3.1367 \(\int \frac {\csc (c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=233 \[ -\frac {\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac {\left (8 a^2-21 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}+\frac {b^6 \log (a+b \sin (c+d x))}{a d \left (a^2-b^2\right )^3}+\frac {5 a+7 b}{16 d (a+b)^2 (1-\sin (c+d x))}+\frac {5 a-7 b}{16 d (a-b)^2 (\sin (c+d x)+1)}+\frac {1}{16 d (a+b) (1-\sin (c+d x))^2}+\frac {1}{16 d (a-b) (\sin (c+d x)+1)^2}+\frac {\log (\sin (c+d x))}{a d} \]

[Out]

-1/16*(8*a^2+21*a*b+15*b^2)*ln(1-sin(d*x+c))/(a+b)^3/d+ln(sin(d*x+c))/a/d-1/16*(8*a^2-21*a*b+15*b^2)*ln(1+sin(
d*x+c))/(a-b)^3/d+b^6*ln(a+b*sin(d*x+c))/a/(a^2-b^2)^3/d+1/16/(a+b)/d/(1-sin(d*x+c))^2+1/16*(5*a+7*b)/(a+b)^2/
d/(1-sin(d*x+c))+1/16/(a-b)/d/(1+sin(d*x+c))^2+1/16*(5*a-7*b)/(a-b)^2/d/(1+sin(d*x+c))

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Rubi [A]  time = 0.37, antiderivative size = 233, 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} \[ \frac {b^6 \log (a+b \sin (c+d x))}{a d \left (a^2-b^2\right )^3}-\frac {\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac {\left (8 a^2-21 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}+\frac {5 a+7 b}{16 d (a+b)^2 (1-\sin (c+d x))}+\frac {5 a-7 b}{16 d (a-b)^2 (\sin (c+d x)+1)}+\frac {1}{16 d (a+b) (1-\sin (c+d x))^2}+\frac {1}{16 d (a-b) (\sin (c+d x)+1)^2}+\frac {\log (\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*a^2 + 21*a*b + 15*b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) + Log[Sin[c + d*x]]/(a*d) - ((8*a^2 - 21*a
*b + 15*b^2)*Log[1 + Sin[c + d*x]])/(16*(a - b)^3*d) + (b^6*Log[a + b*Sin[c + d*x]])/(a*(a^2 - b^2)^3*d) + 1/(
16*(a + b)*d*(1 - Sin[c + d*x])^2) + (5*a + 7*b)/(16*(a + b)^2*d*(1 - Sin[c + d*x])) + 1/(16*(a - b)*d*(1 + Si
n[c + d*x])^2) + (5*a - 7*b)/(16*(a - b)^2*d*(1 + 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 {\csc (c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {b}{x (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b^6 \operatorname {Subst}\left (\int \frac {1}{x (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b^6 \operatorname {Subst}\left (\int \left (\frac {1}{8 b^4 (a+b) (b-x)^3}+\frac {5 a+7 b}{16 b^5 (a+b)^2 (b-x)^2}+\frac {8 a^2+21 a b+15 b^2}{16 b^6 (a+b)^3 (b-x)}+\frac {1}{a b^6 x}+\frac {1}{a (a-b)^3 (a+b)^3 (a+x)}+\frac {1}{8 b^4 (-a+b) (b+x)^3}+\frac {-5 a+7 b}{16 (a-b)^2 b^5 (b+x)^2}+\frac {8 a^2-21 a b+15 b^2}{16 b^6 (-a+b)^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {\log (\sin (c+d x))}{a d}-\frac {\left (8 a^2-21 a b+15 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac {b^6 \log (a+b \sin (c+d x))}{a \left (a^2-b^2\right )^3 d}+\frac {1}{16 (a+b) d (1-\sin (c+d x))^2}+\frac {5 a+7 b}{16 (a+b)^2 d (1-\sin (c+d x))}+\frac {1}{16 (a-b) d (1+\sin (c+d x))^2}+\frac {5 a-7 b}{16 (a-b)^2 d (1+\sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 2.89, size = 220, normalized size = 0.94 \[ \frac {b^6 \left (-\frac {\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sin (c+d x))}{b^6 (a+b)^3}-\frac {\left (8 a^2-21 a b+15 b^2\right ) \log (\sin (c+d x)+1)}{b^6 (a-b)^3}+\frac {-5 a-7 b}{b^6 (a+b)^2 (\sin (c+d x)-1)}+\frac {5 a-7 b}{b^6 (a-b)^2 (\sin (c+d x)+1)}+\frac {1}{b^6 (a+b) (\sin (c+d x)-1)^2}+\frac {1}{b^6 (a-b) (\sin (c+d x)+1)^2}+\frac {16 \log (\sin (c+d x))}{a b^6}+\frac {16 \log (a+b \sin (c+d x))}{a (a-b)^3 (a+b)^3}\right )}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^6*(-(((8*a^2 + 21*a*b + 15*b^2)*Log[1 - Sin[c + d*x]])/(b^6*(a + b)^3)) + (16*Log[Sin[c + d*x]])/(a*b^6) -
((8*a^2 - 21*a*b + 15*b^2)*Log[1 + Sin[c + d*x]])/((a - b)^3*b^6) + (16*Log[a + b*Sin[c + d*x]])/(a*(a - b)^3*
(a + b)^3) + 1/(b^6*(a + b)*(-1 + Sin[c + d*x])^2) + (-5*a - 7*b)/(b^6*(a + b)^2*(-1 + Sin[c + d*x])) + 1/((a
- b)*b^6*(1 + Sin[c + d*x])^2) + (5*a - 7*b)/((a - b)^2*b^6*(1 + Sin[c + d*x]))))/(16*d)

