3.13.47 \(\int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx\) [1247]

3.13.47.1 Optimal result

Integrand size = 27, antiderivative size = 248 \[ \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i (a-i b)^4 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d} f}+\frac {i (a+i b)^4 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d} f}-\frac {2 b^2 \left (40 a b c d-87 a^2 d^2-b^2 \left (8 c^2-15 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 d^3 f}-\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{15 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f} \]

output

-I*(a-I*b)^4*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f/(c-I*d)^(1/2) 
+I*(a+I*b)^4*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/f/(c+I*d)^(1/2) 
-2/15*b^2*(40*a*b*c*d-87*a^2*d^2-b^2*(8*c^2-15*d^2))*(c+d*tan(f*x+e))^(1/2 
)/d^3/f-4/15*b^3*(-7*a*d+2*b*c)*(c+d*tan(f*x+e))^(1/2)*tan(f*x+e)/d^2/f+2/ 
5*b^2*(c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2/d/f
 

3.13.47.2 Mathematica [A] (verified)

Time = 3.53 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.95 \[ \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {-\frac {15 i (a-i b)^4 d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {15 i (a+i b)^4 d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}+\frac {2 b^2 \left (-40 a b c d+87 a^2 d^2+b^2 \left (8 c^2-15 d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{d^2}+\frac {4 b^3 (-2 b c+7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{d}+6 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{15 d f} \]

input

Integrate[(a + b*Tan[e + f*x])^4/Sqrt[c + d*Tan[e + f*x]],x]
 

output

(((-15*I)*(a - I*b)^4*d*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/S 
qrt[c - I*d] + ((15*I)*(a + I*b)^4*d*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt 
[c + I*d]])/Sqrt[c + I*d] + (2*b^2*(-40*a*b*c*d + 87*a^2*d^2 + b^2*(8*c^2 
- 15*d^2))*Sqrt[c + d*Tan[e + f*x]])/d^2 + (4*b^3*(-2*b*c + 7*a*d)*Tan[e + 
 f*x]*Sqrt[c + d*Tan[e + f*x]])/d + 6*b^2*(a + b*Tan[e + f*x])^2*Sqrt[c + 
d*Tan[e + f*x]])/(15*d*f)
 

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

Time = 1.53 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.00, 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, 4120, 27, 3042, 4113, 3042, 4022, 3042, 4020, 25, 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.

\(\displaystyle \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}}dx\)

\(\Big \downarrow \) 4049

\(\displaystyle \frac {2 \int -\frac {(a+b \tan (e+f x)) \left (-5 d a^3+b^2 d a+2 b^2 (2 b c-7 a d) \tan ^2(e+f x)+4 b^3 c-5 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )}{2 \sqrt {c+d \tan (e+f x)}}dx}{5 d}+\frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\int \frac {(a+b \tan (e+f x)) \left (-5 d a^3+b^2 d a+2 b^2 (2 b c-7 a d) \tan ^2(e+f x)+4 b^3 c-5 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}}dx}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\int \frac {(a+b \tan (e+f x)) \left (-5 d a^3+b^2 d a+2 b^2 (2 b c-7 a d) \tan (e+f x)^2+4 b^3 c-5 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}}dx}{5 d}\)

