Integrand size = 23, antiderivative size = 172 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{5/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{5/2} d}-\frac {2 a^2 \tan (c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {4 a^2 \left (a^2+4 b^2\right )}{3 b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]
arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(5/2)/d+arctanh((a+b *tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(5/2)/d-4/3*a^2*(a^2+4*b^2)/b^2/ (a^2+b^2)^2/d/(a+b*tan(d*x+c))^(1/2)-2/3*a^2*tan(d*x+c)/b/(a^2+b^2)/d/(a+b *tan(d*x+c))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.55 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.28 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {-\frac {4 a}{b}-\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )}{i a+b}+\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )}{i a-b}-6 \tan (c+d x)+\frac {3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))}{i a+b}+\frac {3 i \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))}{a+i b}}{3 b d (a+b \tan (c+d x))^{3/2}} \]
((-4*a)/b - (a*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a - I*b)])/(I*a + b) + (a*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x] )/(a + I*b)])/(I*a - b) - 6*Tan[c + d*x] + (3*Hypergeometric2F1[-1/2, 1, 1 /2, (a + b*Tan[c + d*x])/(a - I*b)]*(a + b*Tan[c + d*x]))/(I*a + b) + ((3* I)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a + I*b)]*(a + b* Tan[c + d*x]))/(a + I*b))/(3*b*d*(a + b*Tan[c + d*x])^(3/2))
Time = 0.97 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.22, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 4048, 27, 3042, 4111, 27, 3042, 4022, 3042, 4020, 25, 73, 221}
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.
| \(\displaystyle \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^3}{(a+b \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4048 |
| \(\displaystyle \frac {2 \int \frac {2 a^2-3 b \tan (c+d x) a+\left (2 a^2+3 b^2\right ) \tan ^2(c+d x)}{2 (a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan (c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {2 a^2-3 b \tan (c+d x) a+\left (2 a^2+3 b^2\right ) \tan ^2(c+d x)}{(a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan (c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 a^2-3 b \tan (c+d x) a+\left (2 a^2+3 b^2\right ) \tan (c+d x)^2}{(a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan (c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4111 |
| \(\displaystyle \frac {\frac {\int -\frac {3 \left (2 a b^2+\left (a^2-b^2\right ) \tan (c+d x) b\right )}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {4 a^2 \left (a^2+4 b^2\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan (c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {3 \int \frac {2 a b^2+\left (a^2-b^2\right ) \tan (c+d x) b}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {4 a^2 \left (a^2+4 b^2\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan (c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {3 \int \frac {2 a b^2+\left (a^2-b^2\right ) \tan (c+d x) b}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {4 a^2 \left (a^2+4 b^2\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan (c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle -\frac {2 a^2 \tan (c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a^2 \left (a^2+4 b^2\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {3 \left (\frac {1}{2} i b (a-i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} i b (a+i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx\right )}{a^2+b^2}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 a^2 \tan (c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a^2 \left (a^2+4 b^2\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {3 \left (\frac {1}{2} i b (a-i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} i b (a+i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx\right )}{a^2+b^2}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle -\frac {2 a^2 \tan (c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a^2 \left (a^2+4 b^2\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {3 \left (\frac {b (a-i b)^2 \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {b (a+i b)^2 \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}\right )}{a^2+b^2}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {2 a^2 \tan (c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a^2 \left (a^2+4 b^2\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {3 \left (-\frac {b (a-i b)^2 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {b (a+i b)^2 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}\right )}{a^2+b^2}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {2 a^2 \tan (c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a^2 \left (a^2+4 b^2\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {3 \left (\frac {i (a-i b)^2 \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}-\frac {i (a+i b)^2 \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}\right )}{a^2+b^2}}{3 b \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {2 a^2 \tan (c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {4 a^2 \left (a^2+4 b^2\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {3 \left (\frac {i b (a-i b)^2 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}-\frac {i b (a+i b)^2 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}\right )}{a^2+b^2}}{3 b \left (a^2+b^2\right )}\) |
(-2*a^2*Tan[c + d*x])/(3*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + ((- 3*(((-I)*(a + I*b)^2*b*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]* d) + (I*(a - I*b)^2*b*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d )))/(a^2 + b^2) - (4*a^2*(a^2 + 4*b^2))/(b*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]))/(3*b*(a^2 + b^2))
3.6.48.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1 /(d*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c *(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3*a*b^2*d) *Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*( n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 2] && LtQ [n, -1] && IntegerQ[2*m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2))), x ] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B , C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0 ]
Leaf count of result is larger than twice the leaf count of optimal. \(2164\) vs. \(2(148)=296\).
