3.1319 \(\int \frac {\cot ^5(c+d x) \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=212 \[ \frac {b \csc ^5(c+d x)}{5 a^2 d}+\frac {b^2 \left (a^2-b^2\right )^2 \log (\sin (c+d x))}{a^7 d}-\frac {b^2 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^7 d}+\frac {b \left (a^2-b^2\right )^2 \csc (c+d x)}{a^6 d}-\frac {\left (a^2-b^2\right )^2 \csc ^2(c+d x)}{2 a^5 d}-\frac {b \left (2 a^2-b^2\right ) \csc ^3(c+d x)}{3 a^4 d}+\frac {\left (2 a^2-b^2\right ) \csc ^4(c+d x)}{4 a^3 d}-\frac {\csc ^6(c+d x)}{6 a d} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2837, 12, 894} \[ \frac {\left (2 a^2-b^2\right ) \csc ^4(c+d x)}{4 a^3 d}-\frac {b \left (2 a^2-b^2\right ) \csc ^3(c+d x)}{3 a^4 d}-\frac {\left (a^2-b^2\right )^2 \csc ^2(c+d x)}{2 a^5 d}+\frac {b \left (a^2-b^2\right )^2 \csc (c+d x)}{a^6 d}+\frac {b^2 \left (a^2-b^2\right )^2 \log (\sin (c+d x))}{a^7 d}-\frac {b^2 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^7 d}+\frac {b \csc ^5(c+d x)}{5 a^2 d}-\frac {\csc ^6(c+d x)}{6 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 2.86, size = 165, normalized size = 0.78 \[ \frac {-10 a^6 \csc ^6(c+d x)+12 a^5 b \csc ^5(c+d x)+60 \left (b^3-a^2 b\right )^2 (\log (\sin (c+d x))-\log (a+b \sin (c+d x)))-30 a^2 \left (a^2-b^2\right )^2 \csc ^2(c+d x)+60 a b \left (a^2-b^2\right )^2 \csc (c+d x)+15 a^4 \left (2 a^2-b^2\right ) \csc ^4(c+d x)+20 a^3 b \left (b^2-2 a^2\right ) \csc ^3(c+d x)}{60 a^7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*a*b*(a^2 - b^2)^2*Csc[c + d*x] - 30*a^2*(a^2 - b^2)^2*Csc[c + d*x]^2 + 20*a^3*b*(-2*a^2 + b^2)*Csc[c + d*x
]^3 + 15*a^4*(2*a^2 - b^2)*Csc[c + d*x]^4 + 12*a^5*b*Csc[c + d*x]^5 - 10*a^6*Csc[c + d*x]^6 + 60*(-(a^2*b) + b
^3)^2*(Log[Sin[c + d*x]] - Log[a + b*Sin[c + d*x]]))/(60*a^7*d)

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fricas [B]  time = 0.74, size = 464, normalized size = 2.19 \[ \frac {10 \, a^{6} - 45 \, a^{4} b^{2} + 30 \, a^{2} b^{4} + 30 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{4} - 15 \, {\left (2 \, a^{6} - 7 \, a^{4} b^{2} + 4 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} - 60 \, {\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{6} - a^{4} b^{2} + 2 \, a^{2} b^{4} - b^{6} - 3 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 60 \, {\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{6} - a^{4} b^{2} + 2 \, a^{2} b^{4} - b^{6} - 3 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 4 \, {\left (8 \, a^{5} b - 25 \, a^{3} b^{3} + 15 \, a b^{5} + 15 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )^{4} - 5 \, {\left (4 \, a^{5} b - 11 \, a^{3} b^{3} + 6 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{7} d \cos \left (d x + c\right )^{6} - 3 \, a^{7} d \cos \left (d x + c\right )^{4} + 3 \, a^{7} d \cos \left (d x + c\right )^{2} - a^{7} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/60*(10*a^6 - 45*a^4*b^2 + 30*a^2*b^4 + 30*(a^6 - 2*a^4*b^2 + a^2*b^4)*cos(d*x + c)^4 - 15*(2*a^6 - 7*a^4*b^2
+ 4*a^2*b^4)*cos(d*x + c)^2 - 60*((a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^6 - a^4*b^2 + 2*a^2*b^4 - b^6 - 3*
(a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^4 + 3*(a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2)*log(b*sin(d*x + c)
+ a) + 60*((a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^6 - a^4*b^2 + 2*a^2*b^4 - b^6 - 3*(a^4*b^2 - 2*a^2*b^4 + b
^6)*cos(d*x + c)^4 + 3*(a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2)*log(-1/2*sin(d*x + c)) - 4*(8*a^5*b - 25*a^
3*b^3 + 15*a*b^5 + 15*(a^5*b - 2*a^3*b^3 + a*b^5)*cos(d*x + c)^4 - 5*(4*a^5*b - 11*a^3*b^3 + 6*a*b^5)*cos(d*x
+ c)^2)*sin(d*x + c))/(a^7*d*cos(d*x + c)^6 - 3*a^7*d*cos(d*x + c)^4 + 3*a^7*d*cos(d*x + c)^2 - a^7*d)

