Integrand size = 35, antiderivative size = 209 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {(i A+B-i C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2} f}-\frac {(B-i (A-C)) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(c+i d)^{5/2} f}-\frac {2 \left (c^2 C-B c d+A d^2\right )}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )}{\left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}} \]
-(I*A+B-I*C)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(5/2)/f -(B-I*(A-C))*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(c+I*d)^(5/2)/f -2*(2*c*(A-C)*d-B*(c^2-d^2))/(c^2+d^2)^2/f/(c+d*tan(f*x+e))^(1/2)-2/3*(A*d ^2-B*c*d+C*c^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 = 0.98 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.07 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {2 C \left (c^2+d^2\right )+(B c+(-A+C) d) \left (i (c+i d) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )-(i c+d) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )\right )-3 B \left (i (c+i d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )-(i c+d) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )\right ) (c+d \tan (e+f x))}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}} \]
-1/3*(2*C*(c^2 + d^2) + (B*c + (-A + C)*d)*(I*(c + I*d)*Hypergeometric2F1[ -3/2, 1, -1/2, (c + d*Tan[e + f*x])/(c - I*d)] - (I*c + d)*Hypergeometric2 F1[-3/2, 1, -1/2, (c + d*Tan[e + f*x])/(c + I*d)]) - 3*B*(I*(c + I*d)*Hype rgeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I*d)] - (I*c + d)*Hy pergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c + I*d)])*(c + d*Tan[ e + f*x]))/(d*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2))
Time = 1.12 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.11, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 4111, 3042, 4012, 25, 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.
| \(\displaystyle \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (e+f x)+C \tan (e+f x)^2}{(c+d \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4111 |
| \(\displaystyle \frac {\int \frac {A c-C c+B d+(B c-(A-C) d) \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}}dx}{c^2+d^2}-\frac {2 \left (A d^2-B c d+c^2 C\right )}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {A c-C c+B d+(B c-(A-C) d) \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}}dx}{c^2+d^2}-\frac {2 \left (A d^2-B c d+c^2 C\right )}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {\frac {\int -\frac {C c^2-2 B d c-C d^2-A \left (c^2-d^2\right )+\left (2 c (A-C) d-B \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}-\frac {2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{c^2+d^2}-\frac {2 \left (A d^2-B c d+c^2 C\right )}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {C c^2-2 B d c-C d^2-A \left (c^2-d^2\right )+\left (2 c (A-C) d-B \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}-\frac {2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{c^2+d^2}-\frac {2 \left (A d^2-B c d+c^2 C\right )}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {C c^2-2 B d c-C d^2-A \left (c^2-d^2\right )+\left (2 c (A-C) d-B \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}-\frac {2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{c^2+d^2}-\frac {2 \left (A d^2-B c d+c^2 C\right )}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle -\frac {2 \left (A d^2-B c d+c^2 C\right )}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {1}{2} (c-i d)^2 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {1}{2} (c+i d)^2 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}}{c^2+d^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \left (A d^2-B c d+c^2 C\right )}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {1}{2} (c-i d)^2 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {1}{2} (c+i d)^2 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}}{c^2+d^2}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle -\frac {2 \left (A d^2-B c d+c^2 C\right )}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {\frac {i (c-i d)^2 (A+i B-C) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {i (c+i d)^2 (A-i B-C) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}}{c^2+d^2}}{c^2+d^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {2 \left (A d^2-B c d+c^2 C\right )}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {\frac {i (c+i d)^2 (A-i B-C) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (c-i d)^2 (A+i B-C) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}}{c^2+d^2}}{c^2+d^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {2 \left (A d^2-B c d+c^2 C\right )}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {(c-i d)^2 (A+i B-C) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}-\frac {(c+i d)^2 (A-i B-C) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}}{c^2+d^2}}{c^2+d^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {2 \left (A d^2-B c d+c^2 C\right )}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {-\frac {2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {-\frac {(c-i d)^2 (A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}-\frac {(c+i d)^2 (A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}}{c^2+d^2}}{c^2+d^2}\) |
(-2*(c^2*C - B*c*d + A*d^2))/(3*d*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2) ) + (-((-(((A - I*B - C)*(c + I*d)^2*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/( Sqrt[c - I*d]*f)) - ((A + I*B - C)*(c - I*d)^2*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f))/(c^2 + d^2)) - (2*(2*c*(A - C)*d - B*(c^2 - d^ 2)))/((c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]))/(c^2 + d^2)
3.2.25.3.1 Defintions of rubi rules used
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[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
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_)*((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. \(6787\) vs. \(2(184)=368\).
