\(\int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx\) [1507]

Optimal result

Integrand size = 29, antiderivative size = 171 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {a^3 \csc (c+d x)}{d}-\frac {3 a (a+b) (5 a+3 b) \log (1-\sin (c+d x))}{16 d}+\frac {3 a^2 b \log (\sin (c+d x))}{d}+\frac {3 a (5 a-3 b) (a-b) \log (1+\sin (c+d x))}{16 d}+\frac {b \sec ^4(c+d x) \left (3 a^2+b^2+a \left (3+\frac {a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}+\frac {a b \sec ^2(c+d x) \left (12 a+\left (9+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d} \]

[Out]

-a^3*csc(d*x+c)/d-3/16*a*(a+b)*(5*a+3*b)*ln(1-sin(d*x+c))/d+3*a^2*b*ln(sin(d*x+c))/d+3/16*a*(5*a-3*b)*(a-b)*ln
(1+sin(d*x+c))/d+1/4*b*sec(d*x+c)^4*(3*a^2+b^2+a*(3+a^2/b^2)*b*sin(d*x+c))/d+1/8*a*b*sec(d*x+c)^2*(12*a+(9+7*a
^2/b^2)*b*sin(d*x+c))/d

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2916, 12, 1819, 1816} \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {a^3 \csc (c+d x)}{d}+\frac {b \sec ^4(c+d x) \left (a b \left (\frac {a^2}{b^2}+3\right ) \sin (c+d x)+3 a^2+b^2\right )}{4 d}+\frac {a b \sec ^2(c+d x) \left (b \left (\frac {7 a^2}{b^2}+9\right ) \sin (c+d x)+12 a\right )}{8 d}+\frac {3 a^2 b \log (\sin (c+d x))}{d}-\frac {3 a (a+b) (5 a+3 b) \log (1-\sin (c+d x))}{16 d}+\frac {3 a (5 a-3 b) (a-b) \log (\sin (c+d x)+1)}{16 d} \]

[In]

Int[Csc[c + d*x]^2*Sec[c + d*x]^5*(a + b*Sin[c + d*x])^3,x]

[Out]

-((a^3*Csc[c + d*x])/d) - (3*a*(a + b)*(5*a + 3*b)*Log[1 - Sin[c + d*x]])/(16*d) + (3*a^2*b*Log[Sin[c + d*x]])
/d + (3*a*(5*a - 3*b)*(a - b)*Log[1 + Sin[c + d*x]])/(16*d) + (b*Sec[c + d*x]^4*(3*a^2 + b^2 + a*(3 + a^2/b^2)
*b*Sin[c + d*x]))/(4*d) + (a*b*Sec[c + d*x]^2*(12*a + (9 + (7*a^2)/b^2)*b*Sin[c + d*x]))/(8*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 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1819

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 2916

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/b)*x)^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*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int \frac {b^2 (a+x)^3}{x^2 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^7 \text {Subst}\left (\int \frac {(a+x)^3}{x^2 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b \sec ^4(c+d x) \left (3 a^2+b^2+a \left (3+\frac {a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}-\frac {b^5 \text {Subst}\left (\int \frac {-4 a^3-12 a^2 x-3 a \left (3+\frac {a^2}{b^2}\right ) x^2}{x^2 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d} \\ & = \frac {b \sec ^4(c+d x) \left (3 a^2+b^2+a \left (3+\frac {a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}+\frac {a b \sec ^2(c+d x) \left (12 a+\left (9+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac {b^3 \text {Subst}\left (\int \frac {8 a^3+24 a^2 x+a \left (9+\frac {7 a^2}{b^2}\right ) x^2}{x^2 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = \frac {b \sec ^4(c+d x) \left (3 a^2+b^2+a \left (3+\frac {a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}+\frac {a b \sec ^2(c+d x) \left (12 a+\left (9+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac {b^3 \text {Subst}\left (\int \left (\frac {3 a (a+b) (5 a+3 b)}{2 b^3 (b-x)}+\frac {8 a^3}{b^2 x^2}+\frac {24 a^2}{b^2 x}+\frac {3 a (5 a-3 b) (a-b)}{2 b^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = -\frac {a^3 \csc (c+d x)}{d}-\frac {3 a (a+b) (5 a+3 b) \log (1-\sin (c+d x))}{16 d}+\frac {3 a^2 b \log (\sin (c+d x))}{d}+\frac {3 a (5 a-3 b) (a-b) \log (1+\sin (c+d x))}{16 d}+\frac {b \sec ^4(c+d x) \left (3 a^2+b^2+a \left (3+\frac {a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}+\frac {a b \sec ^2(c+d x) \left (12 a+\left (9+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.99 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.94 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {16 a^3 \csc (c+d x)+3 a (a+b) (5 a+3 b) \log (1-\sin (c+d x))-48 a^2 b \log (\sin (c+d x))-3 a (5 a-3 b) (a-b) \log (1+\sin (c+d x))-\frac {(a+b)^3}{(-1+\sin (c+d x))^2}+\frac {(a+b)^2 (7 a+b)}{-1+\sin (c+d x)}+\frac {(a-b)^3}{(1+\sin (c+d x))^2}+\frac {(a-b)^2 (7 a-b)}{1+\sin (c+d x)}}{16 d} \]

