3.1368 \(\int \frac {\csc ^2(c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=250 \[ -\frac {\left (15 a^2+37 a b+24 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac {\left (15 a^2-37 a b+24 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}-\frac {b^7 \log (a+b \sin (c+d x))}{a^2 d \left (a^2-b^2\right )^3}-\frac {b \log (\sin (c+d x))}{a^2 d}+\frac {7 a+9 b}{16 d (a+b)^2 (1-\sin (c+d x))}-\frac {7 a-9 b}{16 d (a-b)^2 (\sin (c+d x)+1)}+\frac {1}{16 d (a+b) (1-\sin (c+d x))^2}-\frac {1}{16 d (a-b) (\sin (c+d x)+1)^2}-\frac {\csc (c+d x)}{a d} \]

[Out]

-csc(d*x+c)/a/d-1/16*(15*a^2+37*a*b+24*b^2)*ln(1-sin(d*x+c))/(a+b)^3/d-b*ln(sin(d*x+c))/a^2/d+1/16*(15*a^2-37*
a*b+24*b^2)*ln(1+sin(d*x+c))/(a-b)^3/d-b^7*ln(a+b*sin(d*x+c))/a^2/(a^2-b^2)^3/d+1/16/(a+b)/d/(1-sin(d*x+c))^2+
1/16*(7*a+9*b)/(a+b)^2/d/(1-sin(d*x+c))-1/16/(a-b)/d/(1+sin(d*x+c))^2+1/16*(-7*a+9*b)/(a-b)^2/d/(1+sin(d*x+c))

