Integrand size = 21, antiderivative size = 277 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {a (4 a+b) \log (1-\sin (c+d x))}{8 (a+b)^4 d}-\frac {a (4 a-b) \log (1+\sin (c+d x))}{8 (a-b)^4 d}+\frac {a^4 \left (a^2+5 b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4 d}-\frac {4 a^3-5 a^2 b+a b^2+2 b^3}{8 (a-b)^2 (a+b)^3 d (1-\sin (c+d x))}-\frac {4 a^3+5 a^2 b+a b^2-2 b^3}{8 (a-b)^3 (a+b)^2 d (1+\sin (c+d x))}-\frac {a^5}{\left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {\sec ^4(c+d x) \left (a^2+b^2-2 a b \sin (c+d x)\right )}{4 \left (a^2-b^2\right )^2 d} \] Output:
-1/8*a*(4*a+b)*ln(1-sin(d*x+c))/(a+b)^4/d-1/8*a*(4*a-b)*ln(1+sin(d*x+c))/( a-b)^4/d+a^4*(a^2+5*b^2)*ln(a+b*sin(d*x+c))/(a^2-b^2)^4/d-1/8*(4*a^3-5*a^2 *b+a*b^2+2*b^3)/(a-b)^2/(a+b)^3/d/(1-sin(d*x+c))-1/8*(4*a^3+5*a^2*b+a*b^2- 2*b^3)/(a-b)^3/(a+b)^2/d/(1+sin(d*x+c))-a^5/(a^2-b^2)^3/d/(a+b*sin(d*x+c)) +1/4*sec(d*x+c)^4*(a^2+b^2-2*a*b*sin(d*x+c))/(a^2-b^2)^2/d
Time = 4.66 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.74 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-\frac {2 a (4 a+b) \log (1-\sin (c+d x))}{(a+b)^4}-\frac {2 a (4 a-b) \log (1+\sin (c+d x))}{(a-b)^4}+\frac {16 a^4 \left (a^2+5 b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4}+\frac {1}{(a+b)^2 (-1+\sin (c+d x))^2}+\frac {7 a+3 b}{(a+b)^3 (-1+\sin (c+d x))}+\frac {1}{(a-b)^2 (1+\sin (c+d x))^2}+\frac {-7 a+3 b}{(a-b)^3 (1+\sin (c+d x))}-\frac {16 a^5}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))}}{16 d} \] Input:
Integrate[Tan[c + d*x]^5/(a + b*Sin[c + d*x])^2,x]
Output:
((-2*a*(4*a + b)*Log[1 - Sin[c + d*x]])/(a + b)^4 - (2*a*(4*a - b)*Log[1 + Sin[c + d*x]])/(a - b)^4 + (16*a^4*(a^2 + 5*b^2)*Log[a + b*Sin[c + d*x]]) /(a^2 - b^2)^4 + 1/((a + b)^2*(-1 + Sin[c + d*x])^2) + (7*a + 3*b)/((a + b )^3*(-1 + Sin[c + d*x])) + 1/((a - b)^2*(1 + Sin[c + d*x])^2) + (-7*a + 3* b)/((a - b)^3*(1 + Sin[c + d*x])) - (16*a^5)/((a^2 - b^2)^3*(a + b*Sin[c + d*x])))/(16*d)
Time = 1.06 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3200, 601, 27, 2178, 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.
