3.1369 \(\int \frac {\csc ^3(c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=274 \[ -\frac {\left (24 a^2+57 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}-\frac {\left (24 a^2-57 a b+35 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}+\frac {b \csc (c+d x)}{a^2 d}+\frac {\left (3 a^2+b^2\right ) \log (\sin (c+d x))}{a^3 d}+\frac {b^8 \log (a+b \sin (c+d x))}{a^3 d \left (a^2-b^2\right )^3}+\frac {9 a+11 b}{16 d (a+b)^2 (1-\sin (c+d x))}+\frac {9 a-11 b}{16 d (a-b)^2 (\sin (c+d x)+1)}+\frac {1}{16 d (a+b) (1-\sin (c+d x))^2}+\frac {1}{16 d (a-b) (\sin (c+d x)+1)^2}-\frac {\csc ^2(c+d x)}{2 a d} \]

[Out]

b*csc(d*x+c)/a^2/d-1/2*csc(d*x+c)^2/a/d-1/16*(24*a^2+57*a*b+35*b^2)*ln(1-sin(d*x+c))/(a+b)^3/d+(3*a^2+b^2)*ln(
sin(d*x+c))/a^3/d-1/16*(24*a^2-57*a*b+35*b^2)*ln(1+sin(d*x+c))/(a-b)^3/d+b^8*ln(a+b*sin(d*x+c))/a^3/(a^2-b^2)^
3/d+1/16/(a+b)/d/(1-sin(d*x+c))^2+1/16*(9*a+11*b)/(a+b)^2/d/(1-sin(d*x+c))+1/16/(a-b)/d/(1+sin(d*x+c))^2+1/16*
(9*a-11*b)/(a-b)^2/d/(1+sin(d*x+c))

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Rubi [A]  time = 0.47, antiderivative size = 274, 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} \[ \frac {b^8 \log (a+b \sin (c+d x))}{a^3 d \left (a^2-b^2\right )^3}-\frac {\left (24 a^2+57 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac {\left (3 a^2+b^2\right ) \log (\sin (c+d x))}{a^3 d}-\frac {\left (24 a^2-57 a b+35 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}+\frac {b \csc (c+d x)}{a^2 d}+\frac {9 a+11 b}{16 d (a+b)^2 (1-\sin (c+d x))}+\frac {9 a-11 b}{16 d (a-b)^2 (\sin (c+d x)+1)}+\frac {1}{16 d (a+b) (1-\sin (c+d x))^2}+\frac {1}{16 d (a-b) (\sin (c+d x)+1)^2}-\frac {\csc ^2(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^3*Sec[c + d*x]^5)/(a + b*Sin[c + d*x]),x]

[Out]

