Integrand size = 21, antiderivative size = 79 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-3 a b^2 x+\frac {2 b \left (a^2+b^2\right ) \cos (c+d x)}{d}+\frac {a b^2 \cos (c+d x) \sin (c+d x)}{d}+\frac {\sec (c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{d} \]
-3*a*b^2*x+2*b*(a^2+b^2)*cos(d*x+c)/d+a*b^2*cos(d*x+c)*sin(d*x+c)/d+sec(d* x+c)*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^2/d
Time = 0.61 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.86 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {-6 a b^2 (c+d x)+\left (6 a^2 b+3 b^3+b^3 \cos (2 (c+d x))\right ) \sec (c+d x)+2 a \left (a^2+3 b^2\right ) \tan (c+d x)}{2 d} \]
(-6*a*b^2*(c + d*x) + (6*a^2*b + 3*b^3 + b^3*Cos[2*(c + d*x)])*Sec[c + d*x ] + 2*a*(a^2 + 3*b^2)*Tan[c + d*x])/(2*d)
Time = 0.33 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3170, 27, 3042, 3213}
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.
\(\displaystyle \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{\cos (c+d x)^2}dx\) |
\(\Big \downarrow \) 3170 |
\(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d}-\int 2 (a+b \sin (c+d x)) \left (b^2+a \sin (c+d x) b\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d}-2 \int (a+b \sin (c+d x)) \left (b^2+a \sin (c+d x) b\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d}-2 \int (a+b \sin (c+d x)) \left (b^2+a \sin (c+d x) b\right )dx\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d}-2 \left (-\frac {b \left (a^2+b^2\right ) \cos (c+d x)}{d}-\frac {a b^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3}{2} a b^2 x\right )\) |
(Sec[c + d*x]*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^2)/d - 2*((3*a*b^2 *x)/2 - (b*(a^2 + b^2)*Cos[c + d*x])/d - (a*b^2*Cos[c + d*x]*Sin[c + d*x]) /(2*d))
3.5.8.3.1 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 _)])^(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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Time = 1.56 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \(\frac {\cos \left (2 d x +2 c \right ) b^{3}+\left (-6 a \,b^{2} d x +6 a^{2} b +4 b^{3}\right ) \cos \left (d x +c \right )+\left (2 a^{3}+6 a \,b^{2}\right ) \sin \left (d x +c \right )+6 a^{2} b +3 b^{3}}{2 d \cos \left (d x +c \right )}\) | \(85\) |
derivativedivides | \(\frac {a^{3} \tan \left (d x +c \right )+\frac {3 a^{2} b}{\cos \left (d x +c \right )}+3 a \,b^{2} \left (\tan \left (d x +c \right )-d x -c \right )+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )}{d}\) | \(89\) |
default | \(\frac {a^{3} \tan \left (d x +c \right )+\frac {3 a^{2} b}{\cos \left (d x +c \right )}+3 a \,b^{2} \left (\tan \left (d x +c \right )-d x -c \right )+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )}{d}\) | \(89\) |
risch | \(-3 a \,b^{2} x +\frac {b^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {b^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 i a^{3}+6 i a \,b^{2}+6 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{i \left (d x +c \right )}}{d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )}\) | \(103\) |
norman | \(\frac {-\frac {6 a^{2} b +4 b^{3}}{d}+3 a \,b^{2} x -\frac {6 a^{2} b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (9 a^{2} b +4 b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+6 a \,b^{2} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 a \,b^{2} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a \,b^{2} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 a \left (a^{2}+3 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {6 a \left (a^{2}+3 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a \left (a^{2}+3 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (a^{2}+3 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \left (9 a^{2}+2 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\) | \(289\) |
1/2*(cos(2*d*x+2*c)*b^3+(-6*a*b^2*d*x+6*a^2*b+4*b^3)*cos(d*x+c)+(2*a^3+6*a *b^2)*sin(d*x+c)+6*a^2*b+3*b^3)/d/cos(d*x+c)
Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.89 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {3 \, a b^{2} d x \cos \left (d x + c\right ) - b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3} - {\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )} \]
-(3*a*b^2*d*x*cos(d*x + c) - b^3*cos(d*x + c)^2 - 3*a^2*b - b^3 - (a^3 + 3 *a*b^2)*sin(d*x + c))/(d*cos(d*x + c))
\[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \sec ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.89 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {3 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a b^{2} - b^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - a^{3} \tan \left (d x + c\right ) - \frac {3 \, a^{2} b}{\cos \left (d x + c\right )}}{d} \]
-(3*(d*x + c - tan(d*x + c))*a*b^2 - b^3*(1/cos(d*x + c) + cos(d*x + c)) - a^3*tan(d*x + c) - 3*a^2*b/cos(d*x + c))/d
Time = 0.81 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.56 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {3 \, {\left (d x + c\right )} a b^{2} + \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2} b + 2 \, b^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}}{d} \]
-(3*(d*x + c)*a*b^2 + 2*(a^3*tan(1/2*d*x + 1/2*c)^3 + 3*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 3*a^2*b*tan(1/2*d*x + 1/2*c)^2 + a^3*tan(1/2*d*x + 1/2*c) + 3 *a*b^2*tan(1/2*d*x + 1/2*c) + 3*a^2*b + 2*b^3)/(tan(1/2*d*x + 1/2*c)^4 - 1 ))/d
Time = 5.50 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.30 \[ \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^3+6\,a\,b^2\right )+6\,a^2\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^3+6\,a\,b^2\right )+4\,b^3+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-1\right )}-3\,a\,b^2\,x \]