Integrand size = 29, antiderivative size = 213 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{3/2}} \, dx=-\frac {i (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{3/2} f}+\frac {i (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2} f}-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{\left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}} \] Output:
-I*(c-I*d)^(3/2)*arctanh((c-I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a-I*b)^(1/2 )/(c+d*tan(f*x+e))^(1/2))/(a-I*b)^(3/2)/f+I*(c+I*d)^(3/2)*arctanh((c+I*d)^ (1/2)*(a+b*tan(f*x+e))^(1/2)/(a+I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))/(a+I*b) ^(3/2)/f-2*(-a*d+b*c)*(c+d*tan(f*x+e))^(1/2)/(a^2+b^2)/f/(a+b*tan(f*x+e))^ (1/2)
Time = 1.32 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.08 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{3/2}} \, dx=\frac {-\frac {i (-c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(-a+i b)^{3/2}}+\frac {\frac {(i a+b) (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2}}+\frac {2 (-b c+a d) \sqrt {c+d \tan (e+f x)}}{(a+i b) \sqrt {a+b \tan (e+f x)}}}{a-i b}}{f} \] Input:
Integrate[(c + d*Tan[e + f*x])^(3/2)/(a + b*Tan[e + f*x])^(3/2),x]
Output:
(((-I)*(-c + I*d)^(3/2)*ArcTanh[(Sqrt[-c + I*d]*Sqrt[a + b*Tan[e + f*x]])/ (Sqrt[-a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(-a + I*b)^(3/2) + (((I*a + b) *(c + I*d)^(3/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(a + I*b)^(3/2) + (2*(-(b*c) + a*d)*Sqr t[c + d*Tan[e + f*x]])/((a + I*b)*Sqrt[a + b*Tan[e + f*x]]))/(a - I*b))/f
Time = 1.19 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 4050, 27, 3042, 4099, 3042, 4098, 104, 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.
\(\displaystyle \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 4050 |
\(\displaystyle -\frac {2 \int -\frac {2 b c d+a \left (c^2-d^2\right )+\left (2 a c d-b \left (c^2-d^2\right )\right ) \tan (e+f x)}{2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {2 b c d+a \left (c^2-d^2\right )+\left (2 a c d-b \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {2 b c d+a \left (c^2-d^2\right )+\left (2 a c d-b \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}-\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}\) |
\(\Big \downarrow \) 4099 |
\(\displaystyle -\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}+\frac {\frac {1}{2} (a+i b) (c-i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b) (c+i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}+\frac {\frac {1}{2} (a+i b) (c-i d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b) (c+i d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}\) |
\(\Big \downarrow \) 4098 |
\(\displaystyle -\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}+\frac {\frac {(a+i b) (c-i d)^2 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{2 f}+\frac {(a-i b) (c+i d)^2 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{2 f}}{a^2+b^2}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle -\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}+\frac {\frac {(a+i b) (c-i d)^2 \int \frac {1}{i a+b-\frac {(i c+d) (a+b \tan (e+f x))}{c+d \tan (e+f x)}}d\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{f}+\frac {(a-i b) (c+i d)^2 \int \frac {1}{-i a+b+\frac {(i c-d) (a+b \tan (e+f x))}{c+d \tan (e+f x)}}d\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{f}}{a^2+b^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 (b c-a d) \sqrt {c+d \tan (e+f x)}}{f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}+\frac {\frac {i (a-i b) (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a+i b}}-\frac {i (a+i b) (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a-i b}}}{a^2+b^2}\) |
Input:
Int[(c + d*Tan[e + f*x])^(3/2)/(a + b*Tan[e + f*x])^(3/2),x]
Output:
(((-I)*(a + I*b)*(c - I*d)^(3/2)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[a - I*b]*f) + (I* (a - I*b)*(c + I*d)^(3/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]]) /(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[a + I*b]*f))/(a^2 + b^2) - (2*(b*c - a*d)*Sqrt[c + d*Tan[e + f*x]])/((a^2 + b^2)*f*Sqrt[a + b*Tan[ e + f*x]])
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_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*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_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n - 1)/(f*(m + 1)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(a^2 + b^2)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^ (n - 2)*Simp[a*c^2*(m + 1) + a*d^2*(n - 1) + b*c*d*(m - n + 2) - (b*c^2 - 2 *a*c*d - b*d^2)*(m + 1)*Tan[e + f*x] - d*(b*c - a*d)*(m + n)*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] && LtQ[m, -1] && LtQ[1, n, 2] && IntegerQ[ 2*m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[A^2/f Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e + f* x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 + B^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(A + I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T an[e + f*x]), x], x] + Simp[(A - I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d *Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A , B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2 + B^2, 0]
Timed out.