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giac [A]  time = 0.23, size = 301, normalized size = 1.42 \[ \frac {\frac {60 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{7}} - \frac {60 \, {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{7} b} - \frac {147 \, a^{4} b^{2} \sin \left (d x + c\right )^{6} - 294 \, a^{2} b^{4} \sin \left (d x + c\right )^{6} + 147 \, b^{6} \sin \left (d x + c\right )^{6} - 60 \, a^{5} b \sin \left (d x + c\right )^{5} + 120 \, a^{3} b^{3} \sin \left (d x + c\right )^{5} - 60 \, a b^{5} \sin \left (d x + c\right )^{5} + 30 \, a^{6} \sin \left (d x + c\right )^{4} - 60 \, a^{4} b^{2} \sin \left (d x + c\right )^{4} + 30 \, a^{2} b^{4} \sin \left (d x + c\right )^{4} + 40 \, a^{5} b \sin \left (d x + c\right )^{3} - 20 \, a^{3} b^{3} \sin \left (d x + c\right )^{3} - 30 \, a^{6} \sin \left (d x + c\right )^{2} + 15 \, a^{4} b^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{5} b \sin \left (d x + c\right ) + 10 \, a^{6}}{a^{7} \sin \left (d x + c\right )^{6}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/60*(60*(a^4*b^2 - 2*a^2*b^4 + b^6)*log(abs(sin(d*x + c)))/a^7 - 60*(a^4*b^3 - 2*a^2*b^5 + b^7)*log(abs(b*sin
(d*x + c) + a))/(a^7*b) - (147*a^4*b^2*sin(d*x + c)^6 - 294*a^2*b^4*sin(d*x + c)^6 + 147*b^6*sin(d*x + c)^6 -
60*a^5*b*sin(d*x + c)^5 + 120*a^3*b^3*sin(d*x + c)^5 - 60*a*b^5*sin(d*x + c)^5 + 30*a^6*sin(d*x + c)^4 - 60*a^
4*b^2*sin(d*x + c)^4 + 30*a^2*b^4*sin(d*x + c)^4 + 40*a^5*b*sin(d*x + c)^3 - 20*a^3*b^3*sin(d*x + c)^3 - 30*a^
6*sin(d*x + c)^2 + 15*a^4*b^2*sin(d*x + c)^2 - 12*a^5*b*sin(d*x + c) + 10*a^6)/(a^7*sin(d*x + c)^6))/d

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maple [A]  time = 0.55, size = 330, normalized size = 1.56 \[ -\frac {b^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{3} d}+\frac {2 \ln \left (a +b \sin \left (d x +c \right )\right ) b^{4}}{d \,a^{5}}-\frac {b^{6} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,a^{7}}-\frac {1}{6 d a \sin \left (d x +c \right )^{6}}+\frac {1}{2 d a \sin \left (d x +c \right )^{4}}-\frac {b^{2}}{4 d \,a^{3} \sin \left (d x +c \right )^{4}}-\frac {1}{2 d a \sin \left (d x +c \right )^{2}}+\frac {b^{2}}{d \,a^{3} \sin \left (d x +c \right )^{2}}-\frac {b^{4}}{2 d \,a^{5} \sin \left (d x +c \right )^{2}}-\frac {2 b}{3 d \,a^{2} \sin \left (d x +c \right )^{3}}+\frac {b^{3}}{3 d \,a^{4} \sin \left (d x +c \right )^{3}}+\frac {b^{2} \ln \left (\sin \left (d x +c \right )\right )}{a^{3} d}-\frac {2 \ln \left (\sin \left (d x +c \right )\right ) b^{4}}{d \,a^{5}}+\frac {b^{6} \ln \left (\sin \left (d x +c \right )\right )}{d \,a^{7}}+\frac {b}{5 d \,a^{2} \sin \left (d x +c \right )^{5}}+\frac {b}{d \,a^{2} \sin \left (d x +c \right )}-\frac {2 b^{3}}{d \,a^{4} \sin \left (d x +c \right )}+\frac {b^{5}}{d \,a^{6} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^7/(a+b*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.32, size = 206, normalized size = 0.97 \[ -\frac {\frac {60 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{7}} - \frac {60 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{7}} - \frac {12 \, a^{4} b \sin \left (d x + c\right ) + 60 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{5} - 10 \, a^{5} - 30 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{4} - 20 \, {\left (2 \, a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )^{3} + 15 \, {\left (2 \, a^{5} - a^{3} b^{2}\right )} \sin \left (d x + c\right )^{2}}{a^{6} \sin \left (d x + c\right )^{6}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^7/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(60*(a^4*b^2 - 2*a^2*b^4 + b^6)*log(b*sin(d*x + c) + a)/a^7 - 60*(a^4*b^2 - 2*a^2*b^4 + b^6)*log(sin(d*x
+ c))/a^7 - (12*a^4*b*sin(d*x + c) + 60*(a^4*b - 2*a^2*b^3 + b^5)*sin(d*x + c)^5 - 10*a^5 - 30*(a^5 - 2*a^3*b
^2 + a*b^4)*sin(d*x + c)^4 - 20*(2*a^4*b - a^2*b^3)*sin(d*x + c)^3 + 15*(2*a^5 - a^3*b^2)*sin(d*x + c)^2)/(a^6
*sin(d*x + c)^6))/d

