Integrand size = 27, antiderivative size = 195 \[ \int \frac {(a+b \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {i (a-i b)^2 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2} f}+\frac {i (a+i b)^2 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(c+i d)^{5/2} f}-\frac {2 (b c-a d)^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac {4 (b c-a d) (a c+b d)}{\left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}} \]
-I*(a-I*b)^2*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(5/2)/f +I*(a+I*b)^2*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(c+I*d)^(5/2)/f +4*(-a*d+b*c)*(a*c+b*d)/(c^2+d^2)^2/f/(c+d*tan(f*x+e))^(1/2)-2/3*(-a*d+b*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 = 0.20 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.65 \[ \int \frac {(a+b \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {-\frac {2 b^2}{d}-\frac {(a-i b)^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )}{i c+d}+\frac {(a+i b)^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {c+d \tan (e+f x)}{c+i d}\right )}{i c-d}}{3 f (c+d \tan (e+f x))^{3/2}} \]
((-2*b^2)/d - ((a - I*b)^2*Hypergeometric2F1[-3/2, 1, -1/2, (c + d*Tan[e + f*x])/(c - I*d)])/(I*c + d) + ((a + I*b)^2*Hypergeometric2F1[-3/2, 1, -1/ 2, (c + d*Tan[e + f*x])/(c + I*d)])/(I*c - d))/(3*f*(c + d*Tan[e + f*x])^( 3/2))
Time = 1.05 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.11, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 4025, 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.
\(\displaystyle \int \frac {(a+b \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 4025 |
\(\displaystyle \frac {\int \frac {c a^2+2 b d a-b^2 c+\left (-d a^2+2 b c a+b^2 d\right ) \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}}dx}{c^2+d^2}-\frac {2 (b c-a d)^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {c a^2+2 b d a-b^2 c+\left (-d a^2+2 b c a+b^2 d\right ) \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}}dx}{c^2+d^2}-\frac {2 (b c-a d)^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {\frac {\int -\frac {(b (c-d)-a (c+d)) (a (c-d)+b (c+d))-2 (b c-a d) (a c+b d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {4 (b c-a d) (a c+b d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{c^2+d^2}-\frac {2 (b c-a d)^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {4 (b c-a d) (a c+b d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {\int \frac {(b (c-d)-a (c+d)) (a (c-d)+b (c+d))-2 (b c-a d) (a c+b d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}}{c^2+d^2}-\frac {2 (b c-a d)^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {4 (b c-a d) (a c+b d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {\int \frac {(b (c-d)-a (c+d)) (a (c-d)+b (c+d))-2 (b c-a d) (a c+b d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}}{c^2+d^2}-\frac {2 (b c-a d)^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle -\frac {2 (b c-a d)^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 (b c-a d) (a c+b d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {\frac {1}{2} (-b+i a)^2 (c-i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (b+i a)^2 (c+i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}}{c^2+d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 (b c-a d)^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 (b c-a d) (a c+b d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {\frac {1}{2} (-b+i a)^2 (c-i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (b+i a)^2 (c+i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}}{c^2+d^2}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle -\frac {2 (b c-a d)^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 (b c-a d) (a c+b d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {\frac {i (b+i a)^2 (c+i d)^2 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (-b+i a)^2 (c-i d)^2 \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}}{c^2+d^2}}{c^2+d^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2 (b c-a d)^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 (b c-a d) (a c+b d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {\frac {i (-b+i a)^2 (c-i d)^2 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {i (b+i a)^2 (c+i d)^2 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}}{c^2+d^2}}{c^2+d^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {2 (b c-a d)^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 (b c-a d) (a c+b d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {\frac {(-b+i a)^2 (c-i d)^2 \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(b+i a)^2 (c+i d)^2 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}}{c^2+d^2}}{c^2+d^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 (b c-a d)^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {4 (b c-a d) (a c+b d)}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {\frac {(-b+i a)^2 (c-i d)^2 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}+\frac {(b+i a)^2 (c+i d)^2 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}}{c^2+d^2}}{c^2+d^2}\) |
(-2*(b*c - a*d)^2)/(3*d*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) + (-(((( I*a + b)^2*(c + I*d)^2*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]* f) + ((I*a - b)^2*(c - I*d)^2*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f))/(c^2 + d^2)) + (4*(b*c - a*d)*(a*c + b*d))/((c^2 + d^2)*f*Sqrt[ c + d*Tan[e + f*x]]))/(c^2 + d^2)
3.13.61.3.1 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_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*c - a*d)^2*((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[a*c^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(6793\) vs. \(2(169)=338\).
