Integrand size = 19, antiderivative size = 69 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {a \cos (c+d x)}{d}+\frac {a \cos ^3(c+d x)}{3 d}-\frac {b \sin (c+d x)}{d}-\frac {b \sin ^3(c+d x)}{3 d} \] Output:
b*arctanh(sin(d*x+c))/d-a*cos(d*x+c)/d+1/3*a*cos(d*x+c)^3/d-b*sin(d*x+c)/d -1/3*b*sin(d*x+c)^3/d
Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {3 a \cos (c+d x)}{4 d}+\frac {a \cos (3 (c+d x))}{12 d}-\frac {b \sin (c+d x)}{d}-\frac {b \sin ^3(c+d x)}{3 d} \] Input:
Integrate[Sin[c + d*x]^3*(a + b*Tan[c + d*x]),x]
Output:
(b*ArcTanh[Sin[c + d*x]])/d - (3*a*Cos[c + d*x])/(4*d) + (a*Cos[3*(c + d*x )])/(12*d) - (b*Sin[c + d*x])/d - (b*Sin[c + d*x]^3)/(3*d)
Time = 0.27 (sec) , antiderivative size = 69, 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.
\(\displaystyle \int \sin ^3(c+d x) (a+b \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^3 (a+b \tan (c+d x))dx\) |
\(\Big \downarrow \) 4000 |
\(\displaystyle \int \left (a \sin ^3(c+d x)+b \sin ^3(c+d x) \tan (c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a \cos ^3(c+d x)}{3 d}-\frac {a \cos (c+d x)}{d}+\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \sin ^3(c+d x)}{3 d}-\frac {b \sin (c+d x)}{d}\) |
Input:
Int[Sin[c + d*x]^3*(a + b*Tan[c + d*x]),x]
Output:
(b*ArcTanh[Sin[c + d*x]])/d - (a*Cos[c + d*x])/d + (a*Cos[c + d*x]^3)/(3*d ) - (b*Sin[c + d*x])/d - (b*Sin[c + d*x]^3)/(3*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 = 1.94 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.87
method | result | size |
derivativedivides | \(\frac {-\frac {a \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+b \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(60\) |
default | \(\frac {-\frac {a \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+b \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(60\) |
risch | \(\frac {5 i {\mathrm e}^{i \left (d x +c \right )} b}{8 d}-\frac {3 \,{\mathrm e}^{i \left (d x +c \right )} a}{8 d}-\frac {5 i {\mathrm e}^{-i \left (d x +c \right )} b}{8 d}-\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )} a}{8 d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a \cos \left (3 d x +3 c \right )}{12 d}+\frac {b \sin \left (3 d x +3 c \right )}{12 d}\) | \(131\) |
Input:
int(sin(d*x+c)^3*(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/3*a*(2+sin(d*x+c)^2)*cos(d*x+c)+b*(-1/3*sin(d*x+c)^3-sin(d*x+c)+ln (sec(d*x+c)+tan(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {2 \, a \cos \left (d x + c\right )^{3} - 6 \, a \cos \left (d x + c\right ) + 3 \, b \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, b \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (b \cos \left (d x + c\right )^{2} - 4 \, b\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(sin(d*x+c)^3*(a+b*tan(d*x+c)),x, algorithm="fricas")
Output:
1/6*(2*a*cos(d*x + c)^3 - 6*a*cos(d*x + c) + 3*b*log(sin(d*x + c) + 1) - 3 *b*log(-sin(d*x + c) + 1) + 2*(b*cos(d*x + c)^2 - 4*b)*sin(d*x + c))/d
\[ \int \sin ^3(c+d x) (a+b \tan (c+d x)) \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right ) \sin ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(sin(d*x+c)**3*(a+b*tan(d*x+c)),x)
Output:
Integral((a + b*tan(c + d*x))*sin(c + d*x)**3, x)
Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.01 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {2 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a - {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} b}{6 \, d} \] Input:
integrate(sin(d*x+c)^3*(a+b*tan(d*x+c)),x, algorithm="maxima")
Output:
1/6*(2*(cos(d*x + c)^3 - 3*cos(d*x + c))*a - (2*sin(d*x + c)^3 - 3*log(sin (d*x + c) + 1) + 3*log(sin(d*x + c) - 1) + 6*sin(d*x + c))*b)/d
Leaf count of result is larger than twice the leaf count of optimal. 4486 vs. \(2 (65) = 130\).
Time = 0.76 (sec) , antiderivative size = 4486, normalized size of antiderivative = 65.01 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x)) \, dx=\text {Too large to display} \] Input:
integrate(sin(d*x+c)^3*(a+b*tan(d*x+c)),x, algorithm="giac")
Output:
-1/6*(3*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*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 + 4*a*tan(1/2*d*x)^6*tan(1/2*c)^6 + 9*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)/(t an(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)^4 - 9*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 + ta n(1/2*d*x)^2 + tan(1/2*c)^2 + 1))*tan(1/2*d*x)^6*tan(1/2*c)^4 - 12*b*tan(1 /2*d*x)^6*tan(1/2*c)^5 + 9*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)^4*tan(1/2*c)^6 - 9*b*lo g(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(...
Time = 0.85 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.26 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {2\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a\,{\cos \left (c+d\,x\right )}^3}{3\,d}-\frac {a\,\cos \left (c+d\,x\right )}{d}-\frac {4\,b\,\sin \left (c+d\,x\right )}{3\,d}+\frac {b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \] Input:
int(sin(c + d*x)^3*(a + b*tan(c + d*x)),x)
Output:
(2*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (a*cos(c + d*x)^3)/ (3*d) - (a*cos(c + d*x))/d - (4*b*sin(c + d*x))/(3*d) + (b*cos(c + d*x)^2* sin(c + d*x))/(3*d)
Time = 0.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.23 \[ \int \sin ^3(c+d x) (a+b \tan (c+d x)) \, dx=\frac {-\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a -2 \cos \left (d x +c \right ) a -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -\sin \left (d x +c \right )^{3} b -3 \sin \left (d x +c \right ) b +2 a}{3 d} \] Input:
int(sin(d*x+c)^3*(a+b*tan(d*x+c)),x)
Output:
( - cos(c + d*x)*sin(c + d*x)**2*a - 2*cos(c + d*x)*a - 3*log(tan((c + d*x )/2) - 1)*b + 3*log(tan((c + d*x)/2) + 1)*b - sin(c + d*x)**3*b - 3*sin(c + d*x)*b + 2*a)/(3*d)