Time = 0.16 (sec) , antiderivative size = 2165, normalized size of antiderivative = 12.59
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(2165\) |
| default | \(\text {Expression too large to display}\) | \(2165\) |
-6/d*a^2/(a^2+b^2)^2/(a+b*tan(d*x+c))^(1/2)-2/d*b^2/(a^2+b^2)^(5/2)/(2*(a^ 2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d *x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+1/2/d*b^2/(a^2+b^2)^(7/2)*l n(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2 +b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2+1/2/d*b^2/(a^2+b^2)^3*ln((a +b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2 )^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a-1/2/d*b^2/(a^2+b^2)^3*ln(b*tan(d* x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2 ))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a-1/2/d*b^2/(a^2+b^2)^(7/2)*ln((a+b*tan(d *x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2)) *(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2+6/d*b^2/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1 /2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1 /2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^3+5/d*b^4/(a^2+b^2)^(7/2)/(2*(a^2+b^ 2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c ))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+2/d*b^2/(a^2+b^2)^(5/2)/(2*(a^2 +b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2) +2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a-6/d*b^2/(a^2+b^2)^(7/2)/(2*( a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1 /2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^3-5/d*b^4/(a^2+b^2)^(7/2) /(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2...
Leaf count of result is larger than twice the leaf count of optimal. 3324 vs. \(2 (144) = 288\).
Time = 0.31 (sec) , antiderivative size = 3324, normalized size of antiderivative = 19.33 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
-1/6*(3*((a^4*b^4 + 2*a^2*b^6 + b^8)*d*tan(d*x + c)^2 + 2*(a^5*b^3 + 2*a^3 *b^5 + a*b^7)*d*tan(d*x + c) + (a^6*b^2 + 2*a^4*b^4 + a^2*b^6)*d)*sqrt((a^ 5 - 10*a^3*b^2 + 5*a*b^4 + (a^10 + 5*a^8*b^2 + 10*a^6*b^4 + 10*a^4*b^6 + 5 *a^2*b^8 + b^10)*d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^ 2*b^8 + b^10)/((a^20 + 10*a^18*b^2 + 45*a^16*b^4 + 120*a^14*b^6 + 210*a^12 *b^8 + 252*a^10*b^10 + 210*a^8*b^12 + 120*a^6*b^14 + 45*a^4*b^16 + 10*a^2* b^18 + b^20)*d^4)))/((a^10 + 5*a^8*b^2 + 10*a^6*b^4 + 10*a^4*b^6 + 5*a^2*b ^8 + b^10)*d^2))*log((5*a^4 - 10*a^2*b^2 + b^4)*sqrt(b*tan(d*x + c) + a) + ((3*a^12 + 14*a^10*b^2 + 25*a^8*b^4 + 20*a^6*b^6 + 5*a^4*b^8 - 2*a^2*b^10 - b^12)*d^3*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/((a^20 + 10*a^18*b^2 + 45*a^16*b^4 + 120*a^14*b^6 + 210*a^12*b^8 + 2 52*a^10*b^10 + 210*a^8*b^12 + 120*a^6*b^14 + 45*a^4*b^16 + 10*a^2*b^18 + b ^20)*d^4)) - (5*a^7 - 25*a^5*b^2 + 31*a^3*b^4 - 3*a*b^6)*d)*sqrt((a^5 - 10 *a^3*b^2 + 5*a*b^4 + (a^10 + 5*a^8*b^2 + 10*a^6*b^4 + 10*a^4*b^6 + 5*a^2*b ^8 + b^10)*d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/((a^20 + 10*a^18*b^2 + 45*a^16*b^4 + 120*a^14*b^6 + 210*a^12*b^8 + 252*a^10*b^10 + 210*a^8*b^12 + 120*a^6*b^14 + 45*a^4*b^16 + 10*a^2*b^18 + b^20)*d^4)))/((a^10 + 5*a^8*b^2 + 10*a^6*b^4 + 10*a^4*b^6 + 5*a^2*b^8 + b ^10)*d^2))) - 3*((a^4*b^4 + 2*a^2*b^6 + b^8)*d*tan(d*x + c)^2 + 2*(a^5*b^3 + 2*a^3*b^5 + a*b^7)*d*tan(d*x + c) + (a^6*b^2 + 2*a^4*b^4 + a^2*b^6)*...
