Integrand size = 27, antiderivative size = 170 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx=\frac {(c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(i a+b) f}-\frac {(c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(i a-b) f}-\frac {2 (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\sqrt {b} \left (a^2+b^2\right ) f} \]
(c-I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(I*a+b)/f-(c+I *d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(I*a-b)/f-2*(-a*d+ b*c)^(3/2)*arctanh(b^(1/2)*(c+d*tan(f*x+e))^(1/2)/(-a*d+b*c)^(1/2))/(a^2+b ^2)/f/b^(1/2)
Time = 0.54 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.99 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx=\frac {\sqrt {b} (-i a+b) (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {b} (i a+b) (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )-2 (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\sqrt {b} \left (a^2+b^2\right ) f} \]
(Sqrt[b]*((-I)*a + b)*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqr t[c - I*d]] + Sqrt[b]*(I*a + b)*(c + I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] - 2*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*Ta n[e + f*x]])/Sqrt[b*c - a*d]])/(Sqrt[b]*(a^2 + b^2)*f)
Time = 1.12 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.93, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 4056, 3042, 4022, 3042, 4020, 25, 73, 221, 4117, 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.
\(\displaystyle \int \frac {(c+d \tan (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c+d \tan (e+f x))^{3/2}}{a+b \tan (e+f x)}dx\) |
\(\Big \downarrow \) 4056 |
\(\displaystyle \frac {\int \frac {2 b c d+a \left (c^2-d^2\right )+\left (2 a c d-b \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {(b c-a d)^2 \int \frac {\tan ^2(e+f x)+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {2 b c d+a \left (c^2-d^2\right )+\left (2 a c d-b \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {(b c-a d)^2 \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {(b c-a d)^2 \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\frac {1}{2} (a+i b) (c-i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b) (c+i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(b c-a d)^2 \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\frac {1}{2} (a+i b) (c-i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b) (c+i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {(b c-a d)^2 \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\frac {i (a+i b) (c-i d)^2 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (a-i b) (c+i d)^2 \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}}{a^2+b^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {(b c-a d)^2 \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\frac {i (a-i b) (c+i d)^2 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {i (a+i b) (c-i d)^2 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}}{a^2+b^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(b c-a d)^2 \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\frac {(a+i b) (c-i d)^2 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(a-i b) (c+i d)^2 \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}}{a^2+b^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(b c-a d)^2 \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\frac {(a+i b) (c-i d)^{3/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {(a-i b) (c+i d)^{3/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{a^2+b^2}\) |
\(\Big \downarrow \) 4117 |
\(\displaystyle \frac {(b c-a d)^2 \int \frac {1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{f \left (a^2+b^2\right )}+\frac {\frac {(a+i b) (c-i d)^{3/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {(a-i b) (c+i d)^{3/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{a^2+b^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 (b c-a d)^2 \int \frac {1}{a+\frac {b (c+d \tan (e+f x))}{d}-\frac {b c}{d}}d\sqrt {c+d \tan (e+f x)}}{d f \left (a^2+b^2\right )}+\frac {\frac {(a+i b) (c-i d)^{3/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {(a-i b) (c+i d)^{3/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{a^2+b^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\sqrt {b} f \left (a^2+b^2\right )}+\frac {\frac {(a+i b) (c-i d)^{3/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {(a-i b) (c+i d)^{3/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{a^2+b^2}\) |
(((a + I*b)*(c - I*d)^(3/2)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f + ((a - I*b)*(c + I*d)^(3/2)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/f)/(a^2 + b^2) - (2*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/(Sqrt[b]*(a^2 + b^2)*f)
3.13.38.3.1 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[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_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/(c^2 + d^2) Int[Simp[a^2*c - b^2*c + 2*a*b*d + (2*a*b*c - a^2*d + b^2*d)*Tan[e + f*x], x]/Sqrt[a + b*Tan[e + f*x ]], x], x] + Simp[(b*c - a*d)^2/(c^2 + d^2) Int[(1 + Tan[e + f*x]^2)/(Sqr t[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*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]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A/f Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
Leaf count of result is larger than twice the leaf count of optimal. \(1953\) vs. \(2(142)=284\).