| \(\displaystyle \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^5}{(a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3200 |
| \(\displaystyle \frac {\int \frac {b^5 \sin ^5(c+d x)}{(a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 601 |
| \(\displaystyle \frac {\frac {b^4 \left (a^2-2 a b \sin (c+d x)+b^2\right )}{4 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int -\frac {2 \left (-\frac {3 a \sin ^2(c+d x) b^8}{\left (a^2-b^2\right )^2}+\frac {a^3 b^6}{\left (a^2-b^2\right )^2}-2 \sin ^3(c+d x) b^5-\frac {2 a^4 \sin (c+d x) b^5}{\left (a^2-b^2\right )^2}\right )}{(a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {-\frac {3 a \sin ^2(c+d x) b^8}{\left (a^2-b^2\right )^2}+\frac {a^3 b^6}{\left (a^2-b^2\right )^2}-2 \sin ^3(c+d x) b^5-\frac {2 a^4 \sin (c+d x) b^5}{\left (a^2-b^2\right )^2}}{(a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{2 b^2}+\frac {b^4 \left (a^2-2 a b \sin (c+d x)+b^2\right )}{4 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 2178 |
| \(\displaystyle \frac {\frac {\frac {\int -\frac {-\frac {a \left (9 a^2-b^2\right ) \sin ^2(c+d x) b^8}{\left (a^2-b^2\right )^3}+\frac {a^3 \left (7 a^2+b^2\right ) b^6}{\left (a^2-b^2\right )^3}-\frac {2 a^2 \left (2 a^2+b^2\right ) \sin (c+d x) b^5}{\left (a^2-b^2\right )^2}}{(a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^4 \left (2 \left (2 a^4+3 a^2 b^2-b^4\right )-a b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{2 b^2}+\frac {b^4 \left (a^2-2 a b \sin (c+d x)+b^2\right )}{4 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {-\frac {\int \frac {-\frac {a \left (9 a^2-b^2\right ) \sin ^2(c+d x) b^8}{\left (a^2-b^2\right )^3}+\frac {a^3 \left (7 a^2+b^2\right ) b^6}{\left (a^2-b^2\right )^3}-\frac {2 a^2 \left (2 a^2+b^2\right ) \sin (c+d x) b^5}{\left (a^2-b^2\right )^2}}{(a+b \sin (c+d x))^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^4 \left (2 \left (2 a^4+3 a^2 b^2-b^4\right )-a b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{2 b^2}+\frac {b^4 \left (a^2-2 a b \sin (c+d x)+b^2\right )}{4 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 2160 |
| \(\displaystyle \frac {\frac {-\frac {\int \left (-\frac {4 b^4 a^5}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))^2}-\frac {4 b^4 \left (a^2+5 b^2\right ) a^4}{(a-b)^4 (a+b)^4 (a+b \sin (c+d x))}-\frac {b^4 (4 a+b) a}{2 (a+b)^4 (b-b \sin (c+d x))}+\frac {(4 a-b) b^4 a}{2 (a-b)^4 (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 b^2}-\frac {b^4 \left (2 \left (2 a^4+3 a^2 b^2-b^4\right )-a b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{2 b^2}+\frac {b^4 \left (a^2-2 a b \sin (c+d x)+b^2\right )}{4 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^4 \left (a^2-2 a b \sin (c+d x)+b^2\right )}{4 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}+\frac {-\frac {b^4 \left (2 \left (2 a^4+3 a^2 b^2-b^4\right )-a b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^3 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\frac {4 a^5 b^4}{\left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {4 a^4 b^4 \left (a^2+5 b^2\right ) \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^4}+\frac {a b^4 (4 a+b) \log (b-b \sin (c+d x))}{2 (a+b)^4}+\frac {a b^4 (4 a-b) \log (b \sin (c+d x)+b)}{2 (a-b)^4}}{2 b^2}}{2 b^2}}{d}\) |
