)*Csc[c + d*x]^2 + (-40*a^5*b + 30*a^3*b^3)*Csc[c + d*x]^3 + 5*a^4*(4*a^2 - 3
*b^2)*Csc[c + d*x]^4 + 9*a^5*b*Csc[c + d*x]^5 - 6*a^6*Csc[c + d*x]^6 - 60*b^2*(a^4 - 4*a^2*b^2 + 3*b^4)*(Log[S
in[c + d*x]] - Log[a + b*Sin[c + d*x]]) - 60*a*b*(a^4 - 4*a^2*b^2 + 3*b^4)*Csc[c + d*x]*(1 + Log[Sin[c + d*x]]
 - Log[a + b*Sin[c + d*x]]))/(30*a^7*d*(b + a*Csc[c + d*x]))

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fricas [B]  time = 1.00, size = 696, normalized size = 3.08 \[ \frac {16 \, a^{6} - 105 \, a^{4} b^{2} + 90 \, a^{2} b^{4} + 30 \, {\left (a^{6} - 4 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{4} - 5 \, {\left (8 \, a^{6} - 45 \, a^{4} b^{2} + 36 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 60 \, {\left ({\left (a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{6} - a^{4} b^{2} + 4 \, a^{2} b^{4} - 3 \, b^{6} - 3 \, {\left (a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{5} b - 4 \, a^{3} b^{3} + 3 \, a b^{5} + {\left (a^{5} b - 4 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} b - 4 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 60 \, {\left ({\left (a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{6} - a^{4} b^{2} + 4 \, a^{2} b^{4} - 3 \, b^{6} - 3 \, {\left (a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{5} b - 4 \, a^{3} b^{3} + 3 \, a b^{5} + {\left (a^{5} b - 4 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} b - 4 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + {\left (91 \, a^{5} b - 270 \, a^{3} b^{3} + 180 \, a b^{5} + 60 \, {\left (a^{5} b - 4 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} - 10 \, {\left (16 \, a^{5} b - 51 \, a^{3} b^{3} + 36 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{7} b d \cos \left (d x + c\right )^{6} - 3 \, a^{7} b d \cos \left (d x + c\right )^{4} + 3 \, a^{7} b d \cos \left (d x + c\right )^{2} - a^{7} b d - {\left (a^{8} d \cos \left (d x + c\right )^{4} - 2 \, a^{8} d \cos \left (d x + c\right )^{2} + a^{8} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/30*(16*a^6 - 105*a^4*b^2 + 90*a^2*b^4 + 30*(a^6 - 4*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)^4 - 5*(8*a^6 - 45*a^4*
b^2 + 36*a^2*b^4)*cos(d*x + c)^2 + 60*((a^4*b^2 - 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^6 - a^4*b^2 + 4*a^2*b^4 - 3*
b^6 - 3*(a^4*b^2 - 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^4 + 3*(a^4*b^2 - 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^2 - (a^5*b
 - 4*a^3*b^3 + 3*a*b^5 + (a^5*b - 4*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^4 - 2*(a^5*b - 4*a^3*b^3 + 3*a*b^5)*cos(d*
x + c)^2)*sin(d*x + c))*log(b*sin(d*x + c) + a) - 60*((a^4*b^2 - 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^6 - a^4*b^2 +
 4*a^2*b^4 - 3*b^6 - 3*(a^4*b^2 - 4*a^2*b^4 + 3*b^6)*cos(d*x + c)^4 + 3*(a^4*b^2 - 4*a^2*b^4 + 3*b^6)*cos(d*x
+ c)^2 - (a^5*b - 4*a^3*b^3 + 3*a*b^5 + (a^5*b - 4*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^4 - 2*(a^5*b - 4*a^3*b^3 +
3*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*sin(d*x + c)) + (91*a^5*b - 270*a^3*b^3 + 180*a*b^5 + 60*(a^5*b
 - 4*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^4 - 10*(16*a^5*b - 51*a^3*b^3 + 36*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))/(
a^7*b*d*cos(d*x + c)^6 - 3*a^7*b*d*cos(d*x + c)^4 + 3*a^7*b*d*cos(d*x + c)^2 - a^7*b*d - (a^8*d*cos(d*x + c)^4
 - 2*a^8*d*cos(d*x + c)^2 + a^8*d)*sin(d*x + c))

