Integrand size = 23, antiderivative size = 226 \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{5/2} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{5/2} d}-\frac {2 a^2 \tan ^2(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {4 a^3 \left (2 a^2+5 b^2\right )}{3 b^3 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {2 \left (4 a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right ) d} \]
-I*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(5/2)/d+I*arctanh ((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(5/2)/d+4/3*a^3*(2*a^2+5*b^ 2)/b^3/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^(1/2)+2/3*(4*a^2+3*b^2)*(a+b*tan(d*x +c))^(1/2)/b^3/(a^2+b^2)/d-2/3*a^2*tan(d*x+c)^2/b/(a^2+b^2)/d/(a+b*tan(d*x +c))^(3/2)
Time = 3.26 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.36 \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {\frac {3 \left (-b^2\right )^{3/2} \left (a^2-b^2+2 a \sqrt {-b^2}\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {-b^2}}}\right )}{\left (a^2+b^2\right )^2 \sqrt {a-\sqrt {-b^2}}}+\frac {3 \left (-b^2\right )^{3/2} \left (-a^2+b^2+2 a \sqrt {-b^2}\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+\sqrt {-b^2}}}\right )}{\left (a^2+b^2\right )^2 \sqrt {a+\sqrt {-b^2}}}+\frac {2 a^2 \left (8 a^2+9 b^2\right )}{\left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {24 a b \tan (c+d x)}{(a+b \tan (c+d x))^{3/2}}+\frac {6 b^2 \tan ^2(c+d x)}{(a+b \tan (c+d x))^{3/2}}-\frac {12 a b^4}{\left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}}{3 b^3 d} \]
((3*(-b^2)^(3/2)*(a^2 - b^2 + 2*a*Sqrt[-b^2])*ArcTanh[Sqrt[a + b*Tan[c + d *x]]/Sqrt[a - Sqrt[-b^2]]])/((a^2 + b^2)^2*Sqrt[a - Sqrt[-b^2]]) + (3*(-b^ 2)^(3/2)*(-a^2 + b^2 + 2*a*Sqrt[-b^2])*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sq rt[a + Sqrt[-b^2]]])/((a^2 + b^2)^2*Sqrt[a + Sqrt[-b^2]]) + (2*a^2*(8*a^2 + 9*b^2))/((a^2 + b^2)*(a + b*Tan[c + d*x])^(3/2)) + (24*a*b*Tan[c + d*x]) /(a + b*Tan[c + d*x])^(3/2) + (6*b^2*Tan[c + d*x]^2)/(a + b*Tan[c + d*x])^ (3/2) - (12*a*b^4)/((a^2 + b^2)^2*Sqrt[a + b*Tan[c + d*x]]))/(3*b^3*d)
Time = 1.24 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.13, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3042, 4048, 27, 3042, 4118, 3042, 4113, 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.
\(\displaystyle \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^4}{(a+b \tan (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 4048 |
\(\displaystyle \frac {2 \int \frac {\tan (c+d x) \left (4 a^2-3 b \tan (c+d x) a+\left (4 a^2+3 b^2\right ) \tan ^2(c+d x)\right )}{2 (a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\tan (c+d x) \left (4 a^2-3 b \tan (c+d x) a+\left (4 a^2+3 b^2\right ) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\tan (c+d x) \left (4 a^2-3 b \tan (c+d x) a+\left (4 a^2+3 b^2\right ) \tan (c+d x)^2\right )}{(a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4118 |
\(\displaystyle \frac {\frac {\int \frac {-6 a \tan (c+d x) b^3+\left (a^2+b^2\right ) \left (4 a^2+3 b^2\right ) \tan ^2(c+d x)+2 a^2 \left (2 a^2+5 b^2\right )}{\sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}+\frac {4 a^3 \left (2 a^2+5 b^2\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {-6 a \tan (c+d x) b^3+\left (a^2+b^2\right ) \left (4 a^2+3 b^2\right ) \tan (c+d x)^2+2 a^2 \left (2 a^2+5 b^2\right )}{\sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}+\frac {4 a^3 \left (2 a^2+5 b^2\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4113 |
\(\displaystyle \frac {\frac {\int \frac {3 b^2 \left (a^2-b^2\right )-6 a b^3 \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2+b^2\right ) \left (4 a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}+\frac {4 a^3 \left (2 a^2+5 b^2\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {3 b^2 \left (a^2-b^2\right )-6 a b^3 \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2+b^2\right ) \left (4 a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}+\frac {4 a^3 \left (2 a^2+5 b^2\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle -\frac {2 a^2 \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {4 a^3 \left (2 a^2+5 b^2\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {3}{2} b^2 (a-i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {3}{2} b^2 (a+i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2+b^2\right ) \left (4 a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 a^2 \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {4 a^3 \left (2 a^2+5 b^2\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {3}{2} b^2 (a-i b)^2 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {3}{2} b^2 (a+i b)^2 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2+b^2\right ) \left (4 a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle -\frac {2 a^2 \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {4 a^3 \left (2 a^2+5 b^2\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {3 i b^2 (a+i b)^2 \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {3 i b^2 (a-i b)^2 \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (a^2+b^2\right ) \left (4 a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2 a^2 \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {4 a^3 \left (2 a^2+5 b^2\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {-\frac {3 i b^2 (a+i b)^2 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {3 i b^2 (a-i b)^2 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (a^2+b^2\right ) \left (4 a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {2 a^2 \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {4 a^3 \left (2 a^2+5 b^2\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {3 b (a-i b)^2 \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {3 b (a+i b)^2 \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {2 \left (a^2+b^2\right ) \left (4 a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 a^2 \tan ^2(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {4 a^3 \left (2 a^2+5 b^2\right )}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {2 \left (a^2+b^2\right ) \left (4 a^2+3 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}+\frac {3 b^2 (a+i b)^2 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {3 b^2 (a-i b)^2 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}}{b \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\) |
(-2*a^2*Tan[c + d*x]^2)/(3*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + ( (4*a^3*(2*a^2 + 5*b^2))/(b^2*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]) + ((3 *(a + I*b)^2*b^2*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + ( 3*(a - I*b)^2*b^2*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d) + (2*(a^2 + b^2)*(4*a^2 + 3*b^2)*Sqrt[a + b*Tan[c + d*x]])/(b*d))/(b*(a^2 + b^2)))/(3*b*(a^2 + b^2))
3.6.47.3.1 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[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && !LeQ[m, -1]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_. )*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(c^2*C - B*c*d + A*d^2)*((c + d*Tan[e + f*x])^(n + 1)/(d^2*f*(n + 1)*(c^2 + d^2))), x] + Simp[1/(d*(c^2 + d^2)) Int[(c + d*Tan[e + f*x])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b* (c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d + a*C*d) *Tan[e + f*x] + b*C*(c^2 + d^2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && LtQ[n , -1]
Leaf count of result is larger than twice the leaf count of optimal. \(2378\) vs. \(2(196)=392\).