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Rubi [A]  time = 0.40, antiderivative size = 250, 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} \[ -\frac {b^7 \log (a+b \sin (c+d x))}{a^2 d \left (a^2-b^2\right )^3}-\frac {\left (15 a^2+37 a b+24 b^2\right ) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac {\left (15 a^2-37 a b+24 b^2\right ) \log (\sin (c+d x)+1)}{16 d (a-b)^3}-\frac {b \log (\sin (c+d x))}{a^2 d}+\frac {7 a+9 b}{16 d (a+b)^2 (1-\sin (c+d x))}-\frac {7 a-9 b}{16 d (a-b)^2 (\sin (c+d x)+1)}+\frac {1}{16 d (a+b) (1-\sin (c+d x))^2}-\frac {1}{16 d (a-b) (\sin (c+d x)+1)^2}-\frac {\csc (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[c + d*x]/(a*d)) - ((15*a^2 + 37*a*b + 24*b^2)*Log[1 - Sin[c + d*x]])/(16*(a + b)^3*d) - (b*Log[Sin[c + d
*x]])/(a^2*d) + ((15*a^2 - 37*a*b + 24*b^2)*Log[1 + Sin[c + d*x]])/(16*(a - b)^3*d) - (b^7*Log[a + b*Sin[c + d
*x]])/(a^2*(a^2 - b^2)^3*d) + 1/(16*(a + b)*d*(1 - Sin[c + d*x])^2) + (7*a + 9*b)/(16*(a + b)^2*d*(1 - Sin[c +
d*x])) - 1/(16*(a - b)*d*(1 + Sin[c + d*x])^2) - (7*a - 9*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 ^2(c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {b^2}{x^2 (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b^7 \operatorname {Subst}\left (\int \frac {1}{x^2 (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b^7 \operatorname {Subst}\left (\int \left (\frac {1}{8 b^5 (a+b) (b-x)^3}+\frac {7 a+9 b}{16 b^6 (a+b)^2 (b-x)^2}+\frac {15 a^2+37 a b+24 b^2}{16 b^7 (a+b)^3 (b-x)}+\frac {1}{a b^6 x^2}-\frac {1}{a^2 b^6 x}-\frac {1}{a^2 (a-b)^3 (a+b)^3 (a+x)}-\frac {1}{8 b^5 (-a+b) (b+x)^3}+\frac {7 a-9 b}{16 (a-b)^2 b^6 (b+x)^2}+\frac {15 a^2-37 a b+24 b^2}{16 (a-b)^3 b^7 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {\csc (c+d x)}{a d}-\frac {\left (15 a^2+37 a b+24 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}-\frac {b \log (\sin (c+d x))}{a^2 d}+\frac {\left (15 a^2-37 a b+24 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}-\frac {b^7 \log (a+b \sin (c+d x))}{a^2 \left (a^2-b^2\right )^3 d}+\frac {1}{16 (a+b) d (1-\sin (c+d x))^2}+\frac {7 a+9 b}{16 (a+b)^2 d (1-\sin (c+d x))}-\frac {1}{16 (a-b) d (1+\sin (c+d x))^2}-\frac {7 a-9 b}{16 (a-b)^2 d (1+\sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 6.21, size = 257, normalized size = 1.03 \[ \frac {b^7 \left (-\frac {\log (\sin (c+d x))}{a^2 b^6}-\frac {\left (15 a^2+37 a b+24 b^2\right ) \log (1-\sin (c+d x))}{16 b^7 (a+b)^3}+\frac {\left (15 a^2-37 a b+24 b^2\right ) \log (\sin (c+d x)+1)}{16 b^7 (a-b)^3}-\frac {\log (a+b \sin (c+d x))}{a^2 (a-b)^3 (a+b)^3}-\frac {\csc (c+d x)}{a b^7}-\frac {7 a-9 b}{16 b^6 (a-b)^2 (b \sin (c+d x)+b)}+\frac {7 a+9 b}{16 b^6 (a+b)^2 (b-b \sin (c+d x))}+\frac {1}{16 b^5 (a+b) (b-b \sin (c+d x))^2}-\frac {1}{16 b^5 (a-b) (b \sin (c+d x)+b)^2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^7*(-(Csc[c + d*x]/(a*b^7)) - ((15*a^2 + 37*a*b + 24*b^2)*Log[1 - Sin[c + d*x]])/(16*b^7*(a + b)^3) - Log[Si
n[c + d*x]]/(a^2*b^6) + ((15*a^2 - 37*a*b + 24*b^2)*Log[1 + Sin[c + d*x]])/(16*(a - b)^3*b^7) - Log[a + b*Sin[
c + d*x]]/(a^2*(a - b)^3*(a + b)^3) + 1/(16*b^5*(a + b)*(b - b*Sin[c + d*x])^2) + (7*a + 9*b)/(16*b^6*(a + b)^
2*(b - b*Sin[c + d*x])) - 1/(16*(a - b)*b^5*(b + b*Sin[c + d*x])^2) - (7*a - 9*b)/(16*(a - b)^2*b^6*(b + b*Sin
[c + d*x]))))/d

