Integrand size = 21, antiderivative size = 107 \[ \int (a+a \sin (c+d x))^4 \tan ^3(c+d x) \, dx=\frac {16 a^4 \log (1-\sin (c+d x))}{d}+\frac {12 a^4 \sin (c+d x)}{d}+\frac {4 a^4 \sin ^2(c+d x)}{d}+\frac {4 a^4 \sin ^3(c+d x)}{3 d}+\frac {a^4 \sin ^4(c+d x)}{4 d}+\frac {4 a^5}{d (a-a \sin (c+d x))} \] Output:
16*a^4*ln(1-sin(d*x+c))/d+12*a^4*sin(d*x+c)/d+4*a^4*sin(d*x+c)^2/d+4/3*a^4 *sin(d*x+c)^3/d+1/4*a^4*sin(d*x+c)^4/d+4*a^5/d/(a-a*sin(d*x+c))
Time = 0.18 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.71 \[ \int (a+a \sin (c+d x))^4 \tan ^3(c+d x) \, dx=\frac {a^4 \left (192 \log (1-\sin (c+d x))+\frac {48}{1-\sin (c+d x)}+144 \sin (c+d x)+48 \sin ^2(c+d x)+16 \sin ^3(c+d x)+3 \sin ^4(c+d x)\right )}{12 d} \] Input:
Integrate[(a + a*Sin[c + d*x])^4*Tan[c + d*x]^3,x]
Output:
(a^4*(192*Log[1 - Sin[c + d*x]] + 48/(1 - Sin[c + d*x]) + 144*Sin[c + d*x] + 48*Sin[c + d*x]^2 + 16*Sin[c + d*x]^3 + 3*Sin[c + d*x]^4))/(12*d)
Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 99, 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.
| \(\displaystyle \int \tan ^3(c+d x) (a \sin (c+d x)+a)^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^3 (a \sin (c+d x)+a)^4dx\) |
| \(\Big \downarrow \) 3186 |
| \(\displaystyle \frac {\int \frac {a^3 \sin ^3(c+d x) (\sin (c+d x) a+a)^2}{(a-a \sin (c+d x))^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (\frac {4 a^5}{(a-a \sin (c+d x))^2}-\frac {16 a^4}{a-a \sin (c+d x)}+\sin ^3(c+d x) a^3+4 \sin ^2(c+d x) a^3+8 \sin (c+d x) a^3+12 a^3\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {4 a^5}{a-a \sin (c+d x)}+\frac {1}{4} a^4 \sin ^4(c+d x)+\frac {4}{3} a^4 \sin ^3(c+d x)+4 a^4 \sin ^2(c+d x)+12 a^4 \sin (c+d x)+16 a^4 \log (a-a \sin (c+d x))}{d}\) |
Input:
Int[(a + a*Sin[c + d*x])^4*Tan[c + d*x]^3,x]
Output:
(16*a^4*Log[a - a*Sin[c + d*x]] + 12*a^4*Sin[c + d*x] + 4*a^4*Sin[c + d*x] ^2 + (4*a^4*Sin[c + d*x]^3)/3 + (a^4*Sin[c + d*x]^4)/4 + (4*a^5)/(a - a*Si n[c + d*x]))/d
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x) ^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && E qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
Time = 22.78 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.65
| method | result | size |
| derivativedivides | \(\frac {a^{4} \left (\frac {\sin \left (d x +c \right )^{4}}{4}+\frac {4 \sin \left (d x +c \right )^{3}}{3}+4 \sin \left (d x +c \right )^{2}+12 \sin \left (d x +c \right )+16 \ln \left (\sin \left (d x +c \right )-1\right )-\frac {4}{\sin \left (d x +c \right )-1}\right )}{d}\) | \(70\) |
| default | \(\frac {a^{4} \left (\frac {\sin \left (d x +c \right )^{4}}{4}+\frac {4 \sin \left (d x +c \right )^{3}}{3}+4 \sin \left (d x +c \right )^{2}+12 \sin \left (d x +c \right )+16 \ln \left (\sin \left (d x +c \right )-1\right )-\frac {4}{\sin \left (d x +c \right )-1}\right )}{d}\) | \(70\) |
| risch | \(-16 i a^{4} x -\frac {13 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {13 i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {32 i a^{4} c}{d}-\frac {8 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{2} d}+\frac {32 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a^{4} \cos \left (4 d x +4 c \right )}{32 d}-\frac {a^{4} \sin \left (3 d x +3 c \right )}{3 d}-\frac {17 a^{4} \cos \left (2 d x +2 c \right )}{8 d}\) | \(159\) |