Input:
Int[Tan[c + d*x]^5/(a + b*Sin[c + d*x])^2,x]
Output:
((b^4*(a^2 + b^2 - 2*a*b*Sin[c + d*x]))/(4*(a^2 - b^2)^2*(b^2 - b^2*Sin[c + d*x]^2)^2) + (-1/2*(b^4*(2*(2*a^4 + 3*a^2*b^2 - b^4) - a*b*(9*a^2 - b^2) *Sin[c + d*x]))/((a^2 - b^2)^3*(b^2 - b^2*Sin[c + d*x]^2)) - ((a*b^4*(4*a + b)*Log[b - b*Sin[c + d*x]])/(2*(a + b)^4) - (4*a^4*b^4*(a^2 + 5*b^2)*Log [a + b*Sin[c + d*x]])/(a^2 - b^2)^4 + (a*(4*a - b)*b^4*Log[b + b*Sin[c + d *x]])/(2*(a - b)^4) + (4*a^5*b^4)/((a^2 - b^2)^3*(a + b*Sin[c + d*x])))/(2 *b^2))/(2*b^2))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x^2, x], R = Coeff[Po lynomialRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 0], S = Coeff[Polynomia lRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*S - b*R*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(d + e*x )^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*b*(p + 1)*Qx)/(d + e*x)^m + (b*R*( 2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x ] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
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 = 2.03 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.74
| method | result | size |
| derivativedivides | \(\frac {-\frac {a^{5}}{\left (a +b \right )^{3} \left (a -b \right )^{3} \left (a +b \sin \left (d x +c \right )\right )}+\frac {a^{4} \left (a^{2}+5 b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {1}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-3 b +7 a}{16 \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )}-\frac {a \left (4 a -b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{8 \left (a -b \right )^{4}}+\frac {1}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-3 b -7 a}{16 \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )}-\frac {a \left (4 a +b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{8 \left (a +b \right )^{4}}}{d}\) | \(205\) |
| default | \(\frac {-\frac {a^{5}}{\left (a +b \right )^{3} \left (a -b \right )^{3} \left (a +b \sin \left (d x +c \right )\right )}+\frac {a^{4} \left (a^{2}+5 b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4}}+\frac {1}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-3 b +7 a}{16 \left (a -b \right )^{3} \left (1+\sin \left (d x +c \right )\right )}-\frac {a \left (4 a -b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{8 \left (a -b \right )^{4}}+\frac {1}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-3 b -7 a}{16 \left (a +b \right )^{3} \left (\sin \left (d x +c \right )-1\right )}-\frac {a \left (4 a +b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{8 \left (a +b \right )^{4}}}{d}\) | \(205\) |
| risch | \(\text {Expression too large to display}\) | \(1265\) |
Input:
int(tan(d*x+c)^5/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-a^5/(a+b)^3/(a-b)^3/(a+b*sin(d*x+c))+a^4*(a^2+5*b^2)/(a+b)^4/(a-b)^4 *ln(a+b*sin(d*x+c))+1/16/(a-b)^2/(1+sin(d*x+c))^2-1/16*(-3*b+7*a)/(a-b)^3/ (1+sin(d*x+c))-1/8*a*(4*a-b)/(a-b)^4*ln(1+sin(d*x+c))+1/16/(a+b)^2/(sin(d* x+c)-1)^2-1/16*(-3*b-7*a)/(a+b)^3/(sin(d*x+c)-1)-1/8*a*(4*a+b)/(a+b)^4*ln( sin(d*x+c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 555 vs. \(2 (269) = 538\).
