\(\int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx\) [1300]

Optimal result

Integrand size = 29, antiderivative size = 218 \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i (a-i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c-i d} f}+\frac {i (a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c+i d} f}+\frac {2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d} f} \] Output:

-I*(a-I*b)^(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))/(c-I*d)^(1/2)/f+I*(a+I*b)^(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))/(c+I*d) 
^(1/2)/f+2*b^(3/2)*arctanh(d^(1/2)*(a+b*tan(f*x+e))^(1/2)/b^(1/2)/(c+d*tan 
(f*x+e))^(1/2))/d^(1/2)/f
 

Mathematica [A] (verified)

Time = 1.28 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.35 \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {\frac {i (-a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-c+i d}}+\frac {i (a+i b)^{3/2} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right ) \sqrt {c+d \tan (e+f x)}+2 b \sqrt {c+i d} \sqrt {b c-a d} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b c-a d}}\right ) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}{\sqrt {c+i d} \sqrt {d} \sqrt {c+d \tan (e+f x)}}}{f} \] Input:

Integrate[(a + b*Tan[e + f*x])^(3/2)/Sqrt[c + d*Tan[e + f*x]],x]
 

Output:

((I*(-a + I*b)^(3/2)*ArcTanh[(Sqrt[-c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sq 
rt[-a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[-c + I*d] + (I*(a + I*b)^(3/ 
2)*Sqrt[d]*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b] 
*Sqrt[c + d*Tan[e + f*x]])]*Sqrt[c + d*Tan[e + f*x]] + 2*b*Sqrt[c + I*d]*S 
qrt[b*c - a*d]*ArcSinh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/Sqrt[b*c - a*d]] 
*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])/(Sqrt[c + I*d]*Sqrt[d]*Sqrt[c 
 + d*Tan[e + f*x]]))/f
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3042, 4058, 661, 2009}

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.

\(\displaystyle \int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}}dx\)

\(\Big \downarrow \) 4058

\(\displaystyle \frac {\int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)} \left (\tan ^2(e+f x)+1\right )}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 661

\(\displaystyle \frac {\int \left (\frac {b^2}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {a^2+2 b \tan (e+f x) a-b^2}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left (\tan ^2(e+f x)+1\right )}\right )d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d}}-\frac {i (a-i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c-i d}}+\frac {i (a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c+i d}}}{f}\)

Input:

Int[(a + b*Tan[e + f*x])^(3/2)/Sqrt[c + d*Tan[e + f*x]],x]
 

Output:

(((-I)*(a - I*b)^(3/2)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(S 
qrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[c - I*d] + (I*(a + I*b)^(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[c + I*d] + (2*b^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b 
*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[d])/f
 

Defintions of rubi rules used

rule 661

Int[((d_.) + (e_.)*(x_))^(m_)/(Sqrt[(f_.) + (g_.)*(x_)]*((a_.) + (c_.)*(x_) 
^2)), x_Symbol] :> Int[ExpandIntegrand[1/(Sqrt[d + e*x]*Sqrt[f + g*x]), (d 
+ e*x)^(m + 1/2)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && IGt 
Q[m + 1/2, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4058

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S 
imp[ff/f   Subst[Int[(a + b*ff*x)^m*((c + d*ff*x)^n/(1 + ff^2*x^2)), x], x, 
 Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a 
*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
 

Maple [F(-1)]

Timed out.

\[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{\sqrt {c +d \tan \left (f x +e \right )}}d x\]

Input:

int((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(1/2),x)
 

Output:

int((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(1/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 8376 vs. \(2 (162) = 324\).

Time = 8.20 (sec) , antiderivative size = 16813, normalized size of antiderivative = 77.12 \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="fric 
as")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \] Input:

integrate((a+b*tan(f*x+e))**(3/2)/(c+d*tan(f*x+e))**(1/2),x)
 

Output:

Integral((a + b*tan(e + f*x))**(3/2)/sqrt(c + d*tan(e + f*x)), x)
 

Maxima [F]

\[ \int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{\sqrt {d \tan \left (f x + e\right ) + c}} \,d x } \] Input:

integrate((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="maxi 
ma")
 

Output:

integrate((b*tan(f*x + e) + a)^(3/2)/sqrt(d*tan(f*x + e) + c), x)
 

Giac [F]

\[ \int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{\sqrt {d \tan \left (f x + e\right ) + c}} \,d x } \] Input:

integrate((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="giac 
")
 

Output:

integrate((b*tan(f*x + e) + a)^(3/2)/sqrt(d*tan(f*x + e) + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \] Input:

int((a + b*tan(e + f*x))^(3/2)/(c + d*tan(e + f*x))^(1/2),x)
 

Output:

int((a + b*tan(e + f*x))^(3/2)/(c + d*tan(e + f*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {(a+b \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \sqrt {\tan \left (f x +e \right ) b +a}\, \tan \left (f x +e \right )}{d \tan \left (f x +e \right )+c}d x \right ) b +\left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \sqrt {\tan \left (f x +e \right ) b +a}}{d \tan \left (f x +e \right )+c}d x \right ) a \] Input:

int((a+b*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(1/2),x)
 

Output:

int((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x))/(tan( 
e + f*x)*d + c),x)*b + int((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + 
 a))/(tan(e + f*x)*d + c),x)*a