(b*Csc[c + d*x])/(a^2*d) - Csc[c + d*x]^2/(2*a*d) - ((24*a^2 + 57*a*b + 35*b^2)*Log[1 - Sin[c + d*x]])/(16*(a
+ b)^3*d) + ((3*a^2 + b^2)*Log[Sin[c + d*x]])/(a^3*d) - ((24*a^2 - 57*a*b + 35*b^2)*Log[1 + Sin[c + d*x]])/(16
*(a - b)^3*d) + (b^8*Log[a + b*Sin[c + d*x]])/(a^3*(a^2 - b^2)^3*d) + 1/(16*(a + b)*d*(1 - Sin[c + d*x])^2) +
(9*a + 11*b)/(16*(a + b)^2*d*(1 - Sin[c + d*x])) + 1/(16*(a - b)*d*(1 + Sin[c + d*x])^2) + (9*a - 11*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 ^3(c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {b^3}{x^3 (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b^8 \operatorname {Subst}\left (\int \frac {1}{x^3 (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b^8 \operatorname {Subst}\left (\int \left (\frac {1}{8 b^6 (a+b) (b-x)^3}+\frac {9 a+11 b}{16 b^7 (a+b)^2 (b-x)^2}+\frac {24 a^2+57 a b+35 b^2}{16 b^8 (a+b)^3 (b-x)}+\frac {1}{a b^6 x^3}-\frac {1}{a^2 b^6 x^2}+\frac {3 a^2+b^2}{a^3 b^8 x}+\frac {1}{a^3 (a-b)^3 (a+b)^3 (a+x)}+\frac {1}{8 b^6 (-a+b) (b+x)^3}+\frac {-9 a+11 b}{16 (a-b)^2 b^7 (b+x)^2}+\frac {24 a^2-57 a b+35 b^2}{16 b^8 (-a+b)^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b \csc (c+d x)}{a^2 d}-\frac {\csc ^2(c+d x)}{2 a d}-\frac {\left (24 a^2+57 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {\left (3 a^2+b^2\right ) \log (\sin (c+d x))}{a^3 d}-\frac {\left (24 a^2-57 a b+35 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac {b^8 \log (a+b \sin (c+d x))}{a^3 \left (a^2-b^2\right )^3 d}+\frac {1}{16 (a+b) d (1-\sin (c+d x))^2}+\frac {9 a+11 b}{16 (a+b)^2 d (1-\sin (c+d x))}+\frac {1}{16 (a-b) d (1+\sin (c+d x))^2}+\frac {9 a-11 b}{16 (a-b)^2 d (1+\sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 6.26, size = 281, normalized size = 1.03 \[ \frac {b^8 \left (\frac {\log (a+b \sin (c+d x))}{a^3 (a-b)^3 (a+b)^3}+\frac {\csc (c+d x)}{a^2 b^7}-\frac {\left (24 a^2+57 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 b^8 (a+b)^3}-\frac {\left (24 a^2-57 a b+35 b^2\right ) \log (\sin (c+d x)+1)}{16 b^8 (a-b)^3}+\frac {\left (3 a^2+b^2\right ) \log (\sin (c+d x))}{a^3 b^8}-\frac {\csc ^2(c+d x)}{2 a b^8}+\frac {9 a+11 b}{16 b^7 (a+b)^2 (b-b \sin (c+d x))}+\frac {9 a-11 b}{16 b^7 (a-b)^2 (b \sin (c+d x)+b)}+\frac {1}{16 b^6 (a+b) (b-b \sin (c+d x))^2}+\frac {1}{16 b^6 (a-b) (b \sin (c+d x)+b)^2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^3*Sec[c + d*x]^5)/(a + b*Sin[c + d*x]),x]

[Out]

(b^8*(Csc[c + d*x]/(a^2*b^7) - Csc[c + d*x]^2/(2*a*b^8) - ((24*a^2 + 57*a*b + 35*b^2)*Log[1 - Sin[c + d*x]])/(
16*b^8*(a + b)^3) + ((3*a^2 + b^2)*Log[Sin[c + d*x]])/(a^3*b^8) - ((24*a^2 - 57*a*b + 35*b^2)*Log[1 + Sin[c +
d*x]])/(16*(a - b)^3*b^8) + Log[a + b*Sin[c + d*x]]/(a^3*(a - b)^3*(a + b)^3) + 1/(16*b^6*(a + b)*(b - b*Sin[c
+ d*x])^2) + (9*a + 11*b)/(16*b^7*(a + b)^2*(b - b*Sin[c + d*x])) + 1/(16*(a - b)*b^6*(b + b*Sin[c + d*x])^2)
+ (9*a - 11*b)/(16*(a - b)^2*b^7*(b + b*Sin[c + d*x]))))/d