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giac [A]  time = 0.24, size = 332, normalized size = 1.47 \[ -\frac {\frac {60 \, {\left (a^{4} b - 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{7}} - \frac {60 \, {\left (a^{4} b^{2} - 4 \, a^{2} b^{4} + 3 \, b^{6}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{7} b} + \frac {30 \, {\left (2 \, a^{4} b^{2} \sin \left (d x + c\right ) - 8 \, a^{2} b^{4} \sin \left (d x + c\right ) + 6 \, b^{6} \sin \left (d x + c\right ) + 3 \, a^{5} b - 10 \, a^{3} b^{3} + 7 \, a b^{5}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} a^{7}} - \frac {137 \, a^{4} b \sin \left (d x + c\right )^{5} - 548 \, a^{2} b^{3} \sin \left (d x + c\right )^{5} + 411 \, b^{5} \sin \left (d x + c\right )^{5} - 30 \, a^{5} \sin \left (d x + c\right )^{4} + 180 \, a^{3} b^{2} \sin \left (d x + c\right )^{4} - 150 \, a b^{4} \sin \left (d x + c\right )^{4} - 60 \, a^{4} b \sin \left (d x + c\right )^{3} + 60 \, a^{2} b^{3} \sin \left (d x + c\right )^{3} + 20 \, a^{5} \sin \left (d x + c\right )^{2} - 30 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} + 15 \, a^{4} b \sin \left (d x + c\right ) - 6 \, a^{5}}{a^{7} \sin \left (d x + c\right )^{5}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/30*(60*(a^4*b - 4*a^2*b^3 + 3*b^5)*log(abs(sin(d*x + c)))/a^7 - 60*(a^4*b^2 - 4*a^2*b^4 + 3*b^6)*log(abs(b*
sin(d*x + c) + a))/(a^7*b) + 30*(2*a^4*b^2*sin(d*x + c) - 8*a^2*b^4*sin(d*x + c) + 6*b^6*sin(d*x + c) + 3*a^5*
b - 10*a^3*b^3 + 7*a*b^5)/((b*sin(d*x + c) + a)*a^7) - (137*a^4*b*sin(d*x + c)^5 - 548*a^2*b^3*sin(d*x + c)^5
+ 411*b^5*sin(d*x + c)^5 - 30*a^5*sin(d*x + c)^4 + 180*a^3*b^2*sin(d*x + c)^4 - 150*a*b^4*sin(d*x + c)^4 - 60*
a^4*b*sin(d*x + c)^3 + 60*a^2*b^3*sin(d*x + c)^3 + 20*a^5*sin(d*x + c)^2 - 30*a^3*b^2*sin(d*x + c)^2 + 15*a^4*
b*sin(d*x + c) - 6*a^5)/(a^7*sin(d*x + c)^5))/d

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maple [A]  time = 0.90, size = 343, normalized size = 1.52 \[ -\frac {b}{d \,a^{2} \left (a +b \sin \left (d x +c \right )\right )}+\frac {2 b^{3}}{d \,a^{4} \left (a +b \sin \left (d x +c \right )\right )}-\frac {b^{5}}{d \,a^{6} \left (a +b \sin \left (d x +c \right )\right )}+\frac {2 b \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,a^{3}}-\frac {8 b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,a^{5}}+\frac {6 b^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,a^{7}}-\frac {1}{5 d \,a^{2} \sin \left (d x +c \right )^{5}}+\frac {2}{3 d \,a^{2} \sin \left (d x +c \right )^{3}}-\frac {b^{2}}{d \,a^{4} \sin \left (d x +c \right )^{3}}-\frac {1}{d \,a^{2} \sin \left (d x +c \right )}+\frac {6 b^{2}}{d \,a^{4} \sin \left (d x +c \right )}-\frac {5 b^{4}}{d \,a^{6} \sin \left (d x +c \right )}+\frac {b}{2 d \,a^{3} \sin \left (d x +c \right )^{4}}-\frac {2 b}{d \,a^{3} \sin \left (d x +c \right )^{2}}+\frac {2 b^{3}}{d \,a^{5} \sin \left (d x +c \right )^{2}}-\frac {2 b \ln \left (\sin \left (d x +c \right )\right )}{a^{3} d}+\frac {8 b^{3} \ln \left (\sin \left (d x +c \right )\right )}{d \,a^{5}}-\frac {6 b^{5} \ln \left (\sin \left (d x +c \right )\right )}{d \,a^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^6/(a+b*sin(d*x+c))^2,x)

