\(\int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx\) [1508]

Optimal result

Integrand size = 29, antiderivative size = 221 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {3 a^2 b \csc (c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}-\frac {3 (a+b) \left (8 a^2+7 a b+b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {3 a \left (a^2+b^2\right ) \log (\sin (c+d x))}{d}-\frac {3 (a-b) \left (8 a^2-7 a b+b^2\right ) \log (1+\sin (c+d x))}{16 d}+\frac {b^2 \sec ^4(c+d x) \left (a \left (3+\frac {a^2}{b^2}\right )+\left (1+\frac {3 a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}+\frac {b^2 \sec ^2(c+d x) \left (4 a \left (3+\frac {2 a^2}{b^2}\right )+3 \left (1+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 221, 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} \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {a^3 \csc ^2(c+d x)}{2 d}-\frac {3 (a+b) \left (8 a^2+7 a b+b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {3 a \left (a^2+b^2\right ) \log (\sin (c+d x))}{d}-\frac {3 (a-b) \left (8 a^2-7 a b+b^2\right ) \log (\sin (c+d x)+1)}{16 d}+\frac {b^2 \sec ^4(c+d x) \left (b \left (\frac {3 a^2}{b^2}+1\right ) \sin (c+d x)+a \left (\frac {a^2}{b^2}+3\right )\right )}{4 d}+\frac {b^2 \sec ^2(c+d x) \left (3 b \left (\frac {7 a^2}{b^2}+1\right ) \sin (c+d x)+4 a \left (\frac {2 a^2}{b^2}+3\right )\right )}{8 d}-\frac {3 a^2 b \csc (c+d x)}{d} \]

[In]

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

[Out]

(-3*a^2*b*Csc[c + d*x])/d - (a^3*Csc[c + d*x]^2)/(2*d) - (3*(a + b)*(8*a^2 + 7*a*b + b^2)*Log[1 - Sin[c + d*x]
])/(16*d) + (3*a*(a^2 + b^2)*Log[Sin[c + d*x]])/d - (3*(a - b)*(8*a^2 - 7*a*b + b^2)*Log[1 + Sin[c + d*x]])/(1
6*d) + (b^2*Sec[c + d*x]^4*(a*(3 + a^2/b^2) + (1 + (3*a^2)/b^2)*b*Sin[c + d*x]))/(4*d) + (b^2*Sec[c + d*x]^2*(
4*a*(3 + (2*a^2)/b^2) + 3*(1 + (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^3 (a+x)^3}{x^3 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^8 \text {Subst}\left (\int \frac {(a+x)^3}{x^3 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^2 \sec ^4(c+d x) \left (a \left (3+\frac {a^2}{b^2}\right )+\left (1+\frac {3 a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}-\frac {b^6 \text {Subst}\left (\int \frac {-4 a^3-12 a^2 x-4 a \left (3+\frac {a^2}{b^2}\right ) x^2-3 \left (1+\frac {3 a^2}{b^2}\right ) x^3}{x^3 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d} \\ & = \frac {b^2 \sec ^4(c+d x) \left (a \left (3+\frac {a^2}{b^2}\right )+\left (1+\frac {3 a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}+\frac {b^2 \sec ^2(c+d x) \left (4 a \left (3+\frac {2 a^2}{b^2}\right )+3 \left (1+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac {b^4 \text {Subst}\left (\int \frac {8 a^3+24 a^2 x+8 a \left (3+\frac {2 a^2}{b^2}\right ) x^2+3 \left (1+\frac {7 a^2}{b^2}\right ) x^3}{x^3 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = \frac {b^2 \sec ^4(c+d x) \left (a \left (3+\frac {a^2}{b^2}\right )+\left (1+\frac {3 a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}+\frac {b^2 \sec ^2(c+d x) \left (4 a \left (3+\frac {2 a^2}{b^2}\right )+3 \left (1+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac {b^4 \text {Subst}\left (\int \left (\frac {3 (a+b) \left (8 a^2+7 a b+b^2\right )}{2 b^4 (b-x)}+\frac {8 a^3}{b^2 x^3}+\frac {24 a^2}{b^2 x^2}+\frac {24 a \left (a^2+b^2\right )}{b^4 x}+\frac {3 (a-b) \left (-8 a^2+7 a b-b^2\right )}{2 b^4 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = -\frac {3 a^2 b \csc (c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}-\frac {3 (a+b) \left (8 a^2+7 a b+b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {3 a \left (a^2+b^2\right ) \log (\sin (c+d x))}{d}-\frac {3 (a-b) \left (8 a^2-7 a b+b^2\right ) \log (1+\sin (c+d x))}{16 d}+\frac {b^2 \sec ^4(c+d x) \left (a \left (3+\frac {a^2}{b^2}\right )+\left (1+\frac {3 a^2}{b^2}\right ) b \sin (c+d x)\right )}{4 d}+\frac {b^2 \sec ^2(c+d x) \left (4 a \left (3+\frac {2 a^2}{b^2}\right )+3 \left (1+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.12 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.86 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {-48 a^2 b \csc (c+d x)-8 a^3 \csc ^2(c+d x)-3 (a+b) \left (8 a^2+7 a b+b^2\right ) \log (1-\sin (c+d x))+48 a \left (a^2+b^2\right ) \log (\sin (c+d x))-3 (a-b) \left (8 a^2-7 a b+b^2\right ) \log (1+\sin (c+d x))+\frac {(a+b)^3}{(-1+\sin (c+d x))^2}-\frac {3 (a+b)^2 (3 a+b)}{-1+\sin (c+d x)}+\frac {(a-b)^3}{(1+\sin (c+d x))^2}+\frac {3 (a-b)^2 (3 a-b)}{1+\sin (c+d x)}}{16 d} \]