| parts | \(\frac {a^{4} \left (\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {a^{4} \left (\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{6}}{2}+\frac {3 \sin \left (d x +c \right )^{4}}{4}+\frac {3 \sin \left (d x +c \right )^{2}}{2}+3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}+\frac {4 a^{4} \left (\frac {\sin \left (d x +c \right )^{5}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{3}}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {6 a^{4} \left (\frac {\sin \left (d x +c \right )^{6}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{4}}{2}+\sin \left (d x +c \right )^{2}+2 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}+\frac {4 a^{4} \left (\frac {\sin \left (d x +c \right )^{7}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )^{5}}{2}+\frac {5 \sin \left (d x +c \right )^{3}}{6}+\frac {5 \sin \left (d x +c \right )}{2}-\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(284\) |
Input:
int((a+a*sin(d*x+c))^4*tan(d*x+c)^3,x,method=_RETURNVERBOSE)
Output:
a^4/d*(1/4*sin(d*x+c)^4+4/3*sin(d*x+c)^3+4*sin(d*x+c)^2+12*sin(d*x+c)+16*l n(sin(d*x+c)-1)-4/(sin(d*x+c)-1))
Time = 0.09 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.08 \[ \int (a+a \sin (c+d x))^4 \tan ^3(c+d x) \, dx=\frac {104 \, a^{4} \cos \left (d x + c\right )^{4} - 976 \, a^{4} \cos \left (d x + c\right )^{2} + 689 \, a^{4} + 1536 \, {\left (a^{4} \sin \left (d x + c\right ) - a^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (24 \, a^{4} \cos \left (d x + c\right )^{4} - 304 \, a^{4} \cos \left (d x + c\right )^{2} - 1073 \, a^{4}\right )} \sin \left (d x + c\right )}{96 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \] Input:
integrate((a+a*sin(d*x+c))^4*tan(d*x+c)^3,x, algorithm="fricas")
Output:
1/96*(104*a^4*cos(d*x + c)^4 - 976*a^4*cos(d*x + c)^2 + 689*a^4 + 1536*(a^ 4*sin(d*x + c) - a^4)*log(-sin(d*x + c) + 1) + (24*a^4*cos(d*x + c)^4 - 30 4*a^4*cos(d*x + c)^2 - 1073*a^4)*sin(d*x + c))/(d*sin(d*x + c) - d)
\[ \int (a+a \sin (c+d x))^4 \tan ^3(c+d x) \, dx=a^{4} \left (\int 4 \sin {\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int 6 \sin ^{2}{\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int 4 \sin ^{3}{\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int \sin ^{4}{\left (c + d x \right )} \tan ^{3}{\left (c + d x \right )}\, dx + \int \tan ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*sin(d*x+c))**4*tan(d*x+c)**3,x)
Output:
a**4*(Integral(4*sin(c + d*x)*tan(c + d*x)**3, x) + Integral(6*sin(c + d*x )**2*tan(c + d*x)**3, x) + Integral(4*sin(c + d*x)**3*tan(c + d*x)**3, x) + Integral(sin(c + d*x)**4*tan(c + d*x)**3, x) + Integral(tan(c + d*x)**3, x))
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.79 \[ \int (a+a \sin (c+d x))^4 \tan ^3(c+d x) \, dx=\frac {3 \, a^{4} \sin \left (d x + c\right )^{4} + 16 \, a^{4} \sin \left (d x + c\right )^{3} + 48 \, a^{4} \sin \left (d x + c\right )^{2} + 192 \, a^{4} \log \left (\sin \left (d x + c\right ) - 1\right ) + 144 \, a^{4} \sin \left (d x + c\right ) - \frac {48 \, a^{4}}{\sin \left (d x + c\right ) - 1}}{12 \, d} \] Input:
integrate((a+a*sin(d*x+c))^4*tan(d*x+c)^3,x, algorithm="maxima")
Output:
1/12*(3*a^4*sin(d*x + c)^4 + 16*a^4*sin(d*x + c)^3 + 48*a^4*sin(d*x + c)^2 + 192*a^4*log(sin(d*x + c) - 1) + 144*a^4*sin(d*x + c) - 48*a^4/(sin(d*x + c) - 1))/d
Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.98 \[ \int (a+a \sin (c+d x))^4 \tan ^3(c+d x) \, dx=\frac {16 \, a^{4} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {4 \, a^{4}}{d {\left (\sin \left (d x + c\right ) - 1\right )}} + \frac {3 \, a^{4} d^{3} \sin \left (d x + c\right )^{4} + 16 \, a^{4} d^{3} \sin \left (d x + c\right )^{3} + 48 \, a^{4} d^{3} \sin \left (d x + c\right )^{2} + 144 \, a^{4} d^{3} \sin \left (d x + c\right )}{12 \, d^{4}} \] Input:
integrate((a+a*sin(d*x+c))^4*tan(d*x+c)^3,x, algorithm="giac")
Output:
16*a^4*log(abs(sin(d*x + c) - 1))/d - 4*a^4/(d*(sin(d*x + c) - 1)) + 1/12* (3*a^4*d^3*sin(d*x + c)^4 + 16*a^4*d^3*sin(d*x + c)^3 + 48*a^4*d^3*sin(d*x + c)^2 + 144*a^4*d^3*sin(d*x + c))/d^4
Time = 18.25 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.99 \[ \int (a+a \sin (c+d x))^4 \tan ^3(c+d x) \, dx=\frac {32\,a^4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d}+\frac {32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {320\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {340\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {424\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-\frac {340\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {320\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+32\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {16\,a^4\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:
int(tan(c + d*x)^3*(a + a*sin(c + d*x))^4,x)
Output:
(32*a^4*log(tan(c/2 + (d*x)/2) - 1))/d + ((320*a^4*tan(c/2 + (d*x)/2)^3)/3 - 32*a^4*tan(c/2 + (d*x)/2)^2 - (340*a^4*tan(c/2 + (d*x)/2)^4)/3 + (424*a ^4*tan(c/2 + (d*x)/2)^5)/3 - (340*a^4*tan(c/2 + (d*x)/2)^6)/3 + (320*a^4*t an(c/2 + (d*x)/2)^7)/3 - 32*a^4*tan(c/2 + (d*x)/2)^8 + 32*a^4*tan(c/2 + (d *x)/2)^9 + 32*a^4*tan(c/2 + (d*x)/2))/(d*(5*tan(c/2 + (d*x)/2)^2 - 2*tan(c /2 + (d*x)/2) - 8*tan(c/2 + (d*x)/2)^3 + 10*tan(c/2 + (d*x)/2)^4 - 12*tan( c/2 + (d*x)/2)^5 + 10*tan(c/2 + (d*x)/2)^6 - 8*tan(c/2 + (d*x)/2)^7 + 5*ta n(c/2 + (d*x)/2)^8 - 2*tan(c/2 + (d*x)/2)^9 + tan(c/2 + (d*x)/2)^10 + 1)) - (16*a^4*log(tan(c/2 + (d*x)/2)^2 + 1))/d
Time = 0.16 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.36 \[ \int (a+a \sin (c+d x))^4 \tan ^3(c+d x) \, dx=\frac {a^{4} \left (-6 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) \sin \left (d x +c \right )^{2}+6 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )-180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}+180 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )+372 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-372 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+3 \sin \left (d x +c \right )^{6}+16 \sin \left (d x +c \right )^{5}+45 \sin \left (d x +c \right )^{4}+128 \sin \left (d x +c \right )^{3}+6 \sin \left (d x +c \right )^{2} \tan \left (d x +c \right )^{2}-90 \sin \left (d x +c \right )^{2}-192 \sin \left (d x +c \right )-6 \tan \left (d x +c \right )^{2}\right )}{12 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+a*sin(d*x+c))^4*tan(d*x+c)^3,x)
Output:
(a**4*( - 6*log(tan(c + d*x)**2 + 1)*sin(c + d*x)**2 + 6*log(tan(c + d*x)* *2 + 1) - 180*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2 + 180*log(tan(( c + d*x)/2)**2 + 1) + 372*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 372* log(tan((c + d*x)/2) - 1) - 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 12*log(tan((c + d*x)/2) + 1) + 3*sin(c + d*x)**6 + 16*sin(c + d*x)**5 + 4 5*sin(c + d*x)**4 + 128*sin(c + d*x)**3 + 6*sin(c + d*x)**2*tan(c + d*x)** 2 - 90*sin(c + d*x)**2 - 192*sin(c + d*x) - 6*tan(c + d*x)**2))/(12*d*(sin (c + d*x)**2 - 1))