Time = 1.15 (sec) , antiderivative size = 6794, normalized size of antiderivative = 34.84
method | result | size |
parts | \(\text {Expression too large to display}\) | \(6794\) |
derivativedivides | \(\text {Expression too large to display}\) | \(19430\) |
default | \(\text {Expression too large to display}\) | \(19430\) |
Leaf count of result is larger than twice the leaf count of optimal. 10895 vs. \(2 (161) = 322\).
Time = 9.02 (sec) , antiderivative size = 10895, normalized size of antiderivative = 55.87 \[ \int \frac {(a+b \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {(a+b \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{2}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {(a+b \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Time = 23.53 (sec) , antiderivative size = 25298, normalized size of antiderivative = 129.73 \[ \int \frac {(a+b \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
- atan(-(((c + d*tan(e + f*x))^(1/2)*(96*a^2*b^2*d^18*f^3 - 16*b^4*d^18*f^ 3 - 16*a^4*d^18*f^3 + 320*a^4*c^4*d^14*f^3 + 1024*a^4*c^6*d^12*f^3 + 1440* a^4*c^8*d^10*f^3 + 1024*a^4*c^10*d^8*f^3 + 320*a^4*c^12*d^6*f^3 - 16*a^4*c ^16*d^2*f^3 + 320*b^4*c^4*d^14*f^3 + 1024*b^4*c^6*d^12*f^3 + 1440*b^4*c^8* d^10*f^3 + 1024*b^4*c^10*d^8*f^3 + 320*b^4*c^12*d^6*f^3 - 16*b^4*c^16*d^2* f^3 - 1920*a^2*b^2*c^4*d^14*f^3 - 6144*a^2*b^2*c^6*d^12*f^3 - 8640*a^2*b^2 *c^8*d^10*f^3 - 6144*a^2*b^2*c^10*d^8*f^3 - 1920*a^2*b^2*c^12*d^6*f^3 + 96 *a^2*b^2*c^16*d^2*f^3 - 256*a*b^3*c*d^17*f^3 + 256*a^3*b*c*d^17*f^3 - 1280 *a*b^3*c^3*d^15*f^3 - 2304*a*b^3*c^5*d^13*f^3 - 1280*a*b^3*c^7*d^11*f^3 + 1280*a*b^3*c^9*d^9*f^3 + 2304*a*b^3*c^11*d^7*f^3 + 1280*a*b^3*c^13*d^5*f^3 + 256*a*b^3*c^15*d^3*f^3 + 1280*a^3*b*c^3*d^15*f^3 + 2304*a^3*b*c^5*d^13* f^3 + 1280*a^3*b*c^7*d^11*f^3 - 1280*a^3*b*c^9*d^9*f^3 - 2304*a^3*b*c^11*d ^7*f^3 - 1280*a^3*b*c^13*d^5*f^3 - 256*a^3*b*c^15*d^3*f^3) + ((((8*a^4*c^5 *f^2 + 8*b^4*c^5*f^2 - 32*a*b^3*d^5*f^2 + 32*a^3*b*d^5*f^2 + 40*a^4*c*d^4* f^2 + 40*b^4*c*d^4*f^2 - 48*a^2*b^2*c^5*f^2 - 80*a^4*c^3*d^2*f^2 - 80*b^4* c^3*d^2*f^2 + 480*a^2*b^2*c^3*d^2*f^2 - 160*a*b^3*c^4*d*f^2 + 160*a^3*b*c^ 4*d*f^2 + 320*a*b^3*c^2*d^3*f^2 - 240*a^2*b^2*c*d^4*f^2 - 320*a^3*b*c^2*d^ 3*f^2)^2/4 - (a^8 + b^8 + 4*a^2*b^6 + 6*a^4*b^4 + 4*a^6*b^2)*(16*c^10*f^4 + 16*d^10*f^4 + 80*c^2*d^8*f^4 + 160*c^4*d^6*f^4 + 160*c^6*d^4*f^4 + 80*c^ 8*d^2*f^4))^(1/2) - 4*a^4*c^5*f^2 - 4*b^4*c^5*f^2 + 16*a*b^3*d^5*f^2 - ...