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fricas [B]  time = 7.88, size = 640, normalized size = 2.34 \[ -\frac {4 \, a^{8} - 8 \, a^{6} b^{2} + 4 \, a^{4} b^{4} - 8 \, {\left (3 \, a^{8} - 8 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - a^{2} b^{6}\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, a^{8} - 8 \, a^{6} b^{2} + 5 \, a^{4} b^{4}\right )} \cos \left (d x + c\right )^{2} - 16 \, {\left (b^{8} \cos \left (d x + c\right )^{6} - b^{8} \cos \left (d x + c\right )^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 16 \, {\left ({\left (3 \, a^{8} - 8 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - b^{8}\right )} \cos \left (d x + c\right )^{6} - {\left (3 \, a^{8} - 8 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - b^{8}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) + {\left ({\left (24 \, a^{8} + 15 \, a^{7} b - 64 \, a^{6} b^{2} - 42 \, a^{5} b^{3} + 48 \, a^{4} b^{4} + 35 \, a^{3} b^{5}\right )} \cos \left (d x + c\right )^{6} - {\left (24 \, a^{8} + 15 \, a^{7} b - 64 \, a^{6} b^{2} - 42 \, a^{5} b^{3} + 48 \, a^{4} b^{4} + 35 \, a^{3} b^{5}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (24 \, a^{8} - 15 \, a^{7} b - 64 \, a^{6} b^{2} + 42 \, a^{5} b^{3} + 48 \, a^{4} b^{4} - 35 \, a^{3} b^{5}\right )} \cos \left (d x + c\right )^{6} - {\left (24 \, a^{8} - 15 \, a^{7} b - 64 \, a^{6} b^{2} + 42 \, a^{5} b^{3} + 48 \, a^{4} b^{4} - 35 \, a^{3} b^{5}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, a^{7} b - 4 \, a^{5} b^{3} + 2 \, a^{3} b^{5} - {\left (15 \, a^{7} b - 42 \, a^{5} b^{3} + 35 \, a^{3} b^{5} - 8 \, a b^{7}\right )} \cos \left (d x + c\right )^{4} + {\left (5 \, a^{7} b - 14 \, a^{5} b^{3} + 9 \, a^{3} b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left ({\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} d \cos \left (d x + c\right )^{6} - {\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/16*(4*a^8 - 8*a^6*b^2 + 4*a^4*b^4 - 8*(3*a^8 - 8*a^6*b^2 + 6*a^4*b^4 - a^2*b^6)*cos(d*x + c)^4 + 4*(3*a^8 -
8*a^6*b^2 + 5*a^4*b^4)*cos(d*x + c)^2 - 16*(b^8*cos(d*x + c)^6 - b^8*cos(d*x + c)^4)*log(b*sin(d*x + c) + a)
- 16*((3*a^8 - 8*a^6*b^2 + 6*a^4*b^4 - b^8)*cos(d*x + c)^6 - (3*a^8 - 8*a^6*b^2 + 6*a^4*b^4 - b^8)*cos(d*x + c
)^4)*log(-1/2*sin(d*x + c)) + ((24*a^8 + 15*a^7*b - 64*a^6*b^2 - 42*a^5*b^3 + 48*a^4*b^4 + 35*a^3*b^5)*cos(d*x
+ c)^6 - (24*a^8 + 15*a^7*b - 64*a^6*b^2 - 42*a^5*b^3 + 48*a^4*b^4 + 35*a^3*b^5)*cos(d*x + c)^4)*log(sin(d*x
+ c) + 1) + ((24*a^8 - 15*a^7*b - 64*a^6*b^2 + 42*a^5*b^3 + 48*a^4*b^4 - 35*a^3*b^5)*cos(d*x + c)^6 - (24*a^8
- 15*a^7*b - 64*a^6*b^2 + 42*a^5*b^3 + 48*a^4*b^4 - 35*a^3*b^5)*cos(d*x + c)^4)*log(-sin(d*x + c) + 1) - 2*(2*
a^7*b - 4*a^5*b^3 + 2*a^3*b^5 - (15*a^7*b - 42*a^5*b^3 + 35*a^3*b^5 - 8*a*b^7)*cos(d*x + c)^4 + (5*a^7*b - 14*
a^5*b^3 + 9*a^3*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*d*cos(d*x + c)^6 -
(a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*d*cos(d*x + c)^4)