Time = 0.89 (sec) , antiderivative size = 1954, normalized size of antiderivative = 11.49
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1954\) |
default | \(\text {Expression too large to display}\) | \(1954\) |
2/f*d^2/(a^2+b^2)/((a*d-b*c)*b)^(1/2)*arctan(b*(c+d*tan(f*x+e))^(1/2)/((a* d-b*c)*b)^(1/2))*a^2-4/f*d/(a^2+b^2)/((a*d-b*c)*b)^(1/2)*arctan(b*(c+d*tan (f*x+e))^(1/2)/((a*d-b*c)*b)^(1/2))*a*b*c+2/f/(a^2+b^2)/((a*d-b*c)*b)^(1/2 )*arctan(b*(c+d*tan(f*x+e))^(1/2)/((a*d-b*c)*b)^(1/2))*b^2*c^2-1/4/f/(a^2+ b^2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2 )+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*b+1/4/f/d /(a^2+b^2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c )^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c^2+1/2/f/(a^2+b^ 2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+ (c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b*c-1/f*d/(a^2+b^2)/(2*(c^2 +d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2) +2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*a-1/4/f*d/(a^2 +b^2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/ 2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a+1/4/f/(a^2+b^2)*ln(d*t an(f*x+e)+c-(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2) ^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*b+1/f*d^2/(a^2+b^2)/ (2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2 )^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*b+1/f*d^2/(a^2+b^2)/(2* (c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2*(c^2+d^2)^( 1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*b+1/4/f*d/(a^2+b^2)*ln(...
Leaf count of result is larger than twice the leaf count of optimal. 3795 vs. \(2 (136) = 272\).
Time = 3.59 (sec) , antiderivative size = 7610, normalized size of antiderivative = 44.76 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx=\text {Too large to display} \]
\[ \int \frac {(c+d \tan (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx=\int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}{a + b \tan {\left (e + f x \right )}}\, dx \]
Exception generated. \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Timed out. \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx=\text {Timed out} \]
Time = 15.28 (sec) , antiderivative size = 19118, normalized size of antiderivative = 112.46 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx=\text {Too large to display} \]
(atan(-((((((32*(a*b^5*d^15*f^2 + 4*a^5*b*d^15*f^2 - b^6*c*d^14*f^2 - 15*a ^3*b^3*d^15*f^2 + 23*b^6*c^3*d^12*f^2 + 21*b^6*c^5*d^10*f^2 - 3*b^6*c^7*d^ 8*f^2 - 29*a^2*b^4*c^3*d^12*f^2 - 81*a^2*b^4*c^5*d^10*f^2 + a^2*b^4*c^7*d^ 8*f^2 + 59*a^3*b^3*c^2*d^13*f^2 + 75*a^3*b^3*c^4*d^11*f^2 + a^3*b^3*c^6*d^ 9*f^2 - 32*a^4*b^2*c^3*d^12*f^2 - 2*a^4*b^2*c^5*d^10*f^2 - 61*a*b^5*c^2*d^ 13*f^2 - 25*a*b^5*c^4*d^11*f^2 + 37*a*b^5*c^6*d^9*f^2 + 53*a^2*b^4*c*d^14* f^2 - 30*a^4*b^2*c*d^14*f^2 + 4*a^5*b*c^2*d^13*f^2))/f^5 + ((((-b*(a*d - b *c)^3)^(1/2)*((32*(4*a^2*b^6*d^12*f^4 + 8*a^4*b^4*d^12*f^4 + 4*a^6*b^2*d^1 2*f^4 + 12*b^8*c^2*d^10*f^4 + 12*b^8*c^4*d^8*f^4 + 28*a^2*b^6*c^2*d^10*f^4 + 24*a^2*b^6*c^4*d^8*f^4 - 32*a^3*b^5*c^3*d^9*f^4 + 20*a^4*b^4*c^2*d^10*f ^4 + 12*a^4*b^4*c^4*d^8*f^4 - 16*a^5*b^3*c^3*d^9*f^4 + 4*a^6*b^2*c^2*d^10* f^4 - 16*a*b^7*c*d^11*f^4 - 16*a*b^7*c^3*d^9*f^4 - 32*a^3*b^5*c*d^11*f^4 - 16*a^5*b^3*c*d^11*f^4))/f^5 - (32*(-b*(a*d - b*c)^3)^(1/2)*(c + d*tan(e + f*x))^(1/2)*(16*b^9*d^10*f^4 + 16*a^2*b^7*d^10*f^4 - 16*a^4*b^5*d^10*f^4 - 16*a^6*b^3*d^10*f^4 + 24*b^9*c^2*d^8*f^4 + 40*a^2*b^7*c^2*d^8*f^4 + 8*a^ 4*b^5*c^2*d^8*f^4 - 8*a^6*b^3*c^2*d^8*f^4 + 8*a*b^8*c*d^9*f^4 + 24*a^3*b^6 *c*d^9*f^4 + 24*a^5*b^4*c*d^9*f^4 + 8*a^7*b^2*c*d^9*f^4))/(f^4*(b^3*f + a^ 2*b*f))))/(b^3*f + a^2*b*f) - (32*(c + d*tan(e + f*x))^(1/2)*(22*b^7*c*d^1 2*f^2 - 14*a*b^6*d^13*f^2 + 4*a^3*b^4*d^13*f^2 - 14*a^5*b^2*d^13*f^2 + 28* b^7*c^3*d^10*f^2 - 18*b^7*c^5*d^8*f^2 + 24*a^2*b^5*c^3*d^10*f^2 + 12*a^...