Integrand size = 27, antiderivative size = 290 \[ \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {i (a-i b)^4 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2} f}+\frac {i (a+i b)^4 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(c+i d)^{5/2} f}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {4 (b c-a d)^3 \left (2 b c^2+3 a c d+5 b d^2\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}-\frac {2 b^2 \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f} \]
-I*(a-I*b)^4*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(5/2)/f +I*(a+I*b)^4*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(c+I*d)^(5/2)/f +4/3*(-a*d+b*c)^3*(3*a*c*d+2*b*c^2+5*b*d^2)/d^3/(c^2+d^2)^2/f/(c+d*tan(f*x +e))^(1/2)-2/3*b^2*(a*d*(-a*d+2*b*c)-b^2*(4*c^2+3*d^2))*(c+d*tan(f*x+e))^( 1/2)/d^3/(c^2+d^2)/f-2/3*(-a*d+b*c)^2*(a+b*tan(f*x+e))^2/d/(c^2+d^2)/f/(c+ d*tan(f*x+e))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 3.80 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.27 \[ \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {-2 b^2 (c-i d) (c+i d) \left (-20 a b c d+9 a^2 d^2+b^2 \left (8 c^2+d^2\right )\right )+d^2 \left (-4 a^3 b c+4 a b^3 c+a^4 d-6 a^2 b^2 d+b^4 d\right ) \left ((-i c+d) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )+(i c+d) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )\right )-12 b^2 (c-i d) (c+i d) d (2 b c-3 a d) (a+b \tan (e+f x))-6 b^2 (c-i d) (c+i d) d^2 (a+b \tan (e+f x))^2-12 a b \left (a^2-b^2\right ) d^2 \left (i (c+i d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )-(i c+d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )\right ) (c+d \tan (e+f x))}{3 d^3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}} \]
-1/3*(-2*b^2*(c - I*d)*(c + I*d)*(-20*a*b*c*d + 9*a^2*d^2 + b^2*(8*c^2 + d ^2)) + d^2*(-4*a^3*b*c + 4*a*b^3*c + a^4*d - 6*a^2*b^2*d + b^4*d)*(((-I)*c + d)*Hypergeometric2F1[-3/2, 1, -1/2, (c + d*Tan[e + f*x])/(c - I*d)] + ( I*c + d)*Hypergeometric2F1[-3/2, 1, -1/2, (c + d*Tan[e + f*x])/(c + I*d)]) - 12*b^2*(c - I*d)*(c + I*d)*d*(2*b*c - 3*a*d)*(a + b*Tan[e + f*x]) - 6*b ^2*(c - I*d)*(c + I*d)*d^2*(a + b*Tan[e + f*x])^2 - 12*a*b*(a^2 - b^2)*d^2 *(I*(c + I*d)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I* d)] - (I*c + d)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c + I*d)])*(c + d*Tan[e + f*x]))/(d^3*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2) )
Time = 1.99 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.10, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.519, 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.
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4048 |
| \(\displaystyle \frac {2 \int \frac {(a+b \tan (e+f x)) \left (3 c d a^3+10 b d^2 a^2-11 b^2 c d a+4 b^3 c^2-b \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \tan ^2(e+f x)+3 d \left (-d a^3+3 b c a^2+3 b^2 d a-b^3 c\right ) \tan (e+f x)\right )}{2 (c+d \tan (e+f x))^{3/2}}dx}{3 d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(a+b \tan (e+f x)) \left (3 c d a^3+10 b d^2 a^2-11 b^2 c d a+4 b^3 c^2-b \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \tan ^2(e+f x)+3 d \left (-d a^3+3 b c a^2+3 b^2 d a-b^3 c\right ) \tan (e+f x)\right )}{(c+d \tan (e+f x))^{3/2}}dx}{3 d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(a+b \tan (e+f x)) \left (3 c d a^3+10 b d^2 a^2-11 b^2 c d a+4 b^3 c^2-b \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \tan (e+f x)^2+3 d \left (-d a^3+3 b c a^2+3 b^2 d a-b^3 c\right ) \tan (e+f x)\right )}{(c+d \tan (e+f x))^{3/2}}dx}{3 d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4118 |
| \(\displaystyle \frac {\frac {\int \frac {3 d^2 \left (c^2-d^2\right ) a^4+24 b c d^3 a^3-b^2 d^2 \left (17 c^2-19 d^2\right ) a^2-2 b^3 c d \left (c^2+13 d^2\right ) a-b^2 \left (c^2+d^2\right ) \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \tan ^2(e+f x)+2 b^4 \left (2 c^4+5 d^2 c^2\right )+6 d^2 \left (c a^2+2 b d a-b^2 c\right ) \left (-d a^2+2 b c a+b^2 d\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{d \left (c^2+d^2\right )}+\frac {4 \left (3 a c d+2 b c^2+5 b d^2\right ) (b c-a d)^3}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{3 d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {3 d^2 \left (c^2-d^2\right ) a^4+24 b c d^3 a^3-b^2 d^2 \left (17 c^2-19 d^2\right ) a^2-2 b^3 c d \left (c^2+13 d^2\right ) a-b^2 \left (c^2+d^2\right ) \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \tan (e+f x)^2+2 b^4 \left (2 c^4+5 d^2 c^2\right )+6 d^2 \left (c a^2+2 b d a-b^2 c\right ) \left (-d a^2+2 b c a+b^2 d\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{d \left (c^2+d^2\right )}+\frac {4 \left (3 a c d+2 b c^2+5 b d^2\right ) (b c-a d)^3}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{3 d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4113 |
