3.13.44 \(\int \frac {(c+d \tan (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx\) [1244]

3.13.44.1 Optimal result

Integrand size = 27, antiderivative size = 195 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\frac {(c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(i a+b) f}-\frac {(c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(i a-b) f}-\frac {2 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{b^{3/2} \left (a^2+b^2\right ) f}+\frac {2 d^2 \sqrt {c+d \tan (e+f x)}}{b f} \]

output

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

3.13.44.2 Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.02 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\frac {b^{3/2} (-i a+b) (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+b^{3/2} (i a+b) (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )-2 (b c-a d)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )+2 \sqrt {b} \left (a^2+b^2\right ) d^2 \sqrt {c+d \tan (e+f x)}}{b^{3/2} \left (a^2+b^2\right ) f} \]

input

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

output

(b^(3/2)*((-I)*a + b)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqr 
t[c - I*d]] + b^(3/2)*(I*a + b)*(c + I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + 
 f*x]]/Sqrt[c + I*d]] - 2*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*Ta 
n[e + f*x]])/Sqrt[b*c - a*d]] + 2*Sqrt[b]*(a^2 + b^2)*d^2*Sqrt[c + d*Tan[e 
 + f*x]])/(b^(3/2)*(a^2 + b^2)*f)
 

3.13.44.3 Rubi [A] (warning: unable to verify)

Time = 1.58 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.97, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 4049, 27, 3042, 4136, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 d^2 \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\frac {(b c-a d)^3 \int \frac {\tan (e+f x)^2+1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx}{a^2+b^2}+\frac {\frac {b (a+i b) (c-i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {b (a-i b) (c+i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{a^2+b^2}}{b}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {2 d^2 \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\frac {(b c-a d)^3 \int \frac {1}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{f \left (a^2+b^2\right )}+\frac {\frac {b (a+i b) (c-i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {b (a-i b) (c+i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{a^2+b^2}}{b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 d^2 \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\frac {2 (b c-a d)^3 \int \frac {1}{a+\frac {b (c+d \tan (e+f x))}{d}-\frac {b c}{d}}d\sqrt {c+d \tan (e+f x)}}{d f \left (a^2+b^2\right )}+\frac {\frac {b (a+i b) (c-i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {b (a-i b) (c+i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{a^2+b^2}}{b}\)

\(\Big \downarrow \) 221

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

input

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

output

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

3.13.44.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 73

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]
 

rule 221

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]
 

rule 3042

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

rule 4020

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]
 

rule 4022

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]
 

rule 4049

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c 
+ d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) 
 Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n 
- 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ 
e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 
, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) 
)
 

rule 4117

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]
 

rule 4136

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) 
+ (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^ 
n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ 
(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ 
e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 
 C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] & 
&  !GtQ[n, 0] &&  !LeQ[n, -1]
 

3.13.44.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2801\) vs. \(2(165)=330\).

Time = 0.73 (sec) , antiderivative size = 2802, normalized size of antiderivative = 14.37

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2802\)
default	\(\text {Expression too large to display}\)	\(2802\)

input

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

output

2*d^2*(c+d*tan(f*x+e))^(1/2)/b/f+2/f*d/(a^2+b^2)/(2*(c^2+d^2)^(1/2)-2*c)^( 
1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(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)*a*c-6/f*d*b/(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*c^2-2/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*c-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)*(c^2 
+d^2)^(1/2)*a*c^2+1/4/f/d/(a^2+b^2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2) 
^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^ 
(1/2)*(c^2+d^2)^(1/2)*a*c^2+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)*(c^2+d^2)^(1/2)*a-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))*(c^2+d^2)^(1/2)*b+3/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))*a*c^2+6/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*c+3/4/f 
*d/(a^2+b^2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan 
(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c+1/f*d^2/(a...
 

3.13.44.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5642 vs. \(2 (159) = 318\).