Time = 0.15 (sec) , antiderivative size = 2379, normalized size of antiderivative = 10.53
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(2379\) |
default | \(\text {Expression too large to display}\) | \(2379\) |
1/d/b/(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 ^4-1/d/b/(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^6+1/d*b^3/(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^2-1/d/b/(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^4+1/d/b/(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^6+3/4/d*b^3/(a^2+b^2)^(7/2)*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+2/d*b/(a^2+b^2)^3/(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^3-2/d*b/(a^2+b^2)^3/(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^3-4/d*b/(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^4-1/d*b^3/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arct an(((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...
Leaf count of result is larger than twice the leaf count of optimal. 3383 vs. \(2 (190) = 380\).
Time = 0.31 (sec) , antiderivative size = 3383, normalized size of antiderivative = 14.97 \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
1/6*(3*((a^4*b^5 + 2*a^2*b^7 + b^9)*d*tan(d*x + c)^2 + 2*(a^5*b^4 + 2*a^3* b^6 + a*b^8)*d*tan(d*x + c) + (a^6*b^3 + 2*a^4*b^5 + a^2*b^7)*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*b - 10*a^2*b^3 + b^5)*sqrt(b*tan(d*x + c) + a) + ((a^13 + 2*a^11*b^2 - 5*a^9*b^4 - 20*a^7*b^6 - 25*a^5*b^8 - 14*a^3*b^10 - 3*a*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 + 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)) + (15*a^6*b^2 - 35*a^4*b^4 + 13*a^2*b^6 - b^8)*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^5 + 2*a^2*b^7 + b^9)*d*tan(d*x + c)^2 + 2*(a ^5*b^4 + 2*a^3*b^6 + a*b^8)*d*tan(d*x + c) + (a^6*b^3 + 2*a^4*b^5 + a^2...
\[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Time = 13.17 (sec) , antiderivative size = 3739, normalized size of antiderivative = 16.54 \[ \int \frac {\tan ^4(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
(log(16*a*b^15*d^2 - ((-1/(a^5*d^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2* 5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2))^(1/2)*(((-1/(a^5*d^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2))^(1/2)*(8 96*a^6*b^15*d^4 - 160*a^2*b^19*d^4 - 128*a^4*b^17*d^4 - 32*b^21*d^4 + 3136 *a^8*b^13*d^4 + 4928*a^10*b^11*d^4 + 4480*a^12*b^9*d^4 + 2432*a^14*b^7*d^4 + 736*a^16*b^5*d^4 + 96*a^18*b^3*d^4 + ((-1/(a^5*d^2 - b^5*d^2*1i + 5*a*b ^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2))^(1/2)*(a + b*ta n(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))/2))/2 - (a + b*tan(c + d*x))^(1/2)*(320*a^4*b^14*d^3 - 16*b^ 18*d^3 + 1024*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)))/2 + 96*a^3*b^13*d^2 + 240*a^5*b^11*d^2 + 320*a^7*b^9*d^2 + 240*a^9*b^7*d^2 + 96*a^11*b^5*d^2 + 16*a^13*b^3*d^2)*(-1 /(a^5*d^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10 *a^3*b^2*d^2))^(1/2))/2 - log(16*a*b^15*d^2 - ((-1/(4*(a^5*d^2 - b^5*d^2*1 i + 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2)))^(1/2) *(896*a^6*b^15*d^4 - 160*a^2*b^19*d^4 - 128*a^4*b^17*d^4 - 32*b^21*d^4 + 3 136*a^8*b^13*d^4 + 4928*a^10*b^11*d^4 + 4480*a^12*b^9*d^4 + 2432*a^14*b^7* d^4 + 736*a^16*b^5*d^4 + 96*a^18*b^3*d^4 - (-1/(4*(a^5*d^2 - b^5*d^2*1i...