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fricas [A]  time = 4.81, size = 425, normalized size = 1.70 \[ -\frac {16 \, b^{7} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) - 4 \, a^{7} + 8 \, a^{5} b^{2} - 4 \, a^{3} b^{4} + 16 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - {\left (15 \, a^{7} + 8 \, a^{6} b - 42 \, a^{5} b^{2} - 24 \, a^{4} b^{3} + 35 \, a^{3} b^{4} + 24 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + {\left (15 \, a^{7} - 8 \, a^{6} b - 42 \, a^{5} b^{2} + 24 \, a^{4} b^{3} + 35 \, a^{3} b^{4} - 24 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 2 \, {\left (15 \, a^{7} - 42 \, a^{5} b^{2} + 35 \, a^{3} b^{4} - 8 \, a b^{6}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (5 \, a^{7} - 14 \, a^{5} b^{2} + 9 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5} + 2 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 2 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.25, size = 418, normalized size = 1.67 \[ -\frac {\frac {16 \, b^{8} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{8} b - 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} - a^{2} b^{7}} - \frac {{\left (15 \, a^{2} - 37 \, a b + 24 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (15 \, a^{2} + 37 \, a b + 24 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {16 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} + \frac {2 \, {\left (6 \, a^{4} b \sin \left (d x + c\right )^{4} - 18 \, a^{2} b^{3} \sin \left (d x + c\right )^{4} + 18 \, b^{5} \sin \left (d x + c\right )^{4} + 7 \, a^{5} \sin \left (d x + c\right )^{3} - 18 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} + 11 \, a b^{4} \sin \left (d x + c\right )^{3} - 16 \, a^{4} b \sin \left (d x + c\right )^{2} + 48 \, a^{2} b^{3} \sin \left (d x + c\right )^{2} - 44 \, b^{5} \sin \left (d x + c\right )^{2} - 9 \, a^{5} \sin \left (d x + c\right ) + 22 \, a^{3} b^{2} \sin \left (d x + c\right ) - 13 \, a b^{4} \sin \left (d x + c\right ) + 12 \, a^{4} b - 34 \, a^{2} b^{3} + 28 \, b^{5}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}} - \frac {16 \, {\left (b \sin \left (d x + c\right ) - a\right )}}{a^{2} \sin \left (d x + c\right )}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(16*b^8*log(abs(b*sin(d*x + c) + a))/(a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7) - (15*a^2 - 37*a*b + 24*b
^2)*log(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (15*a^2 + 37*a*b + 24*b^2)*log(abs(sin(d*x +
c) - 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 16*b*log(abs(sin(d*x + c)))/a^2 + 2*(6*a^4*b*sin(d*x + c)^4 - 18*a^
2*b^3*sin(d*x + c)^4 + 18*b^5*sin(d*x + c)^4 + 7*a^5*sin(d*x + c)^3 - 18*a^3*b^2*sin(d*x + c)^3 + 11*a*b^4*sin
(d*x + c)^3 - 16*a^4*b*sin(d*x + c)^2 + 48*a^2*b^3*sin(d*x + c)^2 - 44*b^5*sin(d*x + c)^2 - 9*a^5*sin(d*x + c)
+ 22*a^3*b^2*sin(d*x + c) - 13*a*b^4*sin(d*x + c) + 12*a^4*b - 34*a^2*b^3 + 28*b^5)/((a^6 - 3*a^4*b^2 + 3*a^2
*b^4 - b^6)*(sin(d*x + c)^2 - 1)^2) - 16*(b*sin(d*x + c) - a)/(a^2*sin(d*x + c)))/d

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maple [A]  time = 0.48, size = 340, normalized size = 1.36 \[ \frac {1}{2 d \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {7 a}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {9 b}{16 d \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {15 \ln \left (\sin \left (d x +c \right )-1\right ) a^{2}}{16 d \left (a +b \right )^{3}}-\frac {37 \ln \left (\sin \left (d x +c \right )-1\right ) a b}{16 d \left (a +b \right )^{3}}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right ) b^{2}}{2 d \left (a +b \right )^{3}}-\frac {b^{7} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,a^{2} \left (a +b \right )^{3} \left (a -b \right )^{3}}-\frac {1}{d a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2} d}-\frac {1}{2 d \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {7 a}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {9 b}{16 d \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {15 \ln \left (1+\sin \left (d x +c \right )\right ) a^{2}}{16 d \left (a -b \right )^{3}}-\frac {37 \ln \left (1+\sin \left (d x +c \right )\right ) a b}{16 d \left (a -b \right )^{3}}+\frac {3 \ln \left (1+\sin \left (d x +c \right )\right ) b^{2}}{2 d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*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)^2-7/16/d/(a+b)^2/(sin(d*x+c)-1)*a-9/16/d/(a+b)^2/(sin(d*x+c)-1)*b-15/16/d/(a+b)
^3*ln(sin(d*x+c)-1)*a^2-37/16/d/(a+b)^3*ln(sin(d*x+c)-1)*a*b-3/2/d/(a+b)^3*ln(sin(d*x+c)-1)*b^2-1/d*b^7/a^2/(a
+b)^3/(a-b)^3*ln(a+b*sin(d*x+c))-1/d/a/sin(d*x+c)-b*ln(sin(d*x+c))/a^2/d-1/2/d/(8*a-8*b)/(1+sin(d*x+c))^2-7/16
/d/(a-b)^2/(1+sin(d*x+c))*a+9/16/d/(a-b)^2/(1+sin(d*x+c))*b+15/16/d/(a-b)^3*ln(1+sin(d*x+c))*a^2-37/16/d/(a-b)
^3*ln(1+sin(d*x+c))*a*b+3/2/d/(a-b)^3*ln(1+sin(d*x+c))*b^2

