\(\int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx\) [1506]

Optimal result

Integrand size = 27, antiderivative size = 165 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {\left (8 a^3+9 a^2 b-b^3\right ) \log (1-\sin (c+d x))}{16 d}+\frac {a^3 \log (\sin (c+d x))}{d}-\frac {\left (8 a^3-9 a^2 b+b^3\right ) \log (1+\sin (c+d x))}{16 d}+\frac {\sec ^2(c+d x) \left (4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^4(c+d x) \left (a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2916, 12, 1819, 837, 815} \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a^3 \log (\sin (c+d x))}{d}+\frac {\sec ^4(c+d x) \left (b \left (3 a^2+b^2\right ) \sin (c+d x)+a \left (a^2+3 b^2\right )\right )}{4 d}-\frac {\left (8 a^3+9 a^2 b-b^3\right ) \log (1-\sin (c+d x))}{16 d}-\frac {\left (8 a^3-9 a^2 b+b^3\right ) \log (\sin (c+d x)+1)}{16 d}+\frac {\sec ^2(c+d x) \left (4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{8 d} \]

[In]

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

[Out]

-1/16*((8*a^3 + 9*a^2*b - b^3)*Log[1 - Sin[c + d*x]])/d + (a^3*Log[Sin[c + d*x]])/d - ((8*a^3 - 9*a^2*b + b^3)
*Log[1 + Sin[c + d*x]])/(16*d) + (Sec[c + d*x]^2*(4*a^3 + b*(9*a^2 - b^2)*Sin[c + d*x]))/(8*d) + (Sec[c + d*x]
^4*(a*(a^2 + 3*b^2) + b*(3*a^2 + b^2)*Sin[c + d*x]))/(4*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 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 (a+x)^3}{x \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^6 \text {Subst}\left (\int \frac {(a+x)^3}{x \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^4(c+d x) \left (a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 d}-\frac {b^4 \text {Subst}\left (\int \frac {-4 a^3-\left (9 a^2-b^2\right ) x}{x \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d} \\ & = \frac {\sec ^2(c+d x) \left (4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^4(c+d x) \left (a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 d}-\frac {\text {Subst}\left (\int \frac {-8 a^3 b^2-b^2 \left (9 a^2-b^2\right ) x}{x \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = \frac {\sec ^2(c+d x) \left (4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^4(c+d x) \left (a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 d}-\frac {\text {Subst}\left (\int \left (\frac {-8 a^3-9 a^2 b+b^3}{2 (b-x)}-\frac {8 a^3}{x}+\frac {8 a^3-9 a^2 b+b^3}{2 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = -\frac {\left (8 a^3+9 a^2 b-b^3\right ) \log (1-\sin (c+d x))}{16 d}+\frac {a^3 \log (\sin (c+d x))}{d}-\frac {\left (8 a^3-9 a^2 b+b^3\right ) \log (1+\sin (c+d x))}{16 d}+\frac {\sec ^2(c+d x) \left (4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^4(c+d x) \left (a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.95 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {-\left (\left (8 a^3+9 a^2 b-b^3\right ) \log (1-\sin (c+d x))\right )+16 a^3 \log (\sin (c+d x))-\left (8 a^3-9 a^2 b+b^3\right ) \log (1+\sin (c+d x))+\frac {(a+b)^3}{(-1+\sin (c+d x))^2}-\frac {(5 a-b) (a+b)^2}{-1+\sin (c+d x)}+\frac {(a-b)^3}{(1+\sin (c+d x))^2}+\frac {(a-b)^2 (5 a+b)}{1+\sin (c+d x)}}{16 d} \]

[In]

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

[Out]

(-((8*a^3 + 9*a^2*b - b^3)*Log[1 - Sin[c + d*x]]) + 16*a^3*Log[Sin[c + d*x]] - (8*a^3 - 9*a^2*b + b^3)*Log[1 +
Sin[c + d*x]] + (a + b)^3/(-1 + Sin[c + d*x])^2 - ((5*a - b)*(a + b)^2)/(-1 + Sin[c + d*x]) + (a - b)^3/(1 +
Sin[c + d*x])^2 + ((a - b)^2*(5*a + b))/(1 + Sin[c + d*x]))/(16*d)

Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.01

method	result	size
derivativedivides	\(\frac {a^{3} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a^{2} b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a \,b^{2}}{4 \cos \left (d x +c \right )^{4}}+b^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(167\)
default	\(\frac {a^{3} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a^{2} b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a \,b^{2}}{4 \cos \left (d x +c \right )^{4}}+b^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(167\)
parallelrisch	\(\frac {-32 \left (a^{2}+\frac {1}{8} a b -\frac {1}{8} b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-32 \left (a^{2}-\frac {1}{8} a b -\frac {1}{8} b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+32 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-8 a^{3}-24 a \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (-6 a^{3}-6 a \,b^{2}\right ) \cos \left (4 d x +4 c \right )+\left (18 a^{2} b -2 b^{3}\right ) \sin \left (3 d x +3 c \right )+\left (66 a^{2} b +14 b^{3}\right ) \sin \left (d x +c \right )+14 a^{3}+30 a \,b^{2}}{8 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(270\)
risch	\(\frac {i \left (-8 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-9 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}+b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-32 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-48 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-33 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-7 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-8 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+33 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+7 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+9 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{3}}{d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{8 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} b}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{3}}{8 d}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(345\)
norman	\(\frac {\frac {\left (4 a^{3}+6 a \,b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 a^{3}+6 a \,b^{2}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 a^{3}+24 a \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 a^{3}+24 a \,b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (8 a^{3}+18 a \,b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (8 a^{3}+18 a \,b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b \left (21 a^{2}+11 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b \left (15 a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {b \left (15 a^{2}+b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {b \left (27 a^{2}+5 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {b \left (27 a^{2}+5 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {b \left (81 a^{2}+31 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {b \left (81 a^{2}+31 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (8 a^{3}-9 a^{2} b +b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (8 a^{3}+9 a^{2} b -b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\)	\(468\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.05 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {16 \, a^{3} \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (8 \, a^{3} - 9 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (8 \, a^{3} + 9 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, a^{3} \cos \left (d x + c\right )^{2} + 4 \, a^{3} + 12 \, a b^{2} + 2 \, {\left (6 \, a^{2} b + 2 \, b^{3} + {\left (9 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

1/16*(16*a^3*cos(d*x + c)^4*log(1/2*sin(d*x + c)) - (8*a^3 - 9*a^2*b + b^3)*cos(d*x + c)^4*log(sin(d*x + c) +
1) - (8*a^3 + 9*a^2*b - b^3)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 8*a^3*cos(d*x + c)^2 + 4*a^3 + 12*a*b^2 +
2*(6*a^2*b + 2*b^3 + (9*a^2*b - b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.97 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {16 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) - {\left (8 \, a^{3} - 9 \, a^{2} b + b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (8 \, a^{3} + 9 \, a^{2} b - b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (4 \, a^{3} \sin \left (d x + c\right )^{2} + {\left (9 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{3} - 6 \, a^{3} - 6 \, a b^{2} - {\left (15 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]

[In]

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

[Out]

1/16*(16*a^3*log(sin(d*x + c)) - (8*a^3 - 9*a^2*b + b^3)*log(sin(d*x + c) + 1) - (8*a^3 + 9*a^2*b - b^3)*log(s
in(d*x + c) - 1) - 2*(4*a^3*sin(d*x + c)^2 + (9*a^2*b - b^3)*sin(d*x + c)^3 - 6*a^3 - 6*a*b^2 - (15*a^2*b + b^
3)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.06 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {16 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - {\left (8 \, a^{3} - 9 \, a^{2} b + b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (8 \, a^{3} + 9 \, a^{2} b - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (6 \, a^{3} \sin \left (d x + c\right )^{4} - 9 \, a^{2} b \sin \left (d x + c\right )^{3} + b^{3} \sin \left (d x + c\right )^{3} - 16 \, a^{3} \sin \left (d x + c\right )^{2} + 15 \, a^{2} b \sin \left (d x + c\right ) + b^{3} \sin \left (d x + c\right ) + 12 \, a^{3} + 6 \, a b^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

[In]

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

[Out]

1/16*(16*a^3*log(abs(sin(d*x + c))) - (8*a^3 - 9*a^2*b + b^3)*log(abs(sin(d*x + c) + 1)) - (8*a^3 + 9*a^2*b -
b^3)*log(abs(sin(d*x + c) - 1)) + 2*(6*a^3*sin(d*x + c)^4 - 9*a^2*b*sin(d*x + c)^3 + b^3*sin(d*x + c)^3 - 16*a
^3*sin(d*x + c)^2 + 15*a^2*b*sin(d*x + c) + b^3*sin(d*x + c) + 12*a^3 + 6*a*b^2)/(sin(d*x + c)^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 11.83 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.02 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a^3\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {a^3}{2}-\frac {9\,a^2\,b}{16}+\frac {b^3}{16}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {a^3}{2}+\frac {9\,a^2\,b}{16}-\frac {b^3}{16}\right )}{d}+\frac {\frac {3\,a\,b^2}{4}-{\sin \left (c+d\,x\right )}^3\,\left (\frac {9\,a^2\,b}{8}-\frac {b^3}{8}\right )+\frac {3\,a^3}{4}+\sin \left (c+d\,x\right )\,\left (\frac {15\,a^2\,b}{8}+\frac {b^3}{8}\right )-\frac {a^3\,{\sin \left (c+d\,x\right )}^2}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )} \]

[In]

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

[Out]

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