\[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Time = 11.64 (sec) , antiderivative size = 4638, normalized size of antiderivative = 26.97 \[ \int \frac {\tan ^3(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
atan(((1i/(4*(a^5*d^2*1i - b^5*d^2 + a*b^4*d^2*5i - 5*a^4*b*d^2 + 10*a^2*b ^3*d^2 - a^3*b^2*d^2*10i)))^(1/2)*((1i/(4*(a^5*d^2*1i - b^5*d^2 + a*b^4*d^ 2*5i - 5*a^4*b*d^2 + 10*a^2*b^3*d^2 - a^3*b^2*d^2*10i)))^(1/2)*(96*a*b^20* d^4 + 736*a^3*b^18*d^4 + 2432*a^5*b^16*d^4 + 4480*a^7*b^14*d^4 + 4928*a^9* b^12*d^4 + 3136*a^11*b^10*d^4 + 896*a^13*b^8*d^4 - 128*a^15*b^6*d^4 - 160* a^17*b^4*d^4 - 32*a^19*b^2*d^4 + (1i/(4*(a^5*d^2*1i - b^5*d^2 + a*b^4*d^2* 5i - 5*a^4*b*d^2 + 10*a^2*b^3*d^2 - a^3*b^2*d^2*10i)))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^22*d^5 + 640*a^3*b^20*d^5 + 2880*a^5*b^18*d^5 + 7680 *a^7*b^16*d^5 + 13440*a^9*b^14*d^5 + 16128*a^11*b^12*d^5 + 13440*a^13*b^10 *d^5 + 7680*a^15*b^8*d^5 + 2880*a^17*b^6*d^5 + 640*a^19*b^4*d^5 + 64*a^21* b^2*d^5)) + (a + b*tan(c + d*x))^(1/2)*(320*a^4*b^14*d^3 - 16*b^18*d^3 + 1 024*a^6*b^12*d^3 + 1440*a^8*b^10*d^3 + 1024*a^10*b^8*d^3 + 320*a^12*b^6*d^ 3 - 16*a^16*b^2*d^3))*1i - (1i/(4*(a^5*d^2*1i - b^5*d^2 + a*b^4*d^2*5i - 5 *a^4*b*d^2 + 10*a^2*b^3*d^2 - a^3*b^2*d^2*10i)))^(1/2)*((1i/(4*(a^5*d^2*1i - b^5*d^2 + a*b^4*d^2*5i - 5*a^4*b*d^2 + 10*a^2*b^3*d^2 - a^3*b^2*d^2*10i )))^(1/2)*(96*a*b^20*d^4 + 736*a^3*b^18*d^4 + 2432*a^5*b^16*d^4 + 4480*a^7 *b^14*d^4 + 4928*a^9*b^12*d^4 + 3136*a^11*b^10*d^4 + 896*a^13*b^8*d^4 - 12 8*a^15*b^6*d^4 - 160*a^17*b^4*d^4 - 32*a^19*b^2*d^4 - (1i/(4*(a^5*d^2*1i - b^5*d^2 + a*b^4*d^2*5i - 5*a^4*b*d^2 + 10*a^2*b^3*d^2 - a^3*b^2*d^2*10i)) )^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^22*d^5 + 640*a^3*b^20*d^5 + ...