[Out]

-1/d*b/a^2/(a+b*sin(d*x+c))+2/d/a^4*b^3/(a+b*sin(d*x+c))-1/d*b^5/a^6/(a+b*sin(d*x+c))+2/d/a^3*b*ln(a+b*sin(d*x
+c))-8/d*b^3/a^5*ln(a+b*sin(d*x+c))+6/d*b^5/a^7*ln(a+b*sin(d*x+c))-1/5/d/a^2/sin(d*x+c)^5+2/3/d/a^2/sin(d*x+c)
^3-1/d/a^4/sin(d*x+c)^3*b^2-1/d/a^2/sin(d*x+c)+6/d/a^4/sin(d*x+c)*b^2-5/d/a^6/sin(d*x+c)*b^4+1/2/d/a^3*b/sin(d
*x+c)^4-2/d/a^3*b/sin(d*x+c)^2+2/d*b^3/a^5/sin(d*x+c)^2-2*b*ln(sin(d*x+c))/a^3/d+8/d*b^3/a^5*ln(sin(d*x+c))-6/
d*b^5/a^7*ln(sin(d*x+c))

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maxima [A]  time = 0.35, size = 225, normalized size = 1.00 \[ \frac {\frac {9 \, a^{4} b \sin \left (d x + c\right ) - 60 \, {\left (a^{4} b - 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \sin \left (d x + c\right )^{5} - 6 \, a^{5} - 30 \, {\left (a^{5} - 4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \sin \left (d x + c\right )^{4} - 10 \, {\left (4 \, a^{4} b - 3 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )^{3} + 5 \, {\left (4 \, a^{5} - 3 \, a^{3} b^{2}\right )} \sin \left (d x + c\right )^{2}}{a^{6} b \sin \left (d x + c\right )^{6} + a^{7} \sin \left (d x + c\right )^{5}} + \frac {60 \, {\left (a^{4} b - 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{7}} - \frac {60 \, {\left (a^{4} b - 4 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{7}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^6/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/30*((9*a^4*b*sin(d*x + c) - 60*(a^4*b - 4*a^2*b^3 + 3*b^5)*sin(d*x + c)^5 - 6*a^5 - 30*(a^5 - 4*a^3*b^2 + 3*
a*b^4)*sin(d*x + c)^4 - 10*(4*a^4*b - 3*a^2*b^3)*sin(d*x + c)^3 + 5*(4*a^5 - 3*a^3*b^2)*sin(d*x + c)^2)/(a^6*b
*sin(d*x + c)^6 + a^7*sin(d*x + c)^5) + 60*(a^4*b - 4*a^2*b^3 + 3*b^5)*log(b*sin(d*x + c) + a)/a^7 - 60*(a^4*b
 - 4*a^2*b^3 + 3*b^5)*log(sin(d*x + c))/a^7)/d

