Integrand size = 21, antiderivative size = 189 \[ \int \frac {\tan ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {7 \text {arctanh}(\sin (c+d x))}{128 a^2 d}+\frac {a}{192 d (a-a \sin (c+d x))^3}-\frac {1}{32 d (a-a \sin (c+d x))^2}+\frac {a^3}{80 d (a+a \sin (c+d x))^5}-\frac {5 a^2}{64 d (a+a \sin (c+d x))^4}+\frac {19 a}{96 d (a+a \sin (c+d x))^3}-\frac {1}{4 d (a+a \sin (c+d x))^2}+\frac {21}{256 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac {35}{256 d \left (a^2+a^2 \sin (c+d x)\right )} \] Output:
-7/128*arctanh(sin(d*x+c))/a^2/d+1/192*a/d/(a-a*sin(d*x+c))^3-1/32/d/(a-a* sin(d*x+c))^2+1/80*a^3/d/(a+a*sin(d*x+c))^5-5/64*a^2/d/(a+a*sin(d*x+c))^4+ 19/96*a/d/(a+a*sin(d*x+c))^3-1/4/d/(a+a*sin(d*x+c))^2+21/256/d/(a^2-a^2*si n(d*x+c))+35/256/d/(a^2+a^2*sin(d*x+c))
Time = 2.03 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.59 \[ \int \frac {\tan ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {210 \text {arctanh}(\sin (c+d x))-\frac {2 \left (-144-393 \sin (c+d x)+78 \sin ^2(c+d x)+1039 \sin ^3(c+d x)+560 \sin ^4(c+d x)-815 \sin ^5(c+d x)-750 \sin ^6(c+d x)+105 \sin ^7(c+d x)\right )}{(-1+\sin (c+d x))^3 (1+\sin (c+d x))^5}}{3840 a^2 d} \] Input:
Integrate[Tan[c + d*x]^7/(a + a*Sin[c + d*x])^2,x]
Output:
-1/3840*(210*ArcTanh[Sin[c + d*x]] - (2*(-144 - 393*Sin[c + d*x] + 78*Sin[ c + d*x]^2 + 1039*Sin[c + d*x]^3 + 560*Sin[c + d*x]^4 - 815*Sin[c + d*x]^5 - 750*Sin[c + d*x]^6 + 105*Sin[c + d*x]^7))/((-1 + Sin[c + d*x])^3*(1 + S in[c + d*x])^5))/(a^2*d)
Time = 0.35 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.87, 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.
| \(\displaystyle \int \frac {\tan ^7(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^7}{(a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3186 |
| \(\displaystyle \frac {\int \frac {a^7 \sin ^7(c+d x)}{(a-a \sin (c+d x))^4 (\sin (c+d x) a+a)^6}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (-\frac {a^3}{16 (\sin (c+d x) a+a)^6}+\frac {5 a^2}{16 (\sin (c+d x) a+a)^5}+\frac {a}{64 (a-a \sin (c+d x))^4}-\frac {19 a}{32 (\sin (c+d x) a+a)^4}-\frac {1}{16 (a-a \sin (c+d x))^3}+\frac {1}{2 (\sin (c+d x) a+a)^3}-\frac {7}{128 \left (a^2-a^2 \sin ^2(c+d x)\right ) a}+\frac {21}{256 (a-a \sin (c+d x))^2 a}-\frac {35}{256 (\sin (c+d x) a+a)^2 a}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^3}{80 (a \sin (c+d x)+a)^5}-\frac {7 \text {arctanh}(\sin (c+d x))}{128 a^2}-\frac {5 a^2}{64 (a \sin (c+d x)+a)^4}+\frac {a}{192 (a-a \sin (c+d x))^3}+\frac {19 a}{96 (a \sin (c+d x)+a)^3}-\frac {1}{32 (a-a \sin (c+d x))^2}-\frac {1}{4 (a \sin (c+d x)+a)^2}+\frac {21}{256 a (a-a \sin (c+d x))}+\frac {35}{256 a (a \sin (c+d x)+a)}}{d}\) |
Input:
Int[Tan[c + d*x]^7/(a + a*Sin[c + d*x])^2,x]
Output:
((-7*ArcTanh[Sin[c + d*x]])/(128*a^2) + a/(192*(a - a*Sin[c + d*x])^3) - 1 /(32*(a - a*Sin[c + d*x])^2) + 21/(256*a*(a - a*Sin[c + d*x])) + a^3/(80*( a + a*Sin[c + d*x])^5) - (5*a^2)/(64*(a + a*Sin[c + d*x])^4) + (19*a)/(96* (a + a*Sin[c + d*x])^3) - 1/(4*(a + a*Sin[c + d*x])^2) + 35/(256*a*(a + a* Sin[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 = 3.82 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.67