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mupad [B]  time = 12.31, size = 514, normalized size = 2.42 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {1}{64\,a}-\frac {b^2}{64\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {b}{96\,a^2}+\frac {2\,b\,\left (\frac {1}{16\,a}-\frac {b^2}{16\,a^3}\right )}{3\,a}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{384\,a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {b}{32\,a^2}-\frac {2\,b\,\left (\frac {b^2}{16\,a^3}-\frac {5}{64\,a}+\frac {2\,b\,\left (\frac {b}{32\,a^2}+\frac {2\,b\,\left (\frac {1}{16\,a}-\frac {b^2}{16\,a^3}\right )}{a}\right )}{a}\right )}{a}+\frac {2\,b\,\left (\frac {1}{16\,a}-\frac {b^2}{16\,a^3}\right )}{a}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {b^2}{32\,a^3}-\frac {5}{128\,a}+\frac {b\,\left (\frac {b}{32\,a^2}+\frac {2\,b\,\left (\frac {1}{16\,a}-\frac {b^2}{16\,a^3}\right )}{a}\right )}{a}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^4\,b^2-2\,a^2\,b^4+b^6\right )}{a^7\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a^2\,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 (a^4\,b^2-2\,a^2\,b^4+b^6\right )}{a^7\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {10\,a^4\,b}{3}-\frac {8\,a^2\,b^3}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {5\,a^5}{2}-12\,a^3\,b^2+8\,a\,b^4\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (20\,a^4\,b-56\,a^2\,b^3+32\,b^5\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^5-a^3\,b^2\right )+\frac {a^5}{6}-\frac {2\,a^4\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}}{64\,a^6\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^7*(a + b*sin(c + d*x))),x)

[Out]

(tan(c/2 + (d*x)/2)^4*(1/(64*a) - b^2/(64*a^3)))/d - (tan(c/2 + (d*x)/2)^3*(b/(96*a^2) + (2*b*(1/(16*a) - b^2/
(16*a^3)))/(3*a)))/d - tan(c/2 + (d*x)/2)^6/(384*a*d) + (tan(c/2 + (d*x)/2)*(b/(32*a^2) - (2*b*(b^2/(16*a^3) -
5/(64*a) + (2*b*(b/(32*a^2) + (2*b*(1/(16*a) - b^2/(16*a^3)))/a))/a))/a + (2*b*(1/(16*a) - b^2/(16*a^3)))/a))
/d + (tan(c/2 + (d*x)/2)^2*(b^2/(32*a^3) - 5/(128*a) + (b*(b/(32*a^2) + (2*b*(1/(16*a) - b^2/(16*a^3)))/a))/a)
)/d + (log(tan(c/2 + (d*x)/2))*(b^6 - 2*a^2*b^4 + a^4*b^2))/(a^7*d) + (b*tan(c/2 + (d*x)/2)^5)/(160*a^2*d) - (
log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2)*(b^6 - 2*a^2*b^4 + a^4*b^2))/(a^7*d) - (tan(c/2 + (d*
x)/2)^3*((10*a^4*b)/3 - (8*a^2*b^3)/3) + tan(c/2 + (d*x)/2)^4*(8*a*b^4 + (5*a^5)/2 - 12*a^3*b^2) - tan(c/2 + (
d*x)/2)^5*(20*a^4*b + 32*b^5 - 56*a^2*b^3) - tan(c/2 + (d*x)/2)^2*(a^5 - a^3*b^2) + a^5/6 - (2*a^4*b*tan(c/2 +
(d*x)/2))/5)/(64*a^6*d*tan(c/2 + (d*x)/2)^6)

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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)**7/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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