Time = 0.34 (sec) , antiderivative size = 555, normalized size of antiderivative = 2.00 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 \, a^{7} - 6 \, a^{5} b^{2} + 6 \, a^{3} b^{4} - 2 \, a b^{6} - 2 \, {\left (4 \, a^{7} + 5 \, a^{5} b^{2} - 10 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{7} - 9 \, a^{5} b^{2} + 6 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left ({\left (a^{6} b + 5 \, a^{4} b^{3}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (a^{7} + 5 \, a^{5} b^{2}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left ({\left (4 \, a^{6} b + 15 \, a^{5} b^{2} + 20 \, a^{4} b^{3} + 10 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (4 \, a^{7} + 15 \, a^{6} b + 20 \, a^{5} b^{2} + 10 \, a^{4} b^{3} - a^{2} b^{5}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (4 \, a^{6} b - 15 \, a^{5} b^{2} + 20 \, a^{4} b^{3} - 10 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (4 \, a^{7} - 15 \, a^{6} b + 20 \, a^{5} b^{2} - 10 \, a^{4} b^{3} + a^{2} b^{5}\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7} - {\left (5 \, a^{6} b - 12 \, a^{4} b^{3} + 9 \, a^{2} b^{5} - 2 \, b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{8 \, {\left ({\left (a^{8} b - 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - 4 \, a^{2} b^{7} + b^{9}\right )} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (a^{9} - 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} - 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right )^{4}\right )}} \] Input:
integrate(tan(d*x+c)^5/(a+b*sin(d*x+c))^2,x, algorithm="fricas")
Output:
1/8*(2*a^7 - 6*a^5*b^2 + 6*a^3*b^4 - 2*a*b^6 - 2*(4*a^7 + 5*a^5*b^2 - 10*a ^3*b^4 + a*b^6)*cos(d*x + c)^4 - 2*(4*a^7 - 9*a^5*b^2 + 6*a^3*b^4 - a*b^6) *cos(d*x + c)^2 + 8*((a^6*b + 5*a^4*b^3)*cos(d*x + c)^4*sin(d*x + c) + (a^ 7 + 5*a^5*b^2)*cos(d*x + c)^4)*log(b*sin(d*x + c) + a) - ((4*a^6*b + 15*a^ 5*b^2 + 20*a^4*b^3 + 10*a^3*b^4 - a*b^6)*cos(d*x + c)^4*sin(d*x + c) + (4* a^7 + 15*a^6*b + 20*a^5*b^2 + 10*a^4*b^3 - a^2*b^5)*cos(d*x + c)^4)*log(si n(d*x + c) + 1) - ((4*a^6*b - 15*a^5*b^2 + 20*a^4*b^3 - 10*a^3*b^4 + a*b^6 )*cos(d*x + c)^4*sin(d*x + c) + (4*a^7 - 15*a^6*b + 20*a^5*b^2 - 10*a^4*b^ 3 + a^2*b^5)*cos(d*x + c)^4)*log(-sin(d*x + c) + 1) - 2*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 - (5*a^6*b - 12*a^4*b^3 + 9*a^2*b^5 - 2*b^7)*cos(d*x + c)^2)*sin(d*x + c))/((a^8*b - 4*a^6*b^3 + 6*a^4*b^5 - 4*a^2*b^7 + b^9)*d*c os(d*x + c)^4*sin(d*x + c) + (a^9 - 4*a^7*b^2 + 6*a^5*b^4 - 4*a^3*b^6 + a* b^8)*d*cos(d*x + c)^4)
\[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\tan ^{5}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(tan(d*x+c)**5/(a+b*sin(d*x+c))**2,x)
Output:
Integral(tan(c + d*x)**5/(a + b*sin(c + d*x))**2, x)
Time = 0.04 (sec) , antiderivative size = 505, normalized size of antiderivative = 1.82 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {8 \, {\left (a^{6} + 5 \, a^{4} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{8} - 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} - 4 \, a^{2} b^{6} + b^{8}} - \frac {{\left (4 \, a^{2} - a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac {{\left (4 \, a^{2} + a b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac {2 \, {\left (7 \, a^{5} + 6 \, a^{3} b^{2} - a b^{4} + {\left (4 \, a^{5} + 9 \, a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )^{4} + {\left (5 \, a^{4} b - 7 \, a^{2} b^{3} + 2 \, b^{5}\right )} \sin \left (d x + c\right )^{3} - {\left (12 \, a^{5} + 13 \, a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )^{2} - {\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )\right )}}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6} + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )^{5} + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \sin \left (d x + c\right )^{4} - 2 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )^{3} - 2 \, {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} \sin \left (d x + c\right )^{2} + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )}}{8 \, d} \] Input:
integrate(tan(d*x+c)^5/(a+b*sin(d*x+c))^2,x, algorithm="maxima")
Output:
1/8*(8*(a^6 + 5*a^4*b^2)*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) - (4*a^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) - (4*a^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*(7*a^5 + 6*a^3*b^2 - a*b^4 + (4 *a^5 + 9*a^3*b^2 - a*b^4)*sin(d*x + c)^4 + (5*a^4*b - 7*a^2*b^3 + 2*b^5)*s in(d*x + c)^3 - (12*a^5 + 13*a^3*b^2 - a*b^4)*sin(d*x + c)^2 - (4*a^4*b - 5*a^2*b^3 + b^5)*sin(d*x + c))/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6 + (a^6 *b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*sin(d*x + c)^5 + (a^7 - 3*a^5*b^2 + 3*a^ 3*b^4 - a*b^6)*sin(d*x + c)^4 - 2*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*si n(d*x + c)^3 - 2*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*sin(d*x + c)^2 + (a ^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*sin(d*x + c)))/d
Time = 0.17 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.48 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {{\left (a^{6} b + 5 \, a^{4} b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{8} b d - 4 \, a^{6} b^{3} d + 6 \, a^{4} b^{5} d - 4 \, a^{2} b^{7} d + b^{9} d} - \frac {{\left (4 \, a^{2} + a b\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{8 \, {\left (a^{4} d + 4 \, a^{3} b d + 6 \, a^{2} b^{2} d + 4 \, a b^{3} d + b^{4} d\right )}} - \frac {{\left (4 \, a^{2} - a b\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{8 \, {\left (a^{4} d - 4 \, a^{3} b d + 6 \, a^{2} b^{2} d - 4 \, a b^{3} d + b^{4} d\right )}} - \frac {7 \, a^{7} - a^{5} b^{2} - 7 \, a^{3} b^{4} + a b^{6} + {\left (4 \, a^{7} + 5 \, a^{5} b^{2} - 10 \, a^{3} b^{4} + a b^{6}\right )} \sin \left (d x + c\right )^{4} + {\left (5 \, a^{6} b - 12 \, a^{4} b^{3} + 9 \, a^{2} b^{5} - 2 \, b^{7}\right )} \sin \left (d x + c\right )^{3} - {\left (12 \, a^{7} + a^{5} b^{2} - 14 \, a^{3} b^{4} + a b^{6}\right )} \sin \left (d x + c\right )^{2} - {\left (4 \, a^{6} b - 9 \, a^{4} b^{3} + 6 \, a^{2} b^{5} - b^{7}\right )} \sin \left (d x + c\right )}{4 \, {\left (b \sin \left (d x + c\right ) + a\right )} {\left (a + b\right )}^{4} {\left (a - b\right )}^{4} d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(tan(d*x+c)^5/(a+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
(a^6*b + 5*a^4*b^3)*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) - 1/8*(4*a^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) - 1/8*(4*a^ 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*(7*a^7 - a^5*b^2 - 7*a^3*b^4 + a*b^6 + (4*a^7 + 5*a ^5*b^2 - 10*a^3*b^4 + a*b^6)*sin(d*x + c)^4 + (5*a^6*b - 12*a^4*b^3 + 9*a^ 2*b^5 - 2*b^7)*sin(d*x + c)^3 - (12*a^7 + a^5*b^2 - 14*a^3*b^4 + a*b^6)*si n(d*x + c)^2 - (4*a^6*b - 9*a^4*b^3 + 6*a^2*b^5 - b^7)*sin(d*x + c))/((b*s in(d*x + c) + a)*(a + b)^4*(a - b)^4*d*(sin(d*x + c) + 1)^2*(sin(d*x + c) - 1)^2)
