\(\int \frac {(a+b \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx\) [1306]

Optimal result

Integrand size = 29, antiderivative size = 273 \[ \int \frac {(a+b \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i (a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}+\frac {i (a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(c+i d)^{3/2} f}+\frac {2 b^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac {2 (b c-a d)^2 \sqrt {a+b \tan (e+f x)}}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}} \] Output:

-I*(a-I*b)^(5/2)*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)^(3/2)/f+I*(a+I*b)^(5/2)*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) 
^(3/2)/f+2*b^(5/2)*arctanh(d^(1/2)*(a+b*tan(f*x+e))^(1/2)/b^(1/2)/(c+d*tan 
(f*x+e))^(1/2))/d^(3/2)/f-2*(-a*d+b*c)^2*(a+b*tan(f*x+e))^(1/2)/d/(c^2+d^2 
)/f/(c+d*tan(f*x+e))^(1/2)
 

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1503\) vs. \(2(273)=546\).

Time = 6.33 (sec) , antiderivative size = 1503, normalized size of antiderivative = 5.51 \[ \int \frac {(a+b \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

Integrate[(a + b*Tan[e + f*x])^(5/2)/(c + d*Tan[e + f*x])^(3/2),x]
 

Output:

((-1/2*I)*(-a - I*b)*(-((-a - I*b)*((-2*Sqrt[a + I*b]*ArcTanh[(Sqrt[c + I* 
d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(( 
-c - I*d)*Sqrt[c + I*d]) + (2*Sqrt[a + b*Tan[e + f*x]])/((-c - I*d)*Sqrt[c 
 + d*Tan[e + f*x]]))) - (2*(b*c - a*d)*(b/((b^2*c)/(b*c - a*d) - (a*b*d)/( 
b*c - a*d)))^(3/2)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))^2*Sqrt[(b*( 
c + d*Tan[e + f*x]))/(b*c - a*d)]*(-1 - (b*d*(a + b*Tan[e + f*x]))/((b*c - 
 a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))))*(-((b*d*(a + b*Tan[e + 
 f*x]))/((b*c - a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))*(-1 - (b* 
d*(a + b*Tan[e + f*x]))/((b*c - a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - 
 a*d)))))) - (Sqrt[b]*Sqrt[d]*ArcSinh[(Sqrt[b]*Sqrt[d]*Sqrt[a + b*Tan[e + 
f*x]])/(Sqrt[b*c - a*d]*Sqrt[(b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)])]* 
Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b*c - a*d]*Sqrt[(b^2*c)/(b*c - a*d) - (a*b 
*d)/(b*c - a*d)]*Sqrt[1 + (b*d*(a + b*Tan[e + f*x]))/((b*c - a*d)*((b^2*c) 
/(b*c - a*d) - (a*b*d)/(b*c - a*d)))])))/(b*d^2*Sqrt[a + b*Tan[e + f*x]]*S 
qrt[c + d*Tan[e + f*x]]*Sqrt[1 + (b*d*(a + b*Tan[e + f*x]))/((b*c - a*d)*( 
(b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))])))/f - ((I/2)*(-a + I*b)*(-(( 
-a + I*b)*((-2*Sqrt[-a + I*b]*ArcTanh[(Sqrt[-c + I*d]*Sqrt[a + b*Tan[e + f 
*x]])/(Sqrt[-a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/((c - I*d)*Sqrt[-c + I*d 
]) + (2*Sqrt[a + b*Tan[e + f*x]])/((c - I*d)*Sqrt[c + d*Tan[e + f*x]]))) + 
 (2*(b*c - a*d)*(b/((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))^(3/2)*(...
 

Rubi [A] (verified)

Time = 1.36 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 4048, 27, 3042, 4138, 2348, 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.