[In]

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

[Out]

(-48*a^2*b*Csc[c + d*x] - 8*a^3*Csc[c + d*x]^2 - 3*(a + b)*(8*a^2 + 7*a*b + b^2)*Log[1 - Sin[c + d*x]] + 48*a*
(a^2 + b^2)*Log[Sin[c + d*x]] - 3*(a - b)*(8*a^2 - 7*a*b + b^2)*Log[1 + Sin[c + d*x]] + (a + b)^3/(-1 + Sin[c
+ d*x])^2 - (3*(a + b)^2*(3*a + b))/(-1 + Sin[c + d*x]) + (a - b)^3/(1 + Sin[c + d*x])^2 + (3*(a - b)^2*(3*a -
b))/(1 + Sin[c + d*x]))/(16*d)

Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(\frac {a^{3} \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )+3 a^{2} b \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a \,b^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+b^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(217\)
default	\(\frac {a^{3} \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )+3 a^{2} b \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a \,b^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+b^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(217\)
parallelrisch	\(\frac {-12 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) \left (a^{2}+\frac {7}{8} a b +\frac {1}{8} b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-12 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right ) \left (a^{2}-\frac {7}{8} a b +\frac {1}{8} b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+12 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+b^{2}\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {27 \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\cos \left (2 d x +2 c \right )+\frac {2 \cos \left (4 d x +4 c \right )}{9}-\frac {\cos \left (6 d x +6 c \right )}{9}+\frac {2}{27}\right ) a^{3} \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32}-\frac {15 b \left (\cos \left (2 d x +2 c \right )+\frac {3 \cos \left (4 d x +4 c \right )}{8}+\frac {9}{40}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \csc \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-3 b^{2} \left (a \cos \left (2 d x +2 c \right )+\frac {3 \cos \left (4 d x +4 c \right ) a}{4}-\frac {b \sin \left (3 d x +3 c \right )}{4}-\frac {11 b \sin \left (d x +c \right )}{12}-\frac {7 a}{4}\right )}{d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(338\)
risch	\(-\frac {i \left (48 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-144 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+45 a^{2} b \,{\mathrm e}^{11 i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{11 i \left (d x +c \right )}+24 i a^{3} {\mathrm e}^{10 i \left (d x +c \right )}+24 i a \,b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+75 a^{2} b \,{\mathrm e}^{9 i \left (d x +c \right )}+5 b^{3} {\mathrm e}^{9 i \left (d x +c \right )}+48 i a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-16 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-66 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}-30 b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+24 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+24 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+66 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}+30 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+48 i a^{3} {\mathrm e}^{8 i \left (d x +c \right )}+48 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-75 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-5 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-45 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-3 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{3}}{d}+\frac {45 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{8 d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {45 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} b}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a \,b^{2}}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{3}}{8 d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{d}\)	\(577\)

