Integrand size = 19, antiderivative size = 133 \[ \int \sin (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {3 a^2 b \text {arctanh}(\sin (c+d x))}{d}-\frac {3 b^3 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {a^3 \cos (c+d x)}{d}+\frac {3 a b^2 \cos (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}-\frac {3 a^2 b \sin (c+d x)}{d}+\frac {3 b^3 \sin (c+d x)}{2 d}+\frac {b^3 \sin (c+d x) \tan ^2(c+d x)}{2 d} \] Output:
3*a^2*b*arctanh(sin(d*x+c))/d-3/2*b^3*arctanh(sin(d*x+c))/d-a^3*cos(d*x+c) /d+3*a*b^2*cos(d*x+c)/d+3*a*b^2*sec(d*x+c)/d-3*a^2*b*sin(d*x+c)/d+3/2*b^3* sin(d*x+c)/d+1/2*b^3*sin(d*x+c)*tan(d*x+c)^2/d
Time = 3.63 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.81 \[ \int \sin (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {-4 a \left (a^2-3 b^2\right ) \cos (c+d x)+b \left (12 a b-12 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {b^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+24 a b \sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )-\frac {b^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+4 \left (-3 a^2+b^2\right ) \sin (c+d x)\right )}{4 d} \] Input:
Integrate[Sin[c + d*x]*(a + b*Tan[c + d*x])^3,x]
Output:
(-4*a*(a^2 - 3*b^2)*Cos[c + d*x] + b*(12*a*b - 12*a^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*b^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12 *a^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 6*b^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + b^2/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + 24*a* b*Sec[c + d*x]*Sin[(c + d*x)/2]^2 - b^2/(Cos[(c + d*x)/2] + Sin[(c + d*x)/ 2])^2 + 4*(-3*a^2 + b^2)*Sin[c + d*x]))/(4*d)
Time = 0.36 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 4000, 2009}
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.
| \(\displaystyle \int \sin (c+d x) (a+b \tan (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x) (a+b \tan (c+d x))^3dx\) |
| \(\Big \downarrow \) 4000 |
| \(\displaystyle \int \left (a^3 \sin (c+d x)+3 a^2 b \sin (c+d x) \tan (c+d x)+3 a b^2 \sin (c+d x) \tan ^2(c+d x)+b^3 \sin (c+d x) \tan ^3(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^3 \cos (c+d x)}{d}+\frac {3 a^2 b \text {arctanh}(\sin (c+d x))}{d}-\frac {3 a^2 b \sin (c+d x)}{d}+\frac {3 a b^2 \cos (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}-\frac {3 b^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 b^3 \sin (c+d x)}{2 d}+\frac {b^3 \sin (c+d x) \tan ^2(c+d x)}{2 d}\) |
Input:
Int[Sin[c + d*x]*(a + b*Tan[c + d*x])^3,x]
Output:
(3*a^2*b*ArcTanh[Sin[c + d*x]])/d - (3*b^3*ArcTanh[Sin[c + d*x]])/(2*d) - (a^3*Cos[c + d*x])/d + (3*a*b^2*Cos[c + d*x])/d + (3*a*b^2*Sec[c + d*x])/d - (3*a^2*b*Sin[c + d*x])/d + (3*b^3*Sin[c + d*x])/(2*d) + (b^3*Sin[c + d* x]*Tan[c + d*x]^2)/(2*d)
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n _.), x_Symbol] :> Int[Expand[Sin[e + f*x]^m*(a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IGtQ[n, 0]
Time = 2.25 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.08
| method | result | size |
| derivativedivides | \(\frac {-a^{3} \cos \left (d x +c \right )+3 a^{2} b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(144\) |
| default | \(\frac {-a^{3} \cos \left (d x +c \right )+3 a^{2} b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )+b^{3} \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(144\) |
| risch | \(\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{2} b}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} b^{3}}{2 d}-\frac {{\mathrm e}^{i \left (d x +c \right )} a^{3}}{2 d}+\frac {3 \,{\mathrm e}^{i \left (d x +c \right )} a \,b^{2}}{2 d}-\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{2} b}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} b^{3}}{2 d}-\frac {{\mathrm e}^{-i \left (d x +c \right )} a^{3}}{2 d}+\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )} a \,b^{2}}{2 d}-\frac {i b^{2} {\mathrm e}^{i \left (d x +c \right )} \left (6 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}+6 i a -b \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{d}-\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{d}+\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\) | \(309\) |
Input:
int(sin(d*x+c)*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(-a^3*cos(d*x+c)+3*a^2*b*(-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))+3*a*b ^2*(sin(d*x+c)^4/cos(d*x+c)+(2+sin(d*x+c)^2)*cos(d*x+c))+b^3*(1/2*sin(d*x+ c)^5/cos(d*x+c)^2+1/2*sin(d*x+c)^3+3/2*sin(d*x+c)-3/2*ln(sec(d*x+c)+tan(d* x+c))))
Time = 0.11 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.08 \[ \int \sin (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {12 \, a b^{2} \cos \left (d x + c\right ) - 4 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (b^{3} - 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(sin(d*x+c)*(a+b*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/4*(12*a*b^2*cos(d*x + c) - 4*(a^3 - 3*a*b^2)*cos(d*x + c)^3 + 3*(2*a^2*b - b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 3*(2*a^2*b - b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(b^3 - 2*(3*a^2*b - b^3)*cos(d*x + c)^2)* sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int \sin (c+d x) (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \sin {\left (c + d x \right )}\, dx \] Input:
integrate(sin(d*x+c)*(a+b*tan(d*x+c))**3,x)
Output:
Integral((a + b*tan(c + d*x))**3*sin(c + d*x), x)
Time = 0.06 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.96 \[ \int \sin (c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, \sin \left (d x + c\right )\right )} - 12 \, a b^{2} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - 6 \, a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )} + 4 \, a^{3} \cos \left (d x + c\right )}{4 \, d} \] Input:
integrate(sin(d*x+c)*(a+b*tan(d*x+c))^3,x, algorithm="maxima")
Output:
-1/4*(b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + 3*log(sin(d*x + c) + 1) - 3*log(sin(d*x + c) - 1) - 4*sin(d*x + c)) - 12*a*b^2*(1/cos(d*x + c) + co s(d*x + c)) - 6*a^2*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2*s in(d*x + c)) + 4*a^3*cos(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 12476 vs. \(2 (127) = 254\).