[In]

Integrate[Csc[c + d*x]^2*Sec[c + d*x]^5*(a + b*Sin[c + d*x])^3,x]

[Out]

-1/16*(16*a^3*Csc[c + d*x] + 3*a*(a + b)*(5*a + 3*b)*Log[1 - Sin[c + d*x]] - 48*a^2*b*Log[Sin[c + d*x]] - 3*a*
(5*a - 3*b)*(a - b)*Log[1 + Sin[c + d*x]] - (a + b)^3/(-1 + Sin[c + d*x])^2 + ((a + b)^2*(7*a + b))/(-1 + Sin[
c + d*x]) + (a - b)^3/(1 + Sin[c + d*x])^2 + ((a - b)^2*(7*a - b))/(1 + Sin[c + d*x]))/d

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(\frac {a^{3} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{2} b \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b^{3}}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(170\)
default	\(\frac {a^{3} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{2} b \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b^{3}}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(170\)
parallelrisch	\(\frac {-120 \left (a +\frac {3 b}{5}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+120 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -\frac {3 b}{5}\right ) \left (a -b \right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+192 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 \left (\cos \left (2 d x +2 c \right )+\frac {3 \cos \left (4 d x +4 c \right )}{8}+\frac {9}{40}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{3} \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-48 a^{2} b -16 b^{3}\right ) \cos \left (2 d x +2 c \right )+\left (-36 a^{2} b -4 b^{3}\right ) \cos \left (4 d x +4 c \right )+132 \sin \left (d x +c \right ) a \,b^{2}+36 \sin \left (3 d x +3 c \right ) a \,b^{2}+84 a^{2} b +20 b^{3}}{16 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(288\)
risch	\(\frac {-9 i a \,b^{2} {\mathrm e}^{9 i \left (d x +c \right )}-24 i a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-40 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-15 i a^{3} {\mathrm e}^{9 i \left (d x +c \right )}+24 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+66 i a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-15 i a^{3} {\mathrm e}^{i \left (d x +c \right )}+72 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+16 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-18 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}-9 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-72 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-16 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-24 i a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-40 i a^{3} {\mathrm e}^{7 i \left (d x +c \right )}-24 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} b}{d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a \,b^{2}}{8 d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{3}}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b}{d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{8 d}+\frac {3 a^{2} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(428\)

[In]