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fricas [A]  time = 4.11, size = 344, normalized size = 1.48 \[ \frac {16 \, b^{6} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, a^{6} - 8 \, a^{4} b^{2} + 4 \, a^{2} b^{4} + 16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{4} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (8 \, a^{6} + 3 \, a^{5} b - 24 \, a^{4} b^{2} - 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} + 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (8 \, a^{6} - 3 \, a^{5} b - 24 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} - 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 2 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{5} b - 4 \, a^{3} b^{3} + 2 \, a b^{5} + {\left (3 \, a^{5} b - 10 \, a^{3} b^{3} + 7 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*b^6*cos(d*x + c)^4*log(b*sin(d*x + c) + a) + 4*a^6 - 8*a^4*b^2 + 4*a^2*b^4 + 16*(a^6 - 3*a^4*b^2 + 3*
a^2*b^4 - b^6)*cos(d*x + c)^4*log(-1/2*sin(d*x + c)) - (8*a^6 + 3*a^5*b - 24*a^4*b^2 - 10*a^3*b^3 + 24*a^2*b^4
+ 15*a*b^5)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (8*a^6 - 3*a^5*b - 24*a^4*b^2 + 10*a^3*b^3 + 24*a^2*b^4 -
15*a*b^5)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 8*(a^6 - 3*a^4*b^2 + 2*a^2*b^4)*cos(d*x + c)^2 - 2*(2*a^5*b
- 4*a^3*b^3 + 2*a*b^5 + (3*a^5*b - 10*a^3*b^3 + 7*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^7 - 3*a^5*b^2 + 3*a
^3*b^4 - a*b^6)*d*cos(d*x + c)^4)

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giac [A]  time = 0.23, size = 391, normalized size = 1.68 \[ \frac {\frac {16 \, b^{7} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}} - \frac {{\left (8 \, a^{2} - 21 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (8 \, a^{2} + 21 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {16 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} + \frac {2 \, {\left (6 \, a^{5} \sin \left (d x + c\right )^{4} - 18 \, a^{3} b^{2} \sin \left (d x + c\right )^{4} + 18 \, a b^{4} \sin \left (d x + c\right )^{4} + 3 \, a^{4} b \sin \left (d x + c\right )^{3} - 10 \, a^{2} b^{3} \sin \left (d x + c\right )^{3} + 7 \, b^{5} \sin \left (d x + c\right )^{3} - 16 \, a^{5} \sin \left (d x + c\right )^{2} + 48 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} - 44 \, a b^{4} \sin \left (d x + c\right )^{2} - 5 \, a^{4} b \sin \left (d x + c\right ) + 14 \, a^{2} b^{3} \sin \left (d x + c\right ) - 9 \, b^{5} \sin \left (d x + c\right ) + 12 \, a^{5} - 34 \, a^{3} b^{2} + 28 \, a b^{4}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*b^7*log(abs(b*sin(d*x + c) + a))/(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7) - (8*a^2 - 21*a*b + 15*b^2)*
log(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (8*a^2 + 21*a*b + 15*b^2)*log(abs(sin(d*x + c) -
1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 16*log(abs(sin(d*x + c)))/a + 2*(6*a^5*sin(d*x + c)^4 - 18*a^3*b^2*sin(d
*x + c)^4 + 18*a*b^4*sin(d*x + c)^4 + 3*a^4*b*sin(d*x + c)^3 - 10*a^2*b^3*sin(d*x + c)^3 + 7*b^5*sin(d*x + c)^
3 - 16*a^5*sin(d*x + c)^2 + 48*a^3*b^2*sin(d*x + c)^2 - 44*a*b^4*sin(d*x + c)^2 - 5*a^4*b*sin(d*x + c) + 14*a^
2*b^3*sin(d*x + c) - 9*b^5*sin(d*x + c) + 12*a^5 - 34*a^3*b^2 + 28*a*b^4)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)
*(sin(d*x + c)^2 - 1)^2))/d