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mupad [B]  time = 11.90, size = 628, normalized size = 2.78 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {64\,a^2+128\,b^2}{3072\,a^4}+\frac {1}{32\,a^2}-\frac {b^2}{6\,a^4}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {b^2}{2\,a^4}-\frac {4\,b\,\left (\frac {b\,\left (64\,a^2+128\,b^2\right )}{256\,a^5}-\frac {b}{8\,a^3}+\frac {4\,b\,\left (\frac {64\,a^2+128\,b^2}{1024\,a^4}+\frac {3}{32\,a^2}-\frac {b^2}{2\,a^4}\right )}{a}\right )}{a}+\frac {\left (64\,a^2+128\,b^2\right )\,\left (\frac {64\,a^2+128\,b^2}{1024\,a^4}+\frac {3}{32\,a^2}-\frac {b^2}{2\,a^4}\right )}{32\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {23\,a^4\,b}{3}-8\,a^2\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {25\,a^5}{3}-56\,a^3\,b^2+48\,a\,b^4\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (32\,a^4\,b-184\,a^2\,b^3+160\,b^5\right )+\frac {a^5}{5}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {22\,a^5}{15}-2\,a^3\,b^2\right )-\frac {3\,a^4\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (5\,a^6-74\,a^4\,b^2+104\,a^2\,b^4-32\,b^6\right )}{a}}{d\,\left (32\,a^7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,a^7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+64\,b\,a^6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {b\,\left (64\,a^2+128\,b^2\right )}{512\,a^5}-\frac {b}{16\,a^3}+\frac {2\,b\,\left (\frac {64\,a^2+128\,b^2}{1024\,a^4}+\frac {3}{32\,a^2}-\frac {b^2}{2\,a^4}\right )}{a}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^4\,b-8\,a^2\,b^3+6\,b^5\right )}{a^7\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32\,a^3\,d}+\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (2\,a^4\,b-8\,a^2\,b^3+6\,b^5\right )}{a^7\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^6*(a + b*sin(c + d*x))^2),x)

[Out]

(tan(c/2 + (d*x)/2)^3*((64*a^2 + 128*b^2)/(3072*a^4) + 1/(32*a^2) - b^2/(6*a^4)))/d - (tan(c/2 + (d*x)/2)*(b^2
/(2*a^4) - (4*b*((b*(64*a^2 + 128*b^2))/(256*a^5) - b/(8*a^3) + (4*b*((64*a^2 + 128*b^2)/(1024*a^4) + 3/(32*a^
2) - b^2/(2*a^4)))/a))/a + ((64*a^2 + 128*b^2)*((64*a^2 + 128*b^2)/(1024*a^4) + 3/(32*a^2) - b^2/(2*a^4)))/(32
*a^2)))/d - tan(c/2 + (d*x)/2)^5/(160*a^2*d) - (tan(c/2 + (d*x)/2)^3*((23*a^4*b)/3 - 8*a^2*b^3) + tan(c/2 + (d
*x)/2)^4*(48*a*b^4 + (25*a^5)/3 - 56*a^3*b^2) + tan(c/2 + (d*x)/2)^5*(32*a^4*b + 160*b^5 - 184*a^2*b^3) + a^5/
5 - tan(c/2 + (d*x)/2)^2*((22*a^5)/15 - 2*a^3*b^2) - (3*a^4*b*tan(c/2 + (d*x)/2))/5 + (2*tan(c/2 + (d*x)/2)^6*
(5*a^6 - 32*b^6 + 104*a^2*b^4 - 74*a^4*b^2))/a)/(d*(32*a^7*tan(c/2 + (d*x)/2)^5 + 32*a^7*tan(c/2 + (d*x)/2)^7
+ 64*a^6*b*tan(c/2 + (d*x)/2)^6)) - (tan(c/2 + (d*x)/2)^2*((b*(64*a^2 + 128*b^2))/(512*a^5) - b/(16*a^3) + (2*
b*((64*a^2 + 128*b^2)/(1024*a^4) + 3/(32*a^2) - b^2/(2*a^4)))/a))/d - (log(tan(c/2 + (d*x)/2))*(2*a^4*b + 6*b^
5 - 8*a^2*b^3))/(a^7*d) + (b*tan(c/2 + (d*x)/2)^4)/(32*a^3*d) + (log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 +
(d*x)/2)^2)*(2*a^4*b + 6*b^5 - 8*a^2*b^3))/(a^7*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)**5*csc(d*x+c)**6/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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