| \(\displaystyle \frac {\frac {\int \frac {3 \left (\left (c^2-d^2\right ) a^4+8 b c d a^3-6 b^2 \left (c^2-d^2\right ) a^2-8 b^3 c d a+b^4 \left (c^2-d^2\right )\right ) d^2+6 \left (c a^2+2 b d a-b^2 c\right ) \left (-d a^2+2 b c a+b^2 d\right ) \tan (e+f x) d^2}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b^2 \left (c^2+d^2\right ) \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}+\frac {4 \left (3 a c d+2 b c^2+5 b d^2\right ) (b c-a d)^3}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{3 d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {3 \left (\left (c^2-d^2\right ) a^4+8 b c d a^3-6 b^2 \left (c^2-d^2\right ) a^2-8 b^3 c d a+b^4 \left (c^2-d^2\right )\right ) d^2+6 \left (c a^2+2 b d a-b^2 c\right ) \left (-d a^2+2 b c a+b^2 d\right ) \tan (e+f x) d^2}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b^2 \left (c^2+d^2\right ) \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}+\frac {4 \left (3 a c d+2 b c^2+5 b d^2\right ) (b c-a d)^3}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{3 d \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 \left (3 a c d+2 b c^2+5 b d^2\right ) (b c-a d)^3}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {3}{2} d^2 (a-i b)^4 (c+i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {3}{2} d^2 (a+i b)^4 (c-i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b^2 \left (c^2+d^2\right ) \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}}{3 d \left (c^2+d^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 \left (3 a c d+2 b c^2+5 b d^2\right ) (b c-a d)^3}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {3}{2} d^2 (a-i b)^4 (c+i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {3}{2} d^2 (a+i b)^4 (c-i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b^2 \left (c^2+d^2\right ) \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}}{3 d \left (c^2+d^2\right )}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 \left (3 a c d+2 b c^2+5 b d^2\right ) (b c-a d)^3}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {3 i d^2 (a-i b)^4 (c+i d)^2 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {3 i d^2 (a+i b)^4 (c-i d)^2 \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 b^2 \left (c^2+d^2\right ) \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}}{3 d \left (c^2+d^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 \left (3 a c d+2 b c^2+5 b d^2\right ) (b c-a d)^3}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {3 i d^2 (a-i b)^4 (c+i d)^2 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {3 i d^2 (a+i b)^4 (c-i d)^2 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 b^2 \left (c^2+d^2\right ) \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}}{3 d \left (c^2+d^2\right )}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 \left (3 a c d+2 b c^2+5 b d^2\right ) (b c-a d)^3}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {3 d (a-i b)^4 (c+i d)^2 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}+\frac {3 d (a+i b)^4 (c-i d)^2 \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {2 b^2 \left (c^2+d^2\right ) \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}}{3 d \left (c^2+d^2\right )}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 \left (3 a c d+2 b c^2+5 b d^2\right ) (b c-a d)^3}{d^2 f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {3 d^2 (a-i b)^4 (c+i d)^2 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {3 d^2 (a+i b)^4 (c-i d)^2 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}-\frac {2 b^2 \left (c^2+d^2\right ) \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{d \left (c^2+d^2\right )}}{3 d \left (c^2+d^2\right )}\) |
(-2*(b*c - a*d)^2*(a + b*Tan[e + f*x])^2)/(3*d*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) + ((4*(b*c - a*d)^3*(2*b*c^2 + 3*a*c*d + 5*b*d^2))/(d^2*(c^ 2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]) + ((3*(a - I*b)^4*(c + I*d)^2*d^2*Arc Tan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + (3*(a + I*b)^4*(c - I *d)^2*d^2*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f) - (2*b^2*( c^2 + d^2)*(a*d*(2*b*c - a*d) - b^2*(4*c^2 + 3*d^2))*Sqrt[c + d*Tan[e + f* x]])/(d*f))/(d*(c^2 + d^2)))/(3*d*(c^2 + d^2))
3.13.59.3.1 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. \(11347\) vs. \(2(258)=516\).