Time = 41.93 (sec) , antiderivative size = 11304, normalized size of antiderivative = 57.97 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\text {Too large to display} \]

input

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

output

Too large to include
 

3.13.44.6 Sympy [F]

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

input

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

output

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

3.13.44.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\text {Exception raised: ValueError} \]

input

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

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(a*d-b*c>0)', see `assume?` for m 
ore detail
 

3.13.44.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.13.44.9 Mupad [B] (verification not implemented)

Time = 15.51 (sec) , antiderivative size = 27922, normalized size of antiderivative = 143.19 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\text {Too large to display} \]

input

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

output

(2*d^2*(c + d*tan(e + f*x))^(1/2))/(b*f) - atan(((((((32*(4*b^9*c*d^12*f^4 
 - 4*a*b^8*d^13*f^4 - 8*a^3*b^6*d^13*f^4 - 4*a^5*b^4*d^13*f^4 + 16*b^9*c^3 
*d^10*f^4 + 12*b^9*c^5*d^8*f^4 + 40*a^2*b^7*c^3*d^10*f^4 + 24*a^2*b^7*c^5* 
d^8*f^4 - 48*a^3*b^6*c^2*d^11*f^4 - 40*a^3*b^6*c^4*d^9*f^4 + 32*a^4*b^5*c^ 
3*d^10*f^4 + 12*a^4*b^5*c^5*d^8*f^4 - 24*a^5*b^4*c^2*d^11*f^4 - 20*a^5*b^4 
*c^4*d^9*f^4 + 8*a^6*b^3*c^3*d^10*f^4 - 24*a*b^8*c^2*d^11*f^4 - 20*a*b^8*c 
^4*d^9*f^4 + 16*a^2*b^7*c*d^12*f^4 + 20*a^4*b^5*c*d^12*f^4 + 8*a^6*b^3*c*d 
^12*f^4))/(b*f^5) - (32*(c + d*tan(e + f*x))^(1/2)*(-(c*d^4*5i + 5*c^4*d + 
 c^5*1i + d^5 - 10*c^2*d^3 - c^3*d^2*10i)/(4*(a^2*f^2*1i - b^2*f^2*1i + 2* 
a*b*f^2)))^(1/2)*(16*b^10*d^10*f^4 + 16*a^2*b^8*d^10*f^4 - 16*a^4*b^6*d^10 
*f^4 - 16*a^6*b^4*d^10*f^4 + 24*b^10*c^2*d^8*f^4 + 40*a^2*b^8*c^2*d^8*f^4 
+ 8*a^4*b^6*c^2*d^8*f^4 - 8*a^6*b^4*c^2*d^8*f^4 + 8*a*b^9*c*d^9*f^4 + 24*a 
^3*b^7*c*d^9*f^4 + 24*a^5*b^5*c*d^9*f^4 + 8*a^7*b^3*c*d^9*f^4))/(b*f^4))*( 
-(c*d^4*5i + 5*c^4*d + c^5*1i + d^5 - 10*c^2*d^3 - c^3*d^2*10i)/(4*(a^2*f^ 
2*1i - b^2*f^2*1i + 2*a*b*f^2)))^(1/2) + (32*(c + d*tan(e + f*x))^(1/2)*(1 
6*a^7*b*d^15*f^2 - 14*a*b^7*d^15*f^2 - 8*a^8*c*d^14*f^2 + 38*b^8*c*d^14*f^ 
2 + 4*a^3*b^5*d^15*f^2 + 2*a^5*b^3*d^15*f^2 - 10*b^8*c^3*d^12*f^2 - 102*b^ 
8*c^5*d^10*f^2 + 18*b^8*c^7*d^8*f^2 + 100*a^2*b^6*c^3*d^12*f^2 + 36*a^2*b^ 
6*c^5*d^10*f^2 - 12*a^2*b^6*c^7*d^8*f^2 - 60*a^3*b^5*c^2*d^13*f^2 + 140*a^ 
3*b^5*c^4*d^11*f^2 + 44*a^3*b^5*c^6*d^9*f^2 - 170*a^4*b^4*c^3*d^12*f^2 ...