Time = 0.18 (sec) , antiderivative size = 6788, normalized size of antiderivative = 32.48
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(6788\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(20647\) |
| default | \(\text {Expression too large to display}\) | \(20647\) |
Leaf count of result is larger than twice the leaf count of optimal. 13143 vs. \(2 (177) = 354\).
Time = 7.15 (sec) , antiderivative size = 13143, normalized size of antiderivative = 62.89 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Time = 35.92 (sec) , antiderivative size = 14163, normalized size of antiderivative = 67.77 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
(log(96*A^3*c^3*d^13*f^2 - ((((((320*A^4*c^2*d^8*f^4 - 16*A^4*d^10*f^4 - 1 760*A^4*c^4*d^6*f^4 + 1600*A^4*c^6*d^4*f^4 - 400*A^4*c^8*d^2*f^4)^(1/2) - 4*A^2*c^5*f^2 + 40*A^2*c^3*d^2*f^2 - 20*A^2*c*d^4*f^2)/(c^10*f^4 + d^10*f^ 4 + 5*c^2*d^8*f^4 + 10*c^4*d^6*f^4 + 10*c^6*d^4*f^4 + 5*c^8*d^2*f^4))^(1/2 )*(((((320*A^4*c^2*d^8*f^4 - 16*A^4*d^10*f^4 - 1760*A^4*c^4*d^6*f^4 + 1600 *A^4*c^6*d^4*f^4 - 400*A^4*c^8*d^2*f^4)^(1/2) - 4*A^2*c^5*f^2 + 40*A^2*c^3 *d^2*f^2 - 20*A^2*c*d^4*f^2)/(c^10*f^4 + d^10*f^4 + 5*c^2*d^8*f^4 + 10*c^4 *d^6*f^4 + 10*c^6*d^4*f^4 + 5*c^8*d^2*f^4))^(1/2)*(c + d*tan(e + f*x))^(1/ 2)*(64*c*d^22*f^5 + 640*c^3*d^20*f^5 + 2880*c^5*d^18*f^5 + 7680*c^7*d^16*f ^5 + 13440*c^9*d^14*f^5 + 16128*c^11*d^12*f^5 + 13440*c^13*d^10*f^5 + 7680 *c^15*d^8*f^5 + 2880*c^17*d^6*f^5 + 640*c^19*d^4*f^5 + 64*c^21*d^2*f^5))/4 - 32*A*d^21*f^4 - 160*A*c^2*d^19*f^4 - 128*A*c^4*d^17*f^4 + 896*A*c^6*d^1 5*f^4 + 3136*A*c^8*d^13*f^4 + 4928*A*c^10*d^11*f^4 + 4480*A*c^12*d^9*f^4 + 2432*A*c^14*d^7*f^4 + 736*A*c^16*d^5*f^4 + 96*A*c^18*d^3*f^4))/4 - (c + d *tan(e + f*x))^(1/2)*(320*A^2*c^4*d^14*f^3 - 16*A^2*d^18*f^3 + 1024*A^2*c^ 6*d^12*f^3 + 1440*A^2*c^8*d^10*f^3 + 1024*A^2*c^10*d^8*f^3 + 320*A^2*c^12* d^6*f^3 - 16*A^2*c^16*d^2*f^3))*(((320*A^4*c^2*d^8*f^4 - 16*A^4*d^10*f^4 - 1760*A^4*c^4*d^6*f^4 + 1600*A^4*c^6*d^4*f^4 - 400*A^4*c^8*d^2*f^4)^(1/2) - 4*A^2*c^5*f^2 + 40*A^2*c^3*d^2*f^2 - 20*A^2*c*d^4*f^2)/(c^10*f^4 + d^10* f^4 + 5*c^2*d^8*f^4 + 10*c^4*d^6*f^4 + 10*c^6*d^4*f^4 + 5*c^8*d^2*f^4))...