\[\int \frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]
Input:
int((c+d*tan(f*x+e))^(3/2)/(a+b*tan(f*x+e))^(3/2),x)
Output:
int((c+d*tan(f*x+e))^(3/2)/(a+b*tan(f*x+e))^(3/2),x)
Leaf count of result is larger than twice the leaf count of optimal. 14552 vs. \(2 (165) = 330\).
Time = 9.33 (sec) , antiderivative size = 14552, normalized size of antiderivative = 68.32 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate((c+d*tan(f*x+e))^(3/2)/(a+b*tan(f*x+e))^(3/2),x, algorithm="fric as")
Output:
Too large to include
\[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{3/2}} \, dx=\int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((c+d*tan(f*x+e))**(3/2)/(a+b*tan(f*x+e))**(3/2),x)
Output:
Integral((c + d*tan(e + f*x))**(3/2)/(a + b*tan(e + f*x))**(3/2), x)
Exception generated. \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((c+d*tan(f*x+e))^(3/2)/(a+b*tan(f*x+e))^(3/2),x, algorithm="maxi ma")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(((2*b*d+2*a*c)^2>0)', see `assum e?` for mo
\[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{3/2}} \, dx=\int { \frac {{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((c+d*tan(f*x+e))^(3/2)/(a+b*tan(f*x+e))^(3/2),x, algorithm="giac ")
Output:
integrate((d*tan(f*x + e) + c)^(3/2)/(b*tan(f*x + e) + a)^(3/2), x)
Timed out. \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{3/2}} \, dx=\int \frac {{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:
int((c + d*tan(e + f*x))^(3/2)/(a + b*tan(e + f*x))^(3/2),x)
Output:
int((c + d*tan(e + f*x))^(3/2)/(a + b*tan(e + f*x))^(3/2), x)
\[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+b \tan (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:
int((c+d*tan(f*x+e))^(3/2)/(a+b*tan(f*x+e))^(3/2),x)
Output:
(2*sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*d**2 + 2*int((sqrt(ta n(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x))/(tan(e + f*x)**3* b**2*d + 2*tan(e + f*x)**2*a*b*d + tan(e + f*x)**2*b**2*c + tan(e + f*x)*a **2*d + 2*tan(e + f*x)*a*b*c + a**2*c),x)*tan(e + f*x)*a*b*c*d**2*f - 2*in t((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x))/(tan(e + f*x)**3*b**2*d + 2*tan(e + f*x)**2*a*b*d + tan(e + f*x)**2*b**2*c + tan( e + f*x)*a**2*d + 2*tan(e + f*x)*a*b*c + a**2*c),x)*tan(e + f*x)*b**2*c**2 *d*f + 2*int((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f* x))/(tan(e + f*x)**3*b**2*d + 2*tan(e + f*x)**2*a*b*d + tan(e + f*x)**2*b* *2*c + tan(e + f*x)*a**2*d + 2*tan(e + f*x)*a*b*c + a**2*c),x)*a**2*c*d**2 *f - 2*int((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x) )/(tan(e + f*x)**3*b**2*d + 2*tan(e + f*x)**2*a*b*d + tan(e + f*x)**2*b**2 *c + tan(e + f*x)*a**2*d + 2*tan(e + f*x)*a*b*c + a**2*c),x)*a*b*c**2*d*f + int((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a))/(tan(e + f*x)**3 *b**2*d + 2*tan(e + f*x)**2*a*b*d + tan(e + f*x)**2*b**2*c + tan(e + f*x)* a**2*d + 2*tan(e + f*x)*a*b*c + a**2*c),x)*tan(e + f*x)*a*b*c**2*d*f - int ((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a))/(tan(e + f*x)**3*b**2 *d + 2*tan(e + f*x)**2*a*b*d + tan(e + f*x)**2*b**2*c + tan(e + f*x)*a**2* d + 2*tan(e + f*x)*a*b*c + a**2*c),x)*tan(e + f*x)*a*b*d**3*f - int((sqrt( tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a))/(tan(e + f*x)**3*b**2*d +...