[In]

int(csc(d*x+c)^3*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)^2/cos(d*x+c)^4+3/4/sin(d*x+c)^2/cos(d*x+c)^2-3/2/sin(d*x+c)^2+3*ln(tan(d*x+c)))+3*a^2
*b*(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*b^2*(1/4/cos(d*x+c)^4+1/2/cos(d*x+c)^2+ln(tan(d*x+c)))+b^3*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3
/8*ln(sec(d*x+c)+tan(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.52 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {24 \, {\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, a^{3} - 12 \, a b^{2} - 12 \, {\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{2} + 48 \, {\left ({\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{6} - {\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 3 \, {\left ({\left (8 \, a^{3} - 15 \, a^{2} b + 8 \, a b^{2} - b^{3}\right )} \cos \left (d x + c\right )^{6} - {\left (8 \, a^{3} - 15 \, a^{2} b + 8 \, a b^{2} - b^{3}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (8 \, a^{3} + 15 \, a^{2} b + 8 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{6} - {\left (8 \, a^{3} + 15 \, a^{2} b + 8 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, {\left (15 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} - 6 \, a^{2} b - 2 \, b^{3} - {\left (15 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (d \cos \left (d x + c\right )^{6} - d \cos \left (d x + c\right )^{4}\right )}} \]

[In]

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

[Out]

1/16*(24*(a^3 + a*b^2)*cos(d*x + c)^4 - 4*a^3 - 12*a*b^2 - 12*(a^3 + a*b^2)*cos(d*x + c)^2 + 48*((a^3 + a*b^2)
*cos(d*x + c)^6 - (a^3 + a*b^2)*cos(d*x + c)^4)*log(1/2*sin(d*x + c)) - 3*((8*a^3 - 15*a^2*b + 8*a*b^2 - b^3)*
cos(d*x + c)^6 - (8*a^3 - 15*a^2*b + 8*a*b^2 - b^3)*cos(d*x + c)^4)*log(sin(d*x + c) + 1) - 3*((8*a^3 + 15*a^2
*b + 8*a*b^2 + b^3)*cos(d*x + c)^6 - (8*a^3 + 15*a^2*b + 8*a*b^2 + b^3)*cos(d*x + c)^4)*log(-sin(d*x + c) + 1)
+ 2*(3*(15*a^2*b + b^3)*cos(d*x + c)^4 - 6*a^2*b - 2*b^3 - (15*a^2*b + b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*
cos(d*x + c)^6 - d*cos(d*x + c)^4)

Sympy [F(-1)]

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

[In]

integrate(csc(d*x+c)**3*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 = 217, normalized size of antiderivative = 0.98 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {3 \, {\left (8 \, a^{3} - 15 \, a^{2} b + 8 \, a b^{2} - b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (8 \, a^{3} + 15 \, a^{2} b + 8 \, a b^{2} + b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - 48 \, {\left (a^{3} + a b^{2}\right )} \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (3 \, {\left (15 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )^{5} + 12 \, {\left (a^{3} + a b^{2}\right )} \sin \left (d x + c\right )^{4} + 24 \, a^{2} b \sin \left (d x + c\right ) - 5 \, {\left (15 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )^{3} + 4 \, a^{3} - 18 \, {\left (a^{3} + a b^{2}\right )} \sin \left (d x + c\right )^{2}\right )}}{\sin \left (d x + c\right )^{6} - 2 \, \sin \left (d x + c\right )^{4} + \sin \left (d x + c\right )^{2}}}{16 \, d} \]

[In]

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

[Out]

-1/16*(3*(8*a^3 - 15*a^2*b + 8*a*b^2 - b^3)*log(sin(d*x + c) + 1) + 3*(8*a^3 + 15*a^2*b + 8*a*b^2 + b^3)*log(s
in(d*x + c) - 1) - 48*(a^3 + a*b^2)*log(sin(d*x + c)) + 2*(3*(15*a^2*b + b^3)*sin(d*x + c)^5 + 12*(a^3 + a*b^2
)*sin(d*x + c)^4 + 24*a^2*b*sin(d*x + c) - 5*(15*a^2*b + b^3)*sin(d*x + c)^3 + 4*a^3 - 18*(a^3 + a*b^2)*sin(d*
x + c)^2)/(sin(d*x + c)^6 - 2*sin(d*x + c)^4 + sin(d*x + c)^2))/d

