Integrand size = 21, antiderivative size = 232 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {(2 a-b) \log (1-\sin (c+d x))}{4 (a+b)^4 d}+\frac {(2 a+b) \log (1+\sin (c+d x))}{4 (a-b)^4 d}-\frac {a \left (a^4+8 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4 d}+\frac {a^3}{2 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac {a^2 \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {\sec ^2(c+d x) \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 d} \] Output:
1/4*(2*a-b)*ln(1-sin(d*x+c))/(a+b)^4/d+1/4*(2*a+b)*ln(1+sin(d*x+c))/(a-b)^ 4/d-a*(a^4+8*a^2*b^2+3*b^4)*ln(a+b*sin(d*x+c))/(a^2-b^2)^4/d+1/2*a^3/(a^2- b^2)^2/d/(a+b*sin(d*x+c))^2+a^2*(a^2+3*b^2)/(a^2-b^2)^3/d/(a+b*sin(d*x+c)) +1/2*sec(d*x+c)^2*(a*(a^2+3*b^2)-b*(3*a^2+b^2)*sin(d*x+c))/(a^2-b^2)^3/d
Time = 1.71 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.84 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {(2 a-b) \log (1-\sin (c+d x))}{(a+b)^4}+\frac {(2 a+b) \log (1+\sin (c+d x))}{(a-b)^4}-\frac {4 a \left (a^4+8 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4}-\frac {1}{(a+b)^3 (-1+\sin (c+d x))}+\frac {1}{(a-b)^3 (1+\sin (c+d x))}+\frac {2 a^3}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {4 a^2 \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))}}{4 d} \] Input:
Integrate[Tan[c + d*x]^3/(a + b*Sin[c + d*x])^3,x]
Output:
(((2*a - b)*Log[1 - Sin[c + d*x]])/(a + b)^4 + ((2*a + b)*Log[1 + Sin[c + d*x]])/(a - b)^4 - (4*a*(a^4 + 8*a^2*b^2 + 3*b^4)*Log[a + b*Sin[c + d*x]]) /(a^2 - b^2)^4 - 1/((a + b)^3*(-1 + Sin[c + d*x])) + 1/((a - b)^3*(1 + Sin [c + d*x])) + (2*a^3)/((a^2 - b^2)^2*(a + b*Sin[c + d*x])^2) + (4*a^2*(a^2 + 3*b^2))/((a^2 - b^2)^3*(a + b*Sin[c + d*x])))/(4*d)
Time = 0.84 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3200, 601, 25, 2160, 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.
| \(\displaystyle \int \frac {\tan ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^3}{(a+b \sin (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3200 |
| \(\displaystyle \frac {\int \frac {b^3 \sin ^3(c+d x)}{(a+b \sin (c+d x))^3 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 601 |
| \(\displaystyle \frac {\frac {b^2 \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int -\frac {-\frac {\left (3 a^2+b^2\right ) \sin ^3(c+d x) b^7}{\left (a^2-b^2\right )^3}-\frac {a \left (7 a^2-3 b^2\right ) \sin ^2(c+d x) b^6}{\left (a^2-b^2\right )^3}+\frac {a^3 \left (3 a^2+b^2\right ) b^4}{\left (a^2-b^2\right )^3}-\frac {a^2 \left (2 a^4-3 b^2 a^2-3 b^4\right ) \sin (c+d x) b^3}{\left (a^2-b^2\right )^3}}{(a+b \sin (c+d x))^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {-\frac {\left (3 a^2+b^2\right ) \sin ^3(c+d x) b^7}{\left (a^2-b^2\right )^3}-\frac {a \left (7 a^2-3 b^2\right ) \sin ^2(c+d x) b^6}{\left (a^2-b^2\right )^3}+\frac {a^3 \left (3 a^2+b^2\right ) b^4}{\left (a^2-b^2\right )^3}-\frac {a^2 \left (2 a^4-3 b^2 a^2-3 b^4\right ) \sin (c+d x) b^3}{\left (a^2-b^2\right )^3}}{(a+b \sin (c+d x))^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}+\frac {b^2 \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 2160 |
| \(\displaystyle \frac {\frac {\int \left (-\frac {2 b^2 a^3}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^3}-\frac {2 b^2 \left (a^2+3 b^2\right ) a^2}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))^2}-\frac {2 b^2 \left (a^4+8 b^2 a^2+3 b^4\right ) a}{\left (a^2-b^2\right )^4 (a+b \sin (c+d x))}+\frac {b^2 (b-2 a)}{2 (a+b)^4 (b-b \sin (c+d x))}+\frac {b^2 (2 a+b)}{2 (a-b)^4 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 b^2}+\frac {b^2 \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^2 \left (a \left (a^2+3 b^2\right )-b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}+\frac {\frac {2 a^2 b^2 \left (a^2+3 b^2\right )}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {2 a b^2 \left (a^4+8 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4}+\frac {a^3 b^2}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {b^2 (2 a-b) \log (b-b \sin (c+d x))}{2 (a+b)^4}+\frac {b^2 (2 a+b) \log (b \sin (c+d x)+b)}{2 (a-b)^4}}{2 b^2}}{d}\) |