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{80 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {19}{96 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {35}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {7 \ln \left (1+\sin \left (d x +c \right )\right )}{256}-\frac {1}{192 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {21}{256 \left (\sin \left (d x +c \right )-1\right )}+\frac {7 \ln \left (\sin \left (d x +c \right )-1\right )}{256}}{d \,a^{2}}\) | \(127\) |
| default | \(\frac {\frac {1}{80 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {5}{64 \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {19}{96 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {35}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {7 \ln \left (1+\sin \left (d x +c \right )\right )}{256}-\frac {1}{192 \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {21}{256 \left (\sin \left (d x +c \right )-1\right )}+\frac {7 \ln \left (\sin \left (d x +c \right )-1\right )}{256}}{d \,a^{2}}\) | \(127\) |
| risch | \(\frac {i \left (-2084 i {\mathrm e}^{10 i \left (d x +c \right )}+105 \,{\mathrm e}^{15 i \left (d x +c \right )}-4205 \,{\mathrm e}^{7 i \left (d x +c \right )}-105 \,{\mathrm e}^{i \left (d x +c \right )}+4205 \,{\mathrm e}^{9 i \left (d x +c \right )}-2525 \,{\mathrm e}^{3 i \left (d x +c \right )}-2529 \,{\mathrm e}^{5 i \left (d x +c \right )}+2525 \,{\mathrm e}^{13 i \left (d x +c \right )}+2529 \,{\mathrm e}^{11 i \left (d x +c \right )}-1500 i {\mathrm e}^{2 i \left (d x +c \right )}+4520 i {\mathrm e}^{12 i \left (d x +c \right )}-1500 i {\mathrm e}^{14 i \left (d x +c \right )}-2084 i {\mathrm e}^{6 i \left (d x +c \right )}+16560 i {\mathrm e}^{8 i \left (d x +c \right )}+4520 i {\mathrm e}^{4 i \left (d x +c \right )}\right )}{960 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{6} d \,a^{2}}+\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{128 d \,a^{2}}-\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{128 d \,a^{2}}\) | \(254\) |
Input:
int(tan(d*x+c)^7/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d/a^2*(1/80/(1+sin(d*x+c))^5-5/64/(1+sin(d*x+c))^4+19/96/(1+sin(d*x+c))^ 3-1/4/(1+sin(d*x+c))^2+35/256/(1+sin(d*x+c))-7/256*ln(1+sin(d*x+c))-1/192/ (sin(d*x+c)-1)^3-1/32/(sin(d*x+c)-1)^2-21/256/(sin(d*x+c)-1)+7/256*ln(sin( d*x+c)-1))
Time = 0.11 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.15 \[ \int \frac {\tan ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {1500 \, \cos \left (d x + c\right )^{6} - 3380 \, \cos \left (d x + c\right )^{4} + 2104 \, \cos \left (d x + c\right )^{2} - 105 \, {\left (\cos \left (d x + c\right )^{8} - 2 \, \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \, {\left (\cos \left (d x + c\right )^{8} - 2 \, \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (105 \, \cos \left (d x + c\right )^{6} + 500 \, \cos \left (d x + c\right )^{4} - 276 \, \cos \left (d x + c\right )^{2} + 64\right )} \sin \left (d x + c\right ) - 512}{3840 \, {\left (a^{2} d \cos \left (d x + c\right )^{8} - 2 \, a^{2} d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{6}\right )}} \] Input:
integrate(tan(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="fricas")
Output:
1/3840*(1500*cos(d*x + c)^6 - 3380*cos(d*x + c)^4 + 2104*cos(d*x + c)^2 - 105*(cos(d*x + c)^8 - 2*cos(d*x + c)^6*sin(d*x + c) - 2*cos(d*x + c)^6)*lo g(sin(d*x + c) + 1) + 105*(cos(d*x + c)^8 - 2*cos(d*x + c)^6*sin(d*x + c) - 2*cos(d*x + c)^6)*log(-sin(d*x + c) + 1) - 2*(105*cos(d*x + c)^6 + 500*c os(d*x + c)^4 - 276*cos(d*x + c)^2 + 64)*sin(d*x + c) - 512)/(a^2*d*cos(d* x + c)^8 - 2*a^2*d*cos(d*x + c)^6*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^6)
\[ \int \frac {\tan ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\tan ^{7}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(tan(d*x+c)**7/(a+a*sin(d*x+c))**2,x)
Output:
Integral(tan(c + d*x)**7/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2
Time = 0.04 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.07 \[ \int \frac {\tan ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (105 \, \sin \left (d x + c\right )^{7} - 750 \, \sin \left (d x + c\right )^{6} - 815 \, \sin \left (d x + c\right )^{5} + 560 \, \sin \left (d x + c\right )^{4} + 1039 \, \sin \left (d x + c\right )^{3} + 78 \, \sin \left (d x + c\right )^{2} - 393 \, \sin \left (d x + c\right ) - 144\right )}}{a^{2} \sin \left (d x + c\right )^{8} + 2 \, a^{2} \sin \left (d x + c\right )^{7} - 2 \, a^{2} \sin \left (d x + c\right )^{6} - 6 \, a^{2} \sin \left (d x + c\right )^{5} + 6 \, a^{2} \sin \left (d x + c\right )^{3} + 2 \, a^{2} \sin \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - a^{2}} - \frac {105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{3840 \, d} \] Input:
integrate(tan(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="maxima")
Output:
1/3840*(2*(105*sin(d*x + c)^7 - 750*sin(d*x + c)^6 - 815*sin(d*x + c)^5 + 560*sin(d*x + c)^4 + 1039*sin(d*x + c)^3 + 78*sin(d*x + c)^2 - 393*sin(d*x + c) - 144)/(a^2*sin(d*x + c)^8 + 2*a^2*sin(d*x + c)^7 - 2*a^2*sin(d*x + c)^6 - 6*a^2*sin(d*x + c)^5 + 6*a^2*sin(d*x + c)^3 + 2*a^2*sin(d*x + c)^2 - 2*a^2*sin(d*x + c) - a^2) - 105*log(sin(d*x + c) + 1)/a^2 + 105*log(sin( d*x + c) - 1)/a^2)/d
Time = 0.17 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.71 \[ \int \frac {\tan ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {7 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{256 \, a^{2} d} + \frac {7 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{256 \, a^{2} d} + \frac {105 \, \sin \left (d x + c\right )^{7} - 750 \, \sin \left (d x + c\right )^{6} - 815 \, \sin \left (d x + c\right )^{5} + 560 \, \sin \left (d x + c\right )^{4} + 1039 \, \sin \left (d x + c\right )^{3} + 78 \, \sin \left (d x + c\right )^{2} - 393 \, \sin \left (d x + c\right ) - 144}{1920 \, a^{2} d {\left (\sin \left (d x + c\right ) + 1\right )}^{5} {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} \] Input:
integrate(tan(d*x+c)^7/(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
-7/256*log(abs(sin(d*x + c) + 1))/(a^2*d) + 7/256*log(abs(sin(d*x + c) - 1 ))/(a^2*d) + 1/1920*(105*sin(d*x + c)^7 - 750*sin(d*x + c)^6 - 815*sin(d*x + c)^5 + 560*sin(d*x + c)^4 + 1039*sin(d*x + c)^3 + 78*sin(d*x + c)^2 - 3 93*sin(d*x + c) - 144)/(a^2*d*(sin(d*x + c) + 1)^5*(sin(d*x + c) - 1)^3)
Time = 20.59 (sec) , antiderivative size = 444, normalized size of antiderivative = 2.35 \[ \int \frac {\tan ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(tan(c + d*x)^7/(a + a*sin(c + d*x))^2,x)
Output:
((7*tan(c/2 + (d*x)/2))/64 + (7*tan(c/2 + (d*x)/2)^2)/16 + (7*tan(c/2 + (d *x)/2)^3)/192 - (49*tan(c/2 + (d*x)/2)^4)/24 - (693*tan(c/2 + (d*x)/2)^5)/ 320 + (791*tan(c/2 + (d*x)/2)^6)/240 + (1207*tan(c/2 + (d*x)/2)^7)/192 + ( 123*tan(c/2 + (d*x)/2)^8)/4 + (1207*tan(c/2 + (d*x)/2)^9)/192 + (791*tan(c /2 + (d*x)/2)^10)/240 - (693*tan(c/2 + (d*x)/2)^11)/320 - (49*tan(c/2 + (d *x)/2)^12)/24 + (7*tan(c/2 + (d*x)/2)^13)/192 + (7*tan(c/2 + (d*x)/2)^14)/ 16 + (7*tan(c/2 + (d*x)/2)^15)/64)/(d*(36*a^2*tan(c/2 + (d*x)/2)^5 - 20*a^ 2*tan(c/2 + (d*x)/2)^4 - 20*a^2*tan(c/2 + (d*x)/2)^3 + 64*a^2*tan(c/2 + (d *x)/2)^6 - 20*a^2*tan(c/2 + (d*x)/2)^7 - 90*a^2*tan(c/2 + (d*x)/2)^8 - 20* a^2*tan(c/2 + (d*x)/2)^9 + 64*a^2*tan(c/2 + (d*x)/2)^10 + 36*a^2*tan(c/2 + (d*x)/2)^11 - 20*a^2*tan(c/2 + (d*x)/2)^12 - 20*a^2*tan(c/2 + (d*x)/2)^13 + 4*a^2*tan(c/2 + (d*x)/2)^15 + a^2*tan(c/2 + (d*x)/2)^16 + a^2 + 4*a^2*t an(c/2 + (d*x)/2))) - (7*atanh(tan(c/2 + (d*x)/2)))/(64*a^2*d)
Time = 0.20 (sec) , antiderivative size = 482, normalized size of antiderivative = 2.55 \[ \int \frac {\tan ^7(c+d x)}{(a+a \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(tan(d*x+c)^7/(a+a*sin(d*x+c))^2,x)
Output:
(210*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**8 + 420*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**7 - 420*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6 - 126 0*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5 + 1260*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 + 420*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 420*l og(tan((c + d*x)/2) - 1)*sin(c + d*x) - 210*log(tan((c + d*x)/2) - 1) - 21 0*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**8 - 420*log(tan((c + d*x)/2) + 1 )*sin(c + d*x)**7 + 420*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6 + 1260*l og(tan((c + d*x)/2) + 1)*sin(c + d*x)**5 - 1260*log(tan((c + d*x)/2) + 1)* sin(c + d*x)**3 - 420*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 420*log( tan((c + d*x)/2) + 1)*sin(c + d*x) + 210*log(tan((c + d*x)/2) + 1) - 393*s in(c + d*x)**8 - 576*sin(c + d*x)**7 - 714*sin(c + d*x)**6 + 728*sin(c + d *x)**5 + 1120*sin(c + d*x)**4 - 280*sin(c + d*x)**3 - 630*sin(c + d*x)**2 + 105)/(3840*a**2*d*(sin(c + d*x)**8 + 2*sin(c + d*x)**7 - 2*sin(c + d*x)* *6 - 6*sin(c + d*x)**5 + 6*sin(c + d*x)**3 + 2*sin(c + d*x)**2 - 2*sin(c + d*x) - 1))