int(csc(d*x+c)^2*sec(d*x+c)^5*(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^3*(1/4/sin(d*x+c)/cos(d*x+c)^4+5/8/sin(d*x+c)/cos(d*x+c)^2-15/8/sin(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c
)))+3*a^2*b*(1/4/cos(d*x+c)^4+1/2/cos(d*x+c)^2+ln(tan(d*x+c)))+3*a*b^2*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*ta
n(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+1/4*b^3/cos(d*x+c)^4)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.32 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {48 \, a^{2} b \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + 3 \, {\left (5 \, a^{3} - 8 \, a^{2} b + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 3 \, {\left (5 \, a^{3} + 8 \, a^{2} b + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 6 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 4 \, a^{3} + 12 \, a b^{2} + 2 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (6 \, a^{2} b \cos \left (d x + c\right )^{2} + 3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/16*(48*a^2*b*cos(d*x + c)^4*log(1/2*sin(d*x + c))*sin(d*x + c) + 3*(5*a^3 - 8*a^2*b + 3*a*b^2)*cos(d*x + c)^
4*log(sin(d*x + c) + 1)*sin(d*x + c) - 3*(5*a^3 + 8*a^2*b + 3*a*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1)*sin
(d*x + c) - 6*(5*a^3 + 3*a*b^2)*cos(d*x + c)^4 + 4*a^3 + 12*a*b^2 + 2*(5*a^3 + 3*a*b^2)*cos(d*x + c)^2 + 4*(6*
a^2*b*cos(d*x + c)^2 + 3*a^2*b + b^3)*sin(d*x + c))/(d*cos(d*x + c)^4*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)**2*sec(d*x+c)**5*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.10 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {48 \, a^{2} b \log \left (\sin \left (d x + c\right )\right ) + 3 \, {\left (5 \, a^{3} - 8 \, a^{2} b + 3 \, a b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, a^{3} + 8 \, a^{2} b + 3 \, a b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (12 \, a^{2} b \sin \left (d x + c\right )^{3} + 3 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )^{4} + 8 \, a^{3} - 5 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} - 2 \, {\left (9 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{5} - 2 \, \sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )}}{16 \, d} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/16*(48*a^2*b*log(sin(d*x + c)) + 3*(5*a^3 - 8*a^2*b + 3*a*b^2)*log(sin(d*x + c) + 1) - 3*(5*a^3 + 8*a^2*b +
3*a*b^2)*log(sin(d*x + c) - 1) - 2*(12*a^2*b*sin(d*x + c)^3 + 3*(5*a^3 + 3*a*b^2)*sin(d*x + c)^4 + 8*a^3 - 5*(
5*a^3 + 3*a*b^2)*sin(d*x + c)^2 - 2*(9*a^2*b + b^3)*sin(d*x + c))/(sin(d*x + c)^5 - 2*sin(d*x + c)^3 + sin(d*x
+ c)))/d

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.23 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {48 \, a^{2} b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 3 \, {\left (5 \, a^{3} - 8 \, a^{2} b + 3 \, a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (5 \, a^{3} + 8 \, a^{2} b + 3 \, a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {16 \, {\left (3 \, a^{2} b \sin \left (d x + c\right ) + a^{3}\right )}}{\sin \left (d x + c\right )} + \frac {2 \, {\left (18 \, a^{2} b \sin \left (d x + c\right )^{4} - 7 \, a^{3} \sin \left (d x + c\right )^{3} - 9 \, a b^{2} \sin \left (d x + c\right )^{3} - 48 \, a^{2} b \sin \left (d x + c\right )^{2} + 9 \, a^{3} \sin \left (d x + c\right ) + 15 \, a b^{2} \sin \left (d x + c\right ) + 36 \, a^{2} b + 2 \, b^{3}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/16*(48*a^2*b*log(abs(sin(d*x + c))) + 3*(5*a^3 - 8*a^2*b + 3*a*b^2)*log(abs(sin(d*x + c) + 1)) - 3*(5*a^3 +
8*a^2*b + 3*a*b^2)*log(abs(sin(d*x + c) - 1)) - 16*(3*a^2*b*sin(d*x + c) + a^3)/sin(d*x + c) + 2*(18*a^2*b*sin
(d*x + c)^4 - 7*a^3*sin(d*x + c)^3 - 9*a*b^2*sin(d*x + c)^3 - 48*a^2*b*sin(d*x + c)^2 + 9*a^3*sin(d*x + c) + 1
5*a*b^2*sin(d*x + c) + 36*a^2*b + 2*b^3)/(sin(d*x + c)^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.06 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3\,a^2\,b\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^4\,\left (\frac {15\,a^3}{8}+\frac {9\,a\,b^2}{8}\right )-{\sin \left (c+d\,x\right )}^2\,\left (\frac {25\,a^3}{8}+\frac {15\,a\,b^2}{8}\right )+a^3-\sin \left (c+d\,x\right )\,\left (\frac {9\,a^2\,b}{4}+\frac {b^3}{4}\right )+\frac {3\,a^2\,b\,{\sin \left (c+d\,x\right )}^3}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^5-2\,{\sin \left (c+d\,x\right )}^3+\sin \left (c+d\,x\right )\right )}+\frac {3\,a\,\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (a-b\right )\,\left (5\,a-3\,b\right )}{16\,d}-\frac {3\,a\,\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (a+b\right )\,\left (5\,a+3\,b\right )}{16\,d} \]

[In]

int((a + b*sin(c + d*x))^3/(cos(c + d*x)^5*sin(c + d*x)^2),x)

[Out]

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