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giac [B]  time = 0.26, size = 589, normalized size = 2.15 \[ \frac {\frac {16 \, b^{9} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{9} b - 3 \, a^{7} b^{3} + 3 \, a^{5} b^{5} - a^{3} b^{7}} - \frac {{\left (24 \, a^{2} - 57 \, a b + 35 \, 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 (24 \, a^{2} + 57 \, a b + 35 \, 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 \, {\left (3 \, a^{2} + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} + \frac {2 \, {\left (4 \, b^{8} \sin \left (d x + c\right )^{6} + 15 \, a^{7} b \sin \left (d x + c\right )^{5} - 42 \, a^{5} b^{3} \sin \left (d x + c\right )^{5} + 35 \, a^{3} b^{5} \sin \left (d x + c\right )^{5} - 8 \, a b^{7} \sin \left (d x + c\right )^{5} - 12 \, a^{8} \sin \left (d x + c\right )^{4} + 32 \, a^{6} b^{2} \sin \left (d x + c\right )^{4} - 24 \, a^{4} b^{4} \sin \left (d x + c\right )^{4} + 4 \, a^{2} b^{6} \sin \left (d x + c\right )^{4} - 8 \, b^{8} \sin \left (d x + c\right )^{4} - 25 \, a^{7} b \sin \left (d x + c\right )^{3} + 70 \, a^{5} b^{3} \sin \left (d x + c\right )^{3} - 61 \, a^{3} b^{5} \sin \left (d x + c\right )^{3} + 16 \, a b^{7} \sin \left (d x + c\right )^{3} + 18 \, a^{8} \sin \left (d x + c\right )^{2} - 48 \, a^{6} b^{2} \sin \left (d x + c\right )^{2} + 38 \, a^{4} b^{4} \sin \left (d x + c\right )^{2} - 8 \, a^{2} b^{6} \sin \left (d x + c\right )^{2} + 4 \, b^{8} \sin \left (d x + c\right )^{2} + 8 \, a^{7} b \sin \left (d x + c\right ) - 24 \, a^{5} b^{3} \sin \left (d x + c\right ) + 24 \, a^{3} b^{5} \sin \left (d x + c\right ) - 8 \, a b^{7} \sin \left (d x + c\right ) - 4 \, a^{8} + 12 \, a^{6} b^{2} - 12 \, a^{4} b^{4} + 4 \, a^{2} b^{6}\right )}}{{\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} {\left (\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/16*(16*b^9*log(abs(b*sin(d*x + c) + a))/(a^9*b - 3*a^7*b^3 + 3*a^5*b^5 - a^3*b^7) - (24*a^2 - 57*a*b + 35*b^
2)*log(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (24*a^2 + 57*a*b + 35*b^2)*log(abs(sin(d*x + c
) - 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 16*(3*a^2 + b^2)*log(abs(sin(d*x + c)))/a^3 + 2*(4*b^8*sin(d*x + c)^
6 + 15*a^7*b*sin(d*x + c)^5 - 42*a^5*b^3*sin(d*x + c)^5 + 35*a^3*b^5*sin(d*x + c)^5 - 8*a*b^7*sin(d*x + c)^5 -
12*a^8*sin(d*x + c)^4 + 32*a^6*b^2*sin(d*x + c)^4 - 24*a^4*b^4*sin(d*x + c)^4 + 4*a^2*b^6*sin(d*x + c)^4 - 8*
b^8*sin(d*x + c)^4 - 25*a^7*b*sin(d*x + c)^3 + 70*a^5*b^3*sin(d*x + c)^3 - 61*a^3*b^5*sin(d*x + c)^3 + 16*a*b^
7*sin(d*x + c)^3 + 18*a^8*sin(d*x + c)^2 - 48*a^6*b^2*sin(d*x + c)^2 + 38*a^4*b^4*sin(d*x + c)^2 - 8*a^2*b^6*s
in(d*x + c)^2 + 4*b^8*sin(d*x + c)^2 + 8*a^7*b*sin(d*x + c) - 24*a^5*b^3*sin(d*x + c) + 24*a^3*b^5*sin(d*x + c
) - 8*a*b^7*sin(d*x + c) - 4*a^8 + 12*a^6*b^2 - 12*a^4*b^4 + 4*a^2*b^6)/((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^
6)*(sin(d*x + c)^3 - sin(d*x + c))^2))/d

