Integrand size = 21, antiderivative size = 84 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {2 b \left (a^2-b^2\right ) \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{3 d}+\frac {2 a \left (a^2-b^2\right ) \tan (c+d x)}{3 d} \]
2/3*b*(a^2-b^2)*sec(d*x+c)/d+1/3*sec(d*x+c)^3*(b+a*sin(d*x+c))*(a+b*sin(d* x+c))^2/d+2/3*a*(a^2-b^2)*tan(d*x+c)/d
Time = 0.62 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.62 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {\sec ^3(c+d x) \left (24 a^2 b-4 b^3+\left (-9 a^2 b+15 b^3\right ) \cos (c+d x)-12 b^3 \cos (2 (c+d x))-3 a^2 b \cos (3 (c+d x))+5 b^3 \cos (3 (c+d x))+12 a^3 \sin (c+d x)+18 a b^2 \sin (c+d x)+4 a^3 \sin (3 (c+d x))-6 a b^2 \sin (3 (c+d x))\right )}{24 d} \]
(Sec[c + d*x]^3*(24*a^2*b - 4*b^3 + (-9*a^2*b + 15*b^3)*Cos[c + d*x] - 12* b^3*Cos[2*(c + d*x)] - 3*a^2*b*Cos[3*(c + d*x)] + 5*b^3*Cos[3*(c + d*x)] + 12*a^3*Sin[c + d*x] + 18*a*b^2*Sin[c + d*x] + 4*a^3*Sin[3*(c + d*x)] - 6* a*b^2*Sin[3*(c + d*x)]))/(24*d)
Time = 0.41 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.88, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3170, 27, 3042, 3148, 3042, 4254, 24}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{\cos (c+d x)^4}dx\) |
\(\Big \downarrow \) 3170 |
\(\displaystyle \frac {\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{3 d}-\frac {1}{3} \int -2 \left (a^2-b^2\right ) \sec ^2(c+d x) (a+b \sin (c+d x))dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{3} \left (a^2-b^2\right ) \int \sec ^2(c+d x) (a+b \sin (c+d x))dx+\frac {\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} \left (a^2-b^2\right ) \int \frac {a+b \sin (c+d x)}{\cos (c+d x)^2}dx+\frac {\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 3148 |
\(\displaystyle \frac {2}{3} \left (a^2-b^2\right ) \left (a \int \sec ^2(c+d x)dx+\frac {b \sec (c+d x)}{d}\right )+\frac {\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{3} \left (a^2-b^2\right ) \left (a \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {b \sec (c+d x)}{d}\right )+\frac {\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {2}{3} \left (a^2-b^2\right ) \left (\frac {b \sec (c+d x)}{d}-\frac {a \int 1d(-\tan (c+d x))}{d}\right )+\frac {\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{3 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {2}{3} \left (a^2-b^2\right ) \left (\frac {a \tan (c+d x)}{d}+\frac {b \sec (c+d x)}{d}\right )+\frac {\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{3 d}\) |
(Sec[c + d*x]^3*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^2)/(3*d) + (2*(a ^2 - b^2)*((b*Sec[c + d*x])/d + (a*Tan[c + d*x])/d))/3
3.5.9.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-(g*Cos[e + f*x])^(p + 1))*(a + b*Sin[e + f*x ])^(m - 1)*((b + a*Sin[e + f*x])/(f*g*(p + 1))), x] + Simp[1/(g^2*(p + 1)) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(p + 2) + a*b*(m + p + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g }, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (IntegersQ[2*m, 2* p] || IntegerQ[m])
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Result contains complex when optimal does not.
Time = 1.60 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.44
method | result | size |
risch | \(-\frac {2 \left (9 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-6 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-12 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-2 i a^{3}+3 i a \,b^{2}+3 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{3}}\) | \(121\) |
derivativedivides | \(\frac {-a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {a^{2} b}{\cos \left (d x +c \right )^{3}}+\frac {a \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )}{d}\) | \(122\) |
default | \(\frac {-a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {a^{2} b}{\cos \left (d x +c \right )^{3}}+\frac {a \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )}{d}\) | \(122\) |
parallelrisch | \(\frac {-2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}-6 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} b +\frac {4 \left (a^{3}-6 a \,b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}-2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 a^{2} b +\frac {4 b^{3}}{3}}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\) | \(130\) |
norman | \(\frac {-\frac {6 a^{2} b -4 b^{3}}{3 d}-\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (12 a^{2} b +8 b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (18 a^{2} b +4 b^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a \left (a^{2}+6 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a \left (a^{2}+6 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (7 a^{2}+12 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (7 a^{2}+12 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {6 a^{2} b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a^{2} b \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 b \left (15 a^{2}+8 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\) | \(318\) |
-2/3*(9*I*a*b^2*exp(4*I*(d*x+c))+3*b^3*exp(5*I*(d*x+c))-6*I*a^3*exp(2*I*(d *x+c))-12*a^2*b*exp(3*I*(d*x+c))+2*b^3*exp(3*I*(d*x+c))-2*I*a^3+3*I*a*b^2+ 3*b^3*exp(I*(d*x+c)))/d/(1+exp(2*I*(d*x+c)))^3
Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.92 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {3 \, b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3} - {\left (a^{3} + 3 \, a b^{2} + {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \]
-1/3*(3*b^3*cos(d*x + c)^2 - 3*a^2*b - b^3 - (a^3 + 3*a*b^2 + (2*a^3 - 3*a *b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^3)
\[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \sec ^{4}{\left (c + d x \right )}\, dx \]
Time = 0.19 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.95 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3 \, a b^{2} \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} - \frac {{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} b^{3}}{\cos \left (d x + c\right )^{3}} + \frac {3 \, a^{2} b}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]
1/3*(3*a*b^2*tan(d*x + c)^3 + (tan(d*x + c)^3 + 3*tan(d*x + c))*a^3 - (3*c os(d*x + c)^2 - 1)*b^3/cos(d*x + c)^3 + 3*a^2*b/cos(d*x + c)^3)/d
Time = 0.33 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.52 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {2 \, {\left (3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2} b - 2 \, b^{3}\right )}}{3 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} d} \]
-2/3*(3*a^3*tan(1/2*d*x + 1/2*c)^5 + 9*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 2*a^ 3*tan(1/2*d*x + 1/2*c)^3 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*b^3*tan(1/2 *d*x + 1/2*c)^2 + 3*a^3*tan(1/2*d*x + 1/2*c) + 3*a^2*b - 2*b^3)/((tan(1/2* d*x + 1/2*c)^2 - 1)^3*d)
Time = 4.68 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.96 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {a^2\,b+\frac {a^3\,\sin \left (c+d\,x\right )}{3}+\frac {b^3}{3}-{\cos \left (c+d\,x\right )}^2\,\left (-\frac {2\,\sin \left (c+d\,x\right )\,a^3}{3}+\sin \left (c+d\,x\right )\,a\,b^2+b^3\right )+a\,b^2\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^3} \]