Time = 1.12 (sec) , antiderivative size = 11348, normalized size of antiderivative = 39.13
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(11348\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(33803\) |
| default | \(\text {Expression too large to display}\) | \(33803\) |
Leaf count of result is larger than twice the leaf count of optimal. 18118 vs. \(2 (250) = 500\).
Time = 67.54 (sec) , antiderivative size = 18118, normalized size of antiderivative = 62.48 \[ \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{4}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Time = 41.34 (sec) , antiderivative size = 43980, normalized size of antiderivative = 151.66 \[ \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
(2*b^4*(c + d*tan(e + f*x))^(1/2))/(d^3*f) - atan((((-(((8*a^8*c^5*f^2 + 8 *b^8*c^5*f^2 - 64*a*b^7*d^5*f^2 + 64*a^7*b*d^5*f^2 + 40*a^8*c*d^4*f^2 + 40 *b^8*c*d^4*f^2 - 224*a^2*b^6*c^5*f^2 + 560*a^4*b^4*c^5*f^2 - 224*a^6*b^2*c ^5*f^2 + 448*a^3*b^5*d^5*f^2 - 448*a^5*b^3*d^5*f^2 - 80*a^8*c^3*d^2*f^2 - 80*b^8*c^3*d^2*f^2 + 2240*a^2*b^6*c^3*d^2*f^2 - 4480*a^3*b^5*c^2*d^3*f^2 - 5600*a^4*b^4*c^3*d^2*f^2 + 4480*a^5*b^3*c^2*d^3*f^2 + 2240*a^6*b^2*c^3*d^ 2*f^2 - 320*a*b^7*c^4*d*f^2 + 320*a^7*b*c^4*d*f^2 + 640*a*b^7*c^2*d^3*f^2 - 1120*a^2*b^6*c*d^4*f^2 + 2240*a^3*b^5*c^4*d*f^2 + 2800*a^4*b^4*c*d^4*f^2 - 2240*a^5*b^3*c^4*d*f^2 - 1120*a^6*b^2*c*d^4*f^2 - 640*a^7*b*c^2*d^3*f^2 )^2/4 - (16*c^10*f^4 + 16*d^10*f^4 + 80*c^2*d^8*f^4 + 160*c^4*d^6*f^4 + 16 0*c^6*d^4*f^4 + 80*c^8*d^2*f^4)*(a^16 + b^16 + 8*a^2*b^14 + 28*a^4*b^12 + 56*a^6*b^10 + 70*a^8*b^8 + 56*a^10*b^6 + 28*a^12*b^4 + 8*a^14*b^2))^(1/2) + 4*a^8*c^5*f^2 + 4*b^8*c^5*f^2 - 32*a*b^7*d^5*f^2 + 32*a^7*b*d^5*f^2 + 20 *a^8*c*d^4*f^2 + 20*b^8*c*d^4*f^2 - 112*a^2*b^6*c^5*f^2 + 280*a^4*b^4*c^5* f^2 - 112*a^6*b^2*c^5*f^2 + 224*a^3*b^5*d^5*f^2 - 224*a^5*b^3*d^5*f^2 - 40 *a^8*c^3*d^2*f^2 - 40*b^8*c^3*d^2*f^2 + 1120*a^2*b^6*c^3*d^2*f^2 - 2240*a^ 3*b^5*c^2*d^3*f^2 - 2800*a^4*b^4*c^3*d^2*f^2 + 2240*a^5*b^3*c^2*d^3*f^2 + 1120*a^6*b^2*c^3*d^2*f^2 - 160*a*b^7*c^4*d*f^2 + 160*a^7*b*c^4*d*f^2 + 320 *a*b^7*c^2*d^3*f^2 - 560*a^2*b^6*c*d^4*f^2 + 1120*a^3*b^5*c^4*d*f^2 + 1400 *a^4*b^4*c*d^4*f^2 - 1120*a^5*b^3*c^4*d*f^2 - 560*a^6*b^2*c*d^4*f^2 - 3...