Time = 18.83 (sec) , antiderivative size = 755, normalized size of antiderivative = 2.73 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(tan(c + d*x)^5/(a + b*sin(c + d*x))^2,x)
Output:
(log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2)*(a^6 + 5*a^4*b^2 ))/(d*(a^8 + b^8 - 4*a^2*b^6 + 6*a^4*b^4 - 4*a^6*b^2)) - (log(tan(c/2 + (d *x)/2) - 1)*(1/(a + b)^2 - (7*b)/(4*(a + b)^3) + (3*b^2)/(4*(a + b)^4)))/d - (log(tan(c/2 + (d*x)/2) + 1)*((3*b^2)/(4*(a - b)^4) + (7*b)/(4*(a - b)^ 3) + 1/(a - b)^2))/d - ((tan(c/2 + (d*x)/2)^2*(a*b^2 + 2*a^3))/(a^4 + b^4 - 2*a^2*b^2) + (3*tan(c/2 + (d*x)/2)^4*(a*b^2 - 2*a^3))/(a^4 + b^4 - 2*a^2 *b^2) + (3*tan(c/2 + (d*x)/2)^6*(a*b^2 - 2*a^3))/(a^4 + b^4 - 2*a^2*b^2) + (tan(c/2 + (d*x)/2)^8*(a*b^2 + 2*a^3))/(a^4 + b^4 - 2*a^2*b^2) - (b*tan(c /2 + (d*x)/2)^9*(11*a^4 + a^2*b^2))/(2*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2) ) + (8*b*tan(c/2 + (d*x)/2)^3*(2*a^4 + a^2*b^2))/((a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) + (8*b*tan(c/2 + (d*x)/2)^7*(2*a^4 + a^2*b^2))/((a^2 - b^2)*( a^4 + b^4 - 2*a^2*b^2)) - (b*tan(c/2 + (d*x)/2)^5*(13*a^4 - 8*b^4 + 31*a^2 *b^2))/((a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (b*tan(c/2 + (d*x)/2)*(11*a ^4 + a^2*b^2))/(2*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)))/(d*(a + 2*b*tan(c/ 2 + (d*x)/2) - 3*a*tan(c/2 + (d*x)/2)^2 + 2*a*tan(c/2 + (d*x)/2)^4 + 2*a*t an(c/2 + (d*x)/2)^6 - 3*a*tan(c/2 + (d*x)/2)^8 + a*tan(c/2 + (d*x)/2)^10 - 8*b*tan(c/2 + (d*x)/2)^3 + 12*b*tan(c/2 + (d*x)/2)^5 - 8*b*tan(c/2 + (d*x )/2)^7 + 2*b*tan(c/2 + (d*x)/2)^9))
Time = 0.20 (sec) , antiderivative size = 2643, normalized size of antiderivative = 9.54 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(tan(d*x+c)^5/(a+b*sin(d*x+c))^2,x)
Output:
( - 4*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*a**6*b + 15*log(tan((c + d *x)/2) - 1)*sin(c + d*x)**5*a**5*b**2 - 20*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*a**4*b**3 + 10*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*a**3*b **4 - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*a*b**6 - 4*log(tan((c + d* x)/2) - 1)*sin(c + d*x)**4*a**7 + 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x )**4*a**6*b - 20*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**5*b**2 + 10* log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**4*b**3 - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b**5 + 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x) **3*a**6*b - 30*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a**5*b**2 + 40*l og(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a**4*b**3 - 20*log(tan((c + d*x)/ 2) - 1)*sin(c + d*x)**3*a**3*b**4 + 2*log(tan((c + d*x)/2) - 1)*sin(c + d* x)**3*a*b**6 + 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**7 - 30*log(t an((c + d*x)/2) - 1)*sin(c + d*x)**2*a**6*b + 40*log(tan((c + d*x)/2) - 1) *sin(c + d*x)**2*a**5*b**2 - 20*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2* a**4*b**3 + 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**5 - 4*log( tan((c + d*x)/2) - 1)*sin(c + d*x)*a**6*b + 15*log(tan((c + d*x)/2) - 1)*s in(c + d*x)*a**5*b**2 - 20*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**4*b** 3 + 10*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**3*b**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 + 15*log( tan((c + d*x)/2) - 1)*a**6*b - 20*log(tan((c + d*x)/2) - 1)*a**5*b**2 +...