\(\displaystyle \int \frac {(a+b \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 4048

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4138

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

\(\Big \downarrow \) 2348

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 (b c-a d)^2 \sqrt {a+b \tan (e+f x)}}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {2 b^{5/2} \left (c^2+d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d}}-\frac {d (a-i b)^{5/2} (-d+i c) \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c-i d}}+\frac {d (a+i b)^{5/2} (d+i c) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c+i d}}}{d f \left (c^2+d^2\right )}\)

Input:

Int[(a + b*Tan[e + f*x])^(5/2)/(c + d*Tan[e + f*x])^(3/2),x]
 

Output:

(-(((a - I*b)^(5/2)*(I*c - d)*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 - I*d]) + ((a + I 
*b)^(5/2)*d*(I*c + d)*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] + (2*b^(5/2)*(c^2 + 
d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e 
+ f*x]])])/Sqrt[d])/(d*(c^2 + d^2)*f) - (2*(b*c - a*d)^2*Sqrt[a + b*Tan[e 
+ f*x]])/(d*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

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

rule 2348

Int[(Px_)*((c_) + (d_.)*(x_))^(m_.)*((e_) + (f_.)*(x_))^(n_.)*((a_.) + (b_. 
)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(c + d*x)^m*(e + f*x)^ 
n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[P 
x, x] && (IntegerQ[p] || (IntegerQ[2*p] && IntegerQ[m] && ILtQ[n, 0])) && 
!(IGtQ[m, 0] && IGtQ[n, 0])
 

rule 3042

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

rule 4048

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m 
 - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1 
/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + 
f*x])^(n + 1)*Simp[a^2*d*(b*d*(m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c 
*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3*a*b^2*d) 
*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*( 
n + 1)))*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] && GtQ[m, 2] && LtQ 
[n, -1] && IntegerQ[2*m]
 

rule 4138

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), 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*((A + B*ff*x + C*ff^2*x^ 
2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f 
, A, B, C, 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.

\[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{\left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]

Input:

int((a+b*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x)
 

Output:

int((a+b*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 21161 vs. \(2 (213) = 426\).

Time = 42.78 (sec) , antiderivative size = 42383, normalized size of antiderivative = 155.25 \[ \int \frac {(a+b \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="fric 
as")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {(a+b \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((a+b*tan(f*x+e))**(5/2)/(c+d*tan(f*x+e))**(3/2),x)
 

Output:

Integral((a + b*tan(e + f*x))**(5/2)/(c + d*tan(e + f*x))**(3/2), x)
 

Maxima [F]

\[ \int \frac {(a+b \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((a+b*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxi 
ma")
 

Output:

integrate((b*tan(f*x + e) + a)^(5/2)/(d*tan(f*x + e) + c)^(3/2), x)
 

Giac [F]

\[ \int \frac {(a+b \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((a+b*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac 
")
 

Output:

integrate((b*tan(f*x + e) + a)^(5/2)/(d*tan(f*x + e) + c)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

int((a + b*tan(e + f*x))^(5/2)/(c + d*tan(e + f*x))^(3/2),x)
 

Output:

int((a + b*tan(e + f*x))^(5/2)/(c + d*tan(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {(a+b \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {too large to display} \] Input:

int((a+b*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),x)
 

Output:

( - 4*sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*a*b**2 + int((sqrt 
(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x)**2)/(tan(e + f* 
x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*tan(e + f*x)*a*b**2*d**2*f - in 
t((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x)**2)/(tan 
(e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*tan(e + f*x)*b**3*c*d*f 
+ int((sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x)**2)/ 
(tan(e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*a*b**2*c*d*f - int(( 
sqrt(tan(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x)**2)/(tan(e 
+ f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*b**3*c**2*f + 2*int((sqrt(t 
an(e + f*x)*d + c)*sqrt(tan(e + f*x)*b + a)*tan(e + f*x))/(tan(e + f*x)**3 
*b*d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b*c*d + 2*tan(e + f*x 
)*a*c*d + tan(e + f*x)*b*c**2 + a*c**2),x)*tan(e + f*x)*a**3*b*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*d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)**2*b*c*d + 2* 
tan(e + f*x)*a*c*d + tan(e + f*x)*b*c**2 + a*c**2),x)*tan(e + f*x)*a**2*b* 
*2*c*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*d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)* 
*2*b*c*d + 2*tan(e + f*x)*a*c*d + tan(e + f*x)*b*c**2 + a*c**2),x)*a**3*b* 
c*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*d**2 + tan(e + f*x)**2*a*d**2 + 2*tan(e + f*x)*...