Time = 7.55 (sec) , antiderivative size = 12476, normalized size of antiderivative = 93.80 \[ \int \sin (c+d x) (a+b \tan (c+d x))^3 \, dx=\text {Too large to display} \] Input:
integrate(sin(d*x+c)*(a+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/4*(9*pi*a*b^2*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - tan(1/2*d*x)^2 - 4*tan(1 /2*d*x)*tan(1/2*c) - tan(1/2*c)^2 + 1)*tan(1/2*d*x)^6*tan(1/2*c)^6 - 6*a^2 *b*log(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + 2*ta n(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 2*tan(1/2*c) + 1)/(tan(1/2*d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2 *c)^2 + 1))*tan(1/2*d*x)^6*tan(1/2*c)^6 + 3*b^3*log(2*(tan(1/2*d*x)^2*tan( 1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan (1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 2*tan(1/2*c) + 1)/(tan(1/2*d *x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(1/2*d*x)^6*ta n(1/2*c)^6 + 6*a^2*b*log(2*(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2 *tan(1/2*c) - 2*tan(1/2*d*x)*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)/(tan(1/2*d*x)^2*tan(1/2*c)^2 + tan(1/ 2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(1/2*d*x)^6*tan(1/2*c)^6 - 3*b^3*log(2*(t an(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) - 2*tan(1/2*d*x)* tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2* c) + 1)/(tan(1/2*d*x)^2*tan(1/2*c)^2 + tan(1/2*d*x)^2 + tan(1/2*c)^2 + 1)) *tan(1/2*d*x)^6*tan(1/2*c)^6 - 9*pi*a*b^2*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - tan(1/2*d*x)^2 - 4*tan(1/2*d*x)*tan(1/2*c) - tan(1/2*c)^2 + 1)*tan(1/2*d *x)^6*tan(1/2*c)^4 - 72*pi*a*b^2*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - tan(1/2 *d*x)^2 - 4*tan(1/2*d*x)*tan(1/2*c) - tan(1/2*c)^2 + 1)*tan(1/2*d*x)^5*...
Time = 3.13 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.45 \[ \int \sin (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^2\,b-3\,b^3\right )+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-12\,a\,b^2+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (12\,a\,b^2-4\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (6\,a^2\,b-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (12\,a^2\,b-2\,b^3\right )+2\,a^3}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (6\,a^2\,b-3\,b^3\right )}{d} \] Input:
int(sin(c + d*x)*(a + b*tan(c + d*x))^3,x)
Output:
(tan(c/2 + (d*x)/2)*(6*a^2*b - 3*b^3) + 2*a^3*tan(c/2 + (d*x)/2)^4 - 12*a* b^2 + tan(c/2 + (d*x)/2)^2*(12*a*b^2 - 4*a^3) + tan(c/2 + (d*x)/2)^5*(6*a^ 2*b - 3*b^3) - tan(c/2 + (d*x)/2)^3*(12*a^2*b - 2*b^3) + 2*a^3)/(d*(tan(c/ 2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 - tan(c/2 + (d*x)/2)^6 - 1)) + (atan h(tan(c/2 + (d*x)/2))*(6*a^2*b - 3*b^3))/d
Time = 0.21 (sec) , antiderivative size = 340, normalized size of antiderivative = 2.56 \[ \int \sin (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}+6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}+2 \cos \left (d x +c \right ) a^{3}-12 \cos \left (d x +c \right ) a \,b^{2}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2} b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{3}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2} b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{3}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3}-6 \sin \left (d x +c \right )^{3} a^{2} b +2 \sin \left (d x +c \right )^{3} b^{3}+2 \sin \left (d x +c \right )^{2} a^{3}-12 \sin \left (d x +c \right )^{2} a \,b^{2}+6 \sin \left (d x +c \right ) a^{2} b -3 \sin \left (d x +c \right ) b^{3}-2 a^{3}+12 a \,b^{2}}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sin(d*x+c)*(a+b*tan(d*x+c))^3,x)
Output:
( - 2*cos(c + d*x)*sin(c + d*x)**2*a**3 + 6*cos(c + d*x)*sin(c + d*x)**2*a *b**2 + 2*cos(c + d*x)*a**3 - 12*cos(c + d*x)*a*b**2 - 6*log(tan((c + d*x) /2) - 1)*sin(c + d*x)**2*a**2*b + 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x) **2*b**3 + 6*log(tan((c + d*x)/2) - 1)*a**2*b - 3*log(tan((c + d*x)/2) - 1 )*b**3 + 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b - 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**3 - 6*log(tan((c + d*x)/2) + 1)*a**2*b + 3*log(tan((c + d*x)/2) + 1)*b**3 - 6*sin(c + d*x)**3*a**2*b + 2*sin(c + d*x)**3*b**3 + 2*sin(c + d*x)**2*a**3 - 12*sin(c + d*x)**2*a*b**2 + 6*sin( c + d*x)*a**2*b - 3*sin(c + d*x)*b**3 - 2*a**3 + 12*a*b**2)/(2*d*(sin(c + d*x)**2 - 1))