Input:
Int[Tan[c + d*x]^3/(a + b*Sin[c + d*x])^3,x]
Output:
((b^2*(a*(a^2 + 3*b^2) - b*(3*a^2 + b^2)*Sin[c + d*x]))/(2*(a^2 - b^2)^3*( b^2 - b^2*Sin[c + d*x]^2)) + (((2*a - b)*b^2*Log[b - b*Sin[c + d*x]])/(2*( a + b)^4) - (2*a*b^2*(a^4 + 8*a^2*b^2 + 3*b^4)*Log[a + b*Sin[c + d*x]])/(a ^2 - b^2)^4 + (b^2*(2*a + b)*Log[b + b*Sin[c + d*x]])/(2*(a - b)^4) + (a^3 *b^2)/((a^2 - b^2)^2*(a + b*Sin[c + d*x])^2) + (2*a^2*b^2*(a^2 + 3*b^2))/( (a^2 - b^2)^3*(a + b*Sin[c + d*x])))/(2*b^2))/d
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) *((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* (2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
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)/(b^2 - x^2)^((p + 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b ^2, 0] && IntegerQ[(p + 1)/2]
Time = 1.98 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.85
| method | result | size |
| derivativedivides | \(\frac {\frac {a^{3}}{2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{2}}-\frac {a \left (a^{4}+8 b^{2} a^{2}+3 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {a^{2} \left (a^{2}+3 b^{2}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3} \left (a +b \sin \left (d x +c \right )\right )}-\frac {1}{4 \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (2 a -b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{4}}+\frac {1}{4 \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (2 a +b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{4}}}{d}\) | \(197\) |
| default | \(\frac {\frac {a^{3}}{2 \left (a +b \right )^{2} \left (a -b \right )^{2} \left (a +b \sin \left (d x +c \right )\right )^{2}}-\frac {a \left (a^{4}+8 b^{2} a^{2}+3 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {a^{2} \left (a^{2}+3 b^{2}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3} \left (a +b \sin \left (d x +c \right )\right )}-\frac {1}{4 \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (2 a -b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{4}}+\frac {1}{4 \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (2 a +b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{4}}}{d}\) | \(197\) |
| risch | \(\text {Expression too large to display}\) | \(1339\) |
Input:
int(tan(d*x+c)^3/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(1/2*a^3/(a+b)^2/(a-b)^2/(a+b*sin(d*x+c))^2-a*(a^4+8*a^2*b^2+3*b^4)/(a +b)^4/(a-b)^4*ln(a+b*sin(d*x+c))+a^2*(a^2+3*b^2)/(a-b)^3/(a+b)^3/(a+b*sin( d*x+c))-1/4/(a+b)^3/(sin(d*x+c)-1)+1/4*(2*a-b)/(a+b)^4*ln(sin(d*x+c)-1)+1/ 4/(a-b)^3/(1+sin(d*x+c))+1/4*(2*a+b)/(a-b)^4*ln(1+sin(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 788 vs. \(2 (224) = 448\).
Time = 0.42 (sec) , antiderivative size = 788, normalized size of antiderivative = 3.40 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(tan(d*x+c)^3/(a+b*sin(d*x+c))^3,x, algorithm="fricas")
Output:
-1/4*(2*a^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 + 2*(3*a^7 + 7*a^5*b^2 - 11* a^3*b^4 + a*b^6)*cos(d*x + c)^2 + 4*((a^5*b^2 + 8*a^3*b^4 + 3*a*b^6)*cos(d *x + c)^4 - 2*(a^6*b + 8*a^4*b^3 + 3*a^2*b^5)*cos(d*x + c)^2*sin(d*x + c) - (a^7 + 9*a^5*b^2 + 11*a^3*b^4 + 3*a*b^6)*cos(d*x + c)^2)*log(b*sin(d*x + c) + a) - ((2*a^5*b^2 + 9*a^4*b^3 + 16*a^3*b^4 + 14*a^2*b^5 + 6*a*b^6 + b ^7)*cos(d*x + c)^4 - 2*(2*a^6*b + 9*a^5*b^2 + 16*a^4*b^3 + 14*a^3*b^4 + 6* a^2*b^5 + a*b^6)*cos(d*x + c)^2*sin(d*x + c) - (2*a^7 + 9*a^6*b + 18*a^5*b ^2 + 23*a^4*b^3 + 22*a^3*b^4 + 15*a^2*b^5 + 6*a*b^6 + b^7)*cos(d*x + c)^2) *log(sin(d*x + c) + 1) - ((2*a^5*b^2 - 9*a^4*b^3 + 16*a^3*b^4 - 14*a^2*b^5 + 6*a*b^6 - b^7)*cos(d*x + c)^4 - 2*(2*a^6*b - 9*a^5*b^2 + 16*a^4*b^3 - 1 4*a^3*b^4 + 6*a^2*b^5 - a*b^6)*cos(d*x + c)^2*sin(d*x + c) - (2*a^7 - 9*a^ 6*b + 18*a^5*b^2 - 23*a^4*b^3 + 22*a^3*b^4 - 15*a^2*b^5 + 6*a*b^6 - b^7)*c os(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 - (2*a^6*b + 7*a^4*b^3 - 8*a^2*b^5 - b^7)*cos(d*x + c)^2)*sin(d*x + c ))/((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*d*cos(d*x + c)^4 - 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)^2*s in(d*x + c) - (a^10 - 3*a^8*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3*a^2*b^8 + b^10 )*d*cos(d*x + c)^2)