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.09 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {3 \, {\left (8 \, a^{3} - 15 \, a^{2} b + 8 \, a b^{2} - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \, {\left (8 \, a^{3} + 15 \, a^{2} b + 8 \, a b^{2} + b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 48 \, {\left (a^{3} + a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac {2 \, {\left (45 \, a^{2} b \sin \left (d x + c\right )^{5} + 3 \, b^{3} \sin \left (d x + c\right )^{5} + 12 \, a^{3} \sin \left (d x + c\right )^{4} + 12 \, a b^{2} \sin \left (d x + c\right )^{4} - 75 \, a^{2} b \sin \left (d x + c\right )^{3} - 5 \, b^{3} \sin \left (d x + c\right )^{3} - 18 \, a^{3} \sin \left (d x + c\right )^{2} - 18 \, a b^{2} \sin \left (d x + c\right )^{2} + 24 \, a^{2} b \sin \left (d x + c\right ) + 4 \, a^{3}\right )}}{{\left (\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )\right )}^{2}}}{16 \, d} \]

[In]

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

[Out]

-1/16*(3*(8*a^3 - 15*a^2*b + 8*a*b^2 - b^3)*log(abs(sin(d*x + c) + 1)) + 3*(8*a^3 + 15*a^2*b + 8*a*b^2 + b^3)*
log(abs(sin(d*x + c) - 1)) - 48*(a^3 + a*b^2)*log(abs(sin(d*x + c))) + 2*(45*a^2*b*sin(d*x + c)^5 + 3*b^3*sin(
d*x + c)^5 + 12*a^3*sin(d*x + c)^4 + 12*a*b^2*sin(d*x + c)^4 - 75*a^2*b*sin(d*x + c)^3 - 5*b^3*sin(d*x + c)^3
- 18*a^3*sin(d*x + c)^2 - 18*a*b^2*sin(d*x + c)^2 + 24*a^2*b*sin(d*x + c) + 4*a^3)/(sin(d*x + c)^3 - sin(d*x +
c))^2)/d

Mupad [B] (verification not implemented)

Time = 11.64 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.00 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3\,a\,\ln \left (\sin \left (c+d\,x\right )\right )\,\left (a^2+b^2\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^4\,\left (\frac {3\,a^3}{2}+\frac {3\,a\,b^2}{2}\right )-{\sin \left (c+d\,x\right )}^2\,\left (\frac {9\,a^3}{4}+\frac {9\,a\,b^2}{4}\right )+{\sin \left (c+d\,x\right )}^5\,\left (\frac {45\,a^2\,b}{8}+\frac {3\,b^3}{8}\right )-{\sin \left (c+d\,x\right )}^3\,\left (\frac {75\,a^2\,b}{8}+\frac {5\,b^3}{8}\right )+\frac {a^3}{2}+3\,a^2\,b\,\sin \left (c+d\,x\right )}{d\,\left ({\sin \left (c+d\,x\right )}^6-2\,{\sin \left (c+d\,x\right )}^4+{\sin \left (c+d\,x\right )}^2\right )}-\frac {3\,\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (a-b\right )\,\left (8\,a^2-7\,a\,b+b^2\right )}{16\,d}-\frac {3\,\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (a+b\right )\,\left (8\,a^2+7\,a\,b+b^2\right )}{16\,d} \]

[In]

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

[Out]

(3*a*log(sin(c + d*x))*(a^2 + b^2))/d - (sin(c + d*x)^4*((3*a*b^2)/2 + (3*a^3)/2) - sin(c + d*x)^2*((9*a*b^2)/
4 + (9*a^3)/4) + sin(c + d*x)^5*((45*a^2*b)/8 + (3*b^3)/8) - sin(c + d*x)^3*((75*a^2*b)/8 + (5*b^3)/8) + a^3/2
+ 3*a^2*b*sin(c + d*x))/(d*(sin(c + d*x)^2 - 2*sin(c + d*x)^4 + sin(c + d*x)^6)) - (3*log(sin(c + d*x) + 1)*(
a - b)*(8*a^2 - 7*a*b + b^2))/(16*d) - (3*log(sin(c + d*x) - 1)*(a + b)*(7*a*b + 8*a^2 + b^2))/(16*d)