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maple [A]  time = 0.45, size = 321, normalized size = 1.38 \[ \frac {1}{2 d \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {5 a}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {7 b}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) a^{2}}{2 d \left (a +b \right )^{3}}-\frac {21 \ln \left (\sin \left (d x +c \right )-1\right ) a b}{16 d \left (a +b \right )^{3}}-\frac {15 \ln \left (\sin \left (d x +c \right )-1\right ) b^{2}}{16 d \left (a +b \right )^{3}}+\frac {b^{6} \ln \left (a +b \sin \left (d x +c \right )\right )}{d a \left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a d}+\frac {1}{2 d \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5 a}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {7 b}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right ) a^{2}}{2 d \left (a -b \right )^{3}}+\frac {21 \ln \left (1+\sin \left (d x +c \right )\right ) a b}{16 d \left (a -b \right )^{3}}-\frac {15 \ln \left (1+\sin \left (d x +c \right )\right ) b^{2}}{16 d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/(8*a+8*b)/(sin(d*x+c)-1)^2-5/16/d/(a+b)^2/(sin(d*x+c)-1)*a-7/16/d/(a+b)^2/(sin(d*x+c)-1)*b-1/2/d/(a+b)^3
*ln(sin(d*x+c)-1)*a^2-21/16/d/(a+b)^3*ln(sin(d*x+c)-1)*a*b-15/16/d/(a+b)^3*ln(sin(d*x+c)-1)*b^2+1/d/a*b^6/(a+b
)^3/(a-b)^3*ln(a+b*sin(d*x+c))+ln(sin(d*x+c))/a/d+1/2/d/(8*a-8*b)/(1+sin(d*x+c))^2+5/16/d/(a-b)^2/(1+sin(d*x+c
))*a-7/16/d/(a-b)^2/(1+sin(d*x+c))*b-1/2/d/(a-b)^3*ln(1+sin(d*x+c))*a^2+21/16/d/(a-b)^3*ln(1+sin(d*x+c))*a*b-1
5/16/d/(a-b)^3*ln(1+sin(d*x+c))*b^2

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maxima [A]  time = 0.34, size = 299, normalized size = 1.28 \[ \frac {\frac {16 \, b^{6} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}} - \frac {{\left (8 \, a^{2} - 21 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (8 \, a^{2} + 21 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left ({\left (3 \, a^{2} b - 7 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + 6 \, a^{3} - 10 \, a b^{2} - 4 \, {\left (a^{3} - 2 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} - {\left (5 \, a^{2} b - 9 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}} + \frac {16 \, \log \left (\sin \left (d x + c\right )\right )}{a}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*b^6*log(b*sin(d*x + c) + a)/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6) - (8*a^2 - 21*a*b + 15*b^2)*log(sin
(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (8*a^2 + 21*a*b + 15*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*
b + 3*a*b^2 + b^3) + 2*((3*a^2*b - 7*b^3)*sin(d*x + c)^3 + 6*a^3 - 10*a*b^2 - 4*(a^3 - 2*a*b^2)*sin(d*x + c)^2
- (5*a^2*b - 9*b^3)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4 -
2*a^2*b^2 + b^4)*sin(d*x + c)^2) + 16*log(sin(d*x + c))/a)/d

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mupad [B]  time = 12.43, size = 346, normalized size = 1.48 \[ \frac {\ln \left (\sin \left (c+d\,x\right )\right )}{a\,d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {5\,b}{16\,{\left (a+b\right )}^2}+\frac {1}{2\,\left (a+b\right )}+\frac {b^2}{8\,{\left (a+b\right )}^3}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {b^2}{8\,{\left (a-b\right )}^3}-\frac {5\,b}{16\,{\left (a-b\right )}^2}+\frac {1}{2\,\left (a-b\right )}\right )}{d}-\frac {\frac {5\,a\,b^2-3\,a^3}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (2\,a\,b^2-a^3\right )}{2\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (3\,a^2\,b-7\,b^3\right )}{8\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {b\,\sin \left (c+d\,x\right )\,\left (5\,a^2-9\,b^2\right )}{8\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\cos \left (c+d\,x\right )}^2+{\sin \left (c+d\,x\right )}^4-{\sin \left (c+d\,x\right )}^2\right )}-\frac {b^6\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (-a^7+3\,a^5\,b^2-3\,a^3\,b^4+a\,b^6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*sin(c + d*x)*(a + b*sin(c + d*x))),x)

[Out]

log(sin(c + d*x))/(a*d) - (log(sin(c + d*x) - 1)*((5*b)/(16*(a + b)^2) + 1/(2*(a + b)) + b^2/(8*(a + b)^3)))/d
- (log(sin(c + d*x) + 1)*(b^2/(8*(a - b)^3) - (5*b)/(16*(a - b)^2) + 1/(2*(a - b))))/d - ((5*a*b^2 - 3*a^3)/(
4*(a^4 + b^4 - 2*a^2*b^2)) - (sin(c + d*x)^2*(2*a*b^2 - a^3))/(2*(a^4 + b^4 - 2*a^2*b^2)) - (sin(c + d*x)^3*(3
*a^2*b - 7*b^3))/(8*(a^4 + b^4 - 2*a^2*b^2)) + (b*sin(c + d*x)*(5*a^2 - 9*b^2))/(8*(a^4 + b^4 - 2*a^2*b^2)))/(
d*(cos(c + d*x)^2 - sin(c + d*x)^2 + sin(c + d*x)^4)) - (b^6*log(a + b*sin(c + d*x)))/(d*(a*b^6 - a^7 - 3*a^3*
b^4 + 3*a^5*b^2))

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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(csc(d*x+c)*sec(d*x+c)**5/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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