\[ \int \frac {\tan ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \] Input:
integrate(tan(d*x+c)**3/(a+b*sin(d*x+c))**3,x)
Output:
Integral(tan(c + d*x)**3/(a + b*sin(c + d*x))**3, x)
Time = 0.05 (sec) , antiderivative size = 441, normalized size of antiderivative = 1.90 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\frac {4 \, {\left (a^{5} + 8 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} - \frac {{\left (2 \, a + b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac {{\left (2 \, a - b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac {2 \, {\left (4 \, a^{5} + 8 \, a^{3} b^{2} - {\left (2 \, a^{4} b + 9 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{3} - {\left (3 \, a^{5} + 10 \, a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{4} b + 11 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )\right )}}{a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6} - {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} \sin \left (d x + c\right )^{4} - 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \left (d x + c\right )^{3} - {\left (a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}\right )} \sin \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} \sin \left (d x + c\right )}}{4 \, d} \] Input:
integrate(tan(d*x+c)^3/(a+b*sin(d*x+c))^3,x, algorithm="maxima")
Output:
-1/4*(4*(a^5 + 8*a^3*b^2 + 3*a*b^4)*log(b*sin(d*x + c) + a)/(a^8 - 4*a^6*b ^2 + 6*a^4*b^4 - 4*a^2*b^6 + b^8) - (2*a + b)*log(sin(d*x + c) + 1)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - (2*a - b)*log(sin(d*x + c) - 1)/(a ^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - 2*(4*a^5 + 8*a^3*b^2 - (2*a^4* b + 9*a^2*b^3 + b^5)*sin(d*x + c)^3 - (3*a^5 + 10*a^3*b^2 - a*b^4)*sin(d*x + c)^2 + (a^4*b + 11*a^2*b^3)*sin(d*x + c))/(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6 - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*sin(d*x + c)^4 - 2*(a^ 7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*sin(d*x + c)^3 - (a^8 - 4*a^6*b^2 + 6 *a^4*b^4 - 4*a^2*b^6 + b^8)*sin(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3* b^5 - a*b^7)*sin(d*x + c)))/d
Time = 0.16 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.56 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {{\left (a^{5} b + 8 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{8} b d - 4 \, a^{6} b^{3} d + 6 \, a^{4} b^{5} d - 4 \, a^{2} b^{7} d + b^{9} d} + \frac {{\left (2 \, a - b\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{4 \, {\left (a^{4} d + 4 \, a^{3} b d + 6 \, a^{2} b^{2} d + 4 \, a b^{3} d + b^{4} d\right )}} + \frac {{\left (2 \, a + b\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{4 \, {\left (a^{4} d - 4 \, a^{3} b d + 6 \, a^{2} b^{2} d - 4 \, a b^{3} d + b^{4} d\right )}} - \frac {4 \, a^{7} + 4 \, a^{5} b^{2} - 8 \, a^{3} b^{4} - {\left (2 \, a^{6} b + 7 \, a^{4} b^{3} - 8 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )^{3} - {\left (3 \, a^{7} + 7 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + a b^{6}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{6} b + 10 \, a^{4} b^{3} - 11 \, a^{2} b^{5}\right )} \sin \left (d x + c\right )}{2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{2} {\left (a + b\right )}^{4} {\left (a - b\right )}^{4} d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}} \] Input:
integrate(tan(d*x+c)^3/(a+b*sin(d*x+c))^3,x, algorithm="giac")
Output:
-(a^5*b + 8*a^3*b^3 + 3*a*b^5)*log(abs(b*sin(d*x + c) + a))/(a^8*b*d - 4*a ^6*b^3*d + 6*a^4*b^5*d - 4*a^2*b^7*d + b^9*d) + 1/4*(2*a - b)*log(abs(-sin (d*x + c) + 1))/(a^4*d + 4*a^3*b*d + 6*a^2*b^2*d + 4*a*b^3*d + b^4*d) + 1/ 4*(2*a + b)*log(abs(-sin(d*x + c) - 1))/(a^4*d - 4*a^3*b*d + 6*a^2*b^2*d - 4*a*b^3*d + b^4*d) - 1/2*(4*a^7 + 4*a^5*b^2 - 8*a^3*b^4 - (2*a^6*b + 7*a^ 4*b^3 - 8*a^2*b^5 - b^7)*sin(d*x + c)^3 - (3*a^7 + 7*a^5*b^2 - 11*a^3*b^4 + a*b^6)*sin(d*x + c)^2 + (a^6*b + 10*a^4*b^3 - 11*a^2*b^5)*sin(d*x + c))/ ((b*sin(d*x + c) + a)^2*(a + b)^4*(a - b)^4*d*(sin(d*x + c) + 1)*(sin(d*x + c) - 1))
Time = 18.63 (sec) , antiderivative size = 690, normalized size of antiderivative = 2.97 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(tan(c + d*x)^3/(a + b*sin(c + d*x))^3,x)
Output:
(log(tan(c/2 + (d*x)/2) - 1)*(2*a - b))/(2*d*(a + b)^4) - ((2*tan(c/2 + (d *x)/2)^6*(7*a*b^4 - a^5 + 6*a^3*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (4*tan(c/2 + (d*x)/2)^4*(9*a*b^4 + a^5 + 2*a^3*b^2))/(a^6 - b^6 + 3*a^2* b^4 - 3*a^4*b^2) - (tan(c/2 + (d*x)/2)^5*(3*a^4*b - 4*b^5 + 13*a^2*b^3))/( a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) + (2*tan(c/2 + (d*x)/2)^2*(7*a*b^4 - a^ 5 + 6*a^3*b^2))/((a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) + (b*tan(c/2 + (d*x) /2)^7*(7*a^4 + 5*a^2*b^2))/(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) - (b*tan(c/ 2 + (d*x)/2)^3*(3*a^4 - 4*b^4 + 13*a^2*b^2))/((a^2 - b^2)*(a^4 + b^4 - 2*a ^2*b^2)) + (a*tan(c/2 + (d*x)/2)*(5*a*b^3 + 7*a^3*b))/((a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)))/(d*(a^2*tan(c/2 + (d*x)/2)^8 - tan(c/2 + (d*x)/2)^4*(2* a^2 + 8*b^2) + 4*b^2*tan(c/2 + (d*x)/2)^2 + 4*b^2*tan(c/2 + (d*x)/2)^6 + a ^2 - 4*a*b*tan(c/2 + (d*x)/2)^3 - 4*a*b*tan(c/2 + (d*x)/2)^5 + 4*a*b*tan(c /2 + (d*x)/2)^7 + 4*a*b*tan(c/2 + (d*x)/2))) + (log(tan(c/2 + (d*x)/2) + 1 )*(2*a + b))/(2*d*(a - b)^4) - (log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2)*(3*a*b^4 + a^5 + 8*a^3*b^2))/(d*(a^8 + b^8 - 4*a^2*b^6 + 6* a^4*b^4 - 4*a^6*b^2))
Time = 0.20 (sec) , antiderivative size = 2813, normalized size of antiderivative = 12.12 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(tan(d*x+c)^3/(a+b*sin(d*x+c))^3,x)
Output:
(4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**5*b**2 - 18*log(tan((c + d *x)/2) - 1)*sin(c + d*x)**4*a**4*b**3 + 32*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3*b**4 - 28*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b **5 + 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**6 - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b**7 + 8*log(tan((c + d*x)/2) - 1)*sin(c + d *x)**3*a**6*b - 36*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a**5*b**2 + 6 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a**4*b**3 - 56*log(tan((c + d* x)/2) - 1)*sin(c + d*x)**3*a**3*b**4 + 24*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a**2*b**5 - 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a*b**6 + 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**7 - 18*log(tan((c + d*x)/2 ) - 1)*sin(c + d*x)**2*a**6*b + 28*log(tan((c + d*x)/2) - 1)*sin(c + d*x)* *2*a**5*b**2 - 10*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4*b**3 - 20 *log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*b**4 + 26*log(tan((c + d*x )/2) - 1)*sin(c + d*x)**2*a**2*b**5 - 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**6 + 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**7 - 8*log (tan((c + d*x)/2) - 1)*sin(c + d*x)*a**6*b + 36*log(tan((c + d*x)/2) - 1)* sin(c + d*x)*a**5*b**2 - 64*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**4*b* *3 + 56*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**3*b**4 - 24*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**2*b**5 + 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a*b**6 - 4*log(tan((c + d*x)/2) - 1)*a**7 + 18*log(tan((c + d*x)/...