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maple [A]  time = 0.55, size = 371, normalized size = 1.35 \[ \frac {1}{2 d \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {9 a}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {11 b}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) a^{2}}{2 d \left (a +b \right )^{3}}-\frac {57 \ln \left (\sin \left (d x +c \right )-1\right ) a b}{16 d \left (a +b \right )^{3}}-\frac {35 \ln \left (\sin \left (d x +c \right )-1\right ) b^{2}}{16 d \left (a +b \right )^{3}}+\frac {b^{8} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,a^{3} \left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {1}{2 d a \sin \left (d x +c \right )^{2}}+\frac {3 \ln \left (\sin \left (d x +c \right )\right )}{a d}+\frac {b^{2} \ln \left (\sin \left (d x +c \right )\right )}{a^{3} d}+\frac {b}{d \,a^{2} \sin \left (d x +c \right )}+\frac {1}{2 d \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {9 a}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {11 b}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) a^{2}}{2 d \left (a -b \right )^{3}}+\frac {57 \ln \left (1+\sin \left (d x +c \right )\right ) a b}{16 d \left (a -b \right )^{3}}-\frac {35 \ln \left (1+\sin \left (d x +c \right )\right ) b^{2}}{16 d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*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)^2-9/16/d/(a+b)^2/(sin(d*x+c)-1)*a-11/16/d/(a+b)^2/(sin(d*x+c)-1)*b-3/2/d/(a+b)^
3*ln(sin(d*x+c)-1)*a^2-57/16/d/(a+b)^3*ln(sin(d*x+c)-1)*a*b-35/16/d/(a+b)^3*ln(sin(d*x+c)-1)*b^2+1/d/a^3*b^8/(
a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c))-1/2/d/a/sin(d*x+c)^2+3*ln(sin(d*x+c))/a/d+b^2*ln(sin(d*x+c))/a^3/d+1/d/a^2*b
/sin(d*x+c)+1/2/d/(8*a-8*b)/(1+sin(d*x+c))^2+9/16/d/(a-b)^2/(1+sin(d*x+c))*a-11/16/d/(a-b)^2/(1+sin(d*x+c))*b-
3/2/d/(a-b)^3*ln(1+sin(d*x+c))*a^2+57/16/d/(a-b)^3*ln(1+sin(d*x+c))*a*b-35/16/d/(a-b)^3*ln(1+sin(d*x+c))*b^2

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maxima [A]  time = 0.35, size = 422, normalized size = 1.54 \[ \frac {\frac {16 \, b^{8} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}} - \frac {{\left (24 \, a^{2} - 57 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (24 \, a^{2} + 57 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left ({\left (15 \, a^{4} b - 27 \, a^{2} b^{3} + 8 \, b^{5}\right )} \sin \left (d x + c\right )^{5} - 4 \, a^{5} + 8 \, a^{3} b^{2} - 4 \, a b^{4} - 4 \, {\left (3 \, a^{5} - 5 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{4} - {\left (25 \, a^{4} b - 45 \, a^{2} b^{3} + 16 \, b^{5}\right )} \sin \left (d x + c\right )^{3} + 2 \, {\left (9 \, a^{5} - 15 \, a^{3} b^{2} + 4 \, a b^{4}\right )} \sin \left (d x + c\right )^{2} + 8 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \sin \left (d x + c\right )^{6} - 2 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \sin \left (d x + c\right )^{4} + {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \sin \left (d x + c\right )^{2}} + \frac {16 \, {\left (3 \, a^{2} + b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{3}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/16*(16*b^8*log(b*sin(d*x + c) + a)/(a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6) - (24*a^2 - 57*a*b + 35*b^2)*log(
sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (24*a^2 + 57*a*b + 35*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*
a^2*b + 3*a*b^2 + b^3) + 2*((15*a^4*b - 27*a^2*b^3 + 8*b^5)*sin(d*x + c)^5 - 4*a^5 + 8*a^3*b^2 - 4*a*b^4 - 4*(
3*a^5 - 5*a^3*b^2 + a*b^4)*sin(d*x + c)^4 - (25*a^4*b - 45*a^2*b^3 + 16*b^5)*sin(d*x + c)^3 + 2*(9*a^5 - 15*a^
3*b^2 + 4*a*b^4)*sin(d*x + c)^2 + 8*(a^4*b - 2*a^2*b^3 + b^5)*sin(d*x + c))/((a^6 - 2*a^4*b^2 + a^2*b^4)*sin(d
*x + c)^6 - 2*(a^6 - 2*a^4*b^2 + a^2*b^4)*sin(d*x + c)^4 + (a^6 - 2*a^4*b^2 + a^2*b^4)*sin(d*x + c)^2) + 16*(3
*a^2 + b^2)*log(sin(d*x + c))/a^3)/d