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maxima [A]  time = 0.35, size = 361, normalized size = 1.44 \[ -\frac {\frac {16 \, b^{7} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}} - \frac {{\left (15 \, a^{2} - 37 \, a b + 24 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (15 \, a^{2} + 37 \, a b + 24 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left ({\left (15 \, a^{4} - 27 \, a^{2} b^{2} + 8 \, b^{4}\right )} \sin \left (d x + c\right )^{4} + 8 \, a^{4} - 16 \, a^{2} b^{2} + 8 \, b^{4} - 4 \, {\left (a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )^{3} - {\left (25 \, a^{4} - 45 \, a^{2} b^{2} + 16 \, b^{4}\right )} \sin \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{3} b - 5 \, a b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{5} - 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{3} + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )} + \frac {16 \, b \log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(16*b^7*log(b*sin(d*x + c) + a)/(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6) - (15*a^2 - 37*a*b + 24*b^2)*log
(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (15*a^2 + 37*a*b + 24*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3
*a^2*b + 3*a*b^2 + b^3) + 2*((15*a^4 - 27*a^2*b^2 + 8*b^4)*sin(d*x + c)^4 + 8*a^4 - 16*a^2*b^2 + 8*b^4 - 4*(a^
3*b - 2*a*b^3)*sin(d*x + c)^3 - (25*a^4 - 45*a^2*b^2 + 16*b^4)*sin(d*x + c)^2 + 2*(3*a^3*b - 5*a*b^3)*sin(d*x
+ c))/((a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c)^5 - 2*(a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c)^3 + (a^5 - 2*a^3*b^
2 + a*b^4)*sin(d*x + c)) + 16*b*log(sin(d*x + c))/a^2)/d

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mupad [B]  time = 12.47, size = 373, normalized size = 1.49 \[ \frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {b^2}{8\,{\left (a-b\right )}^3}-\frac {7\,b}{16\,{\left (a-b\right )}^2}+\frac {15}{16\,\left (a-b\right )}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {7\,b}{16\,{\left (a+b\right )}^2}+\frac {15}{16\,\left (a+b\right )}+\frac {b^2}{8\,{\left (a+b\right )}^3}\right )}{d}-\frac {\frac {1}{a}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (a^2\,b-2\,b^3\right )}{2\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {\sin \left (c+d\,x\right )\,\left (3\,a^2\,b-5\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\sin \left (c+d\,x\right )}^4\,\left (15\,a^4-27\,a^2\,b^2+8\,b^4\right )}{8\,a\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (25\,a^4-45\,a^2\,b^2+16\,b^4\right )}{8\,a\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\sin \left (c+d\,x\right )}^5-2\,{\sin \left (c+d\,x\right )}^3+\sin \left (c+d\,x\right )\right )}-\frac {b\,\ln \left (\sin \left (c+d\,x\right )\right )}{a^2\,d}-\frac {b^7\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^8-3\,a^6\,b^2+3\,a^4\,b^4-a^2\,b^6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*sin(c + d*x)^2*(a + b*sin(c + d*x))),x)

[Out]

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

[Out]

Timed out

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