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mupad [B]  time = 12.70, size = 412, normalized size = 1.50 \[ \frac {\ln \left (\sin \left (c+d\,x\right )\right )\,\left (3\,a^2+b^2\right )}{a^3\,d}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {b^2}{8\,{\left (a-b\right )}^3}-\frac {9\,b}{16\,{\left (a-b\right )}^2}+\frac {3}{2\,\left (a-b\right )}\right )}{d}-\frac {\frac {1}{2\,a}-\frac {b\,\sin \left (c+d\,x\right )}{a^2}+\frac {{\sin \left (c+d\,x\right )}^4\,\left (3\,a^4-5\,a^2\,b^2+b^4\right )}{2\,a\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (9\,a^4-15\,a^2\,b^2+4\,b^4\right )}{4\,a\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\sin \left (c+d\,x\right )}^5\,\left (15\,a^4\,b-27\,a^2\,b^3+8\,b^5\right )}{8\,a^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\sin \left (c+d\,x\right )}^3\,\left (25\,a^4\,b-45\,a^2\,b^3+16\,b^5\right )}{8\,a^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\sin \left (c+d\,x\right )}^6-2\,{\sin \left (c+d\,x\right )}^4+{\sin \left (c+d\,x\right )}^2\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {9\,b}{16\,{\left (a+b\right )}^2}+\frac {3}{2\,\left (a+b\right )}+\frac {b^2}{8\,{\left (a+b\right )}^3}\right )}{d}+\frac {b^8\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^9-3\,a^7\,b^2+3\,a^5\,b^4-a^3\,b^6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*sin(c + d*x)^3*(a + b*sin(c + d*x))),x)

[Out]

(log(sin(c + d*x))*(3*a^2 + b^2))/(a^3*d) - (log(sin(c + d*x) + 1)*(b^2/(8*(a - b)^3) - (9*b)/(16*(a - b)^2) +
3/(2*(a - b))))/d - (1/(2*a) - (b*sin(c + d*x))/a^2 + (sin(c + d*x)^4*(3*a^4 + b^4 - 5*a^2*b^2))/(2*a*(a^4 +
b^4 - 2*a^2*b^2)) - (sin(c + d*x)^2*(9*a^4 + 4*b^4 - 15*a^2*b^2))/(4*a*(a^4 + b^4 - 2*a^2*b^2)) - (sin(c + d*x
)^5*(15*a^4*b + 8*b^5 - 27*a^2*b^3))/(8*a^2*(a^4 + b^4 - 2*a^2*b^2)) + (sin(c + d*x)^3*(25*a^4*b + 16*b^5 - 45
*a^2*b^3))/(8*a^2*(a^4 + b^4 - 2*a^2*b^2)))/(d*(sin(c + d*x)^2 - 2*sin(c + d*x)^4 + sin(c + d*x)^6)) - (log(si
n(c + d*x) - 1)*((9*b)/(16*(a + b)^2) + 3/(2*(a + b)) + b^2/(8*(a + b)^3)))/d + (b^8*log(a + b*sin(c + d*x)))/
(d*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*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)**3*sec(d*x+c)**5/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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