\(\Big \downarrow \) 4120

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {2 \int \frac {15 d^2 a^4-3 b^2 d^2 a^2-40 b^3 c d a+60 b \left (a^2-b^2\right ) d^2 \tan (e+f x) a+8 b^4 c^2-b^2 \left (-\left (\left (8 c^2-15 d^2\right ) b^2\right )+40 a c d b-87 a^2 d^2\right ) \tan ^2(e+f x)}{2 \sqrt {c+d \tan (e+f x)}}dx}{3 d}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\int \frac {15 d^2 a^4-3 b^2 d^2 a^2-40 b^3 c d a+60 b \left (a^2-b^2\right ) d^2 \tan (e+f x) a+8 b^4 c^2-b^2 \left (-\left (\left (8 c^2-15 d^2\right ) b^2\right )+40 a c d b-87 a^2 d^2\right ) \tan ^2(e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\int \frac {15 d^2 a^4-3 b^2 d^2 a^2-40 b^3 c d a+60 b \left (a^2-b^2\right ) d^2 \tan (e+f x) a+8 b^4 c^2-b^2 \left (-\left (\left (8 c^2-15 d^2\right ) b^2\right )+40 a c d b-87 a^2 d^2\right ) \tan (e+f x)^2}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}}{5 d}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\int \frac {15 \left (a^4-6 b^2 a^2+b^4\right ) d^2+60 a b \left (a^2-b^2\right ) \tan (e+f x) d^2}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\int \frac {15 \left (a^4-6 b^2 a^2+b^4\right ) d^2+60 a b \left (a^2-b^2\right ) \tan (e+f x) d^2}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{5 d}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\frac {15}{2} d^2 (a-i b)^4 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {15}{2} d^2 (a+i b)^4 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\frac {15}{2} d^2 (a-i b)^4 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {15}{2} d^2 (a+i b)^4 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{5 d}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\frac {15 i d^2 (a-i b)^4 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {15 i d^2 (a+i b)^4 \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{5 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {-\frac {15 i d^2 (a-i b)^4 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {15 i d^2 (a+i b)^4 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{5 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {\frac {15 d (a-i b)^4 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}+\frac {15 d (a+i b)^4 \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}}{3 d}}{5 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {4 b^3 (2 b c-7 a d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {-\frac {2 b^2 \left (-87 a^2 d^2+40 a b c d-\left (b^2 \left (8 c^2-15 d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{d f}+\frac {15 d^2 (a-i b)^4 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {15 d^2 (a+i b)^4 \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}}{3 d}}{5 d}\)

input

Int[(a + b*Tan[e + f*x])^4/Sqrt[c + d*Tan[e + f*x]],x]
 

output

(2*b^2*(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]])/(5*d*f) - ((4*b^3* 
(2*b*c - 7*a*d)*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(3*d*f) - ((15*(a - 
 I*b)^4*d^2*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + (15*(a 
 + I*b)^4*d^2*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f) - (2*b 
^2*(40*a*b*c*d - 87*a^2*d^2 - b^2*(8*c^2 - 15*d^2))*Sqrt[c + d*Tan[e + f*x 
]])/(d*f))/(3*d))/(5*d)
 

3.13.47.3.1 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 4113

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

rule 4120

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

3.13.47.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(5513\) vs. \(2(216)=432\).

Time = 1.14 (sec) , antiderivative size = 5514, normalized size of antiderivative = 22.23

method	result	size
parts	\(\text {Expression too large to display}\)	\(5514\)
derivativedivides	\(\text {Expression too large to display}\)	\(10033\)
default	\(\text {Expression too large to display}\)	\(10033\)

input

int((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
 

output

result too large to display
 

3.13.47.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4951 vs. \(2 (208) = 416\).

Time = 0.73 (sec) , antiderivative size = 4951, normalized size of antiderivative = 19.96 \[ \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]

input

integrate((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(1/2),x, algorithm="fricas")
 

output

1/30*(15*d^3*f*sqrt(-((c^2 + d^2)*f^2*sqrt(-(64*(a^14*b^2 - 14*a^12*b^4 + 
63*a^10*b^6 - 100*a^8*b^8 + 63*a^6*b^10 - 14*a^4*b^12 + a^2*b^14)*c^2 - 16 
*(a^15*b - 35*a^13*b^3 + 273*a^11*b^5 - 715*a^9*b^7 + 715*a^7*b^9 - 273*a^ 
5*b^11 + 35*a^3*b^13 - a*b^15)*c*d + (a^16 - 56*a^14*b^2 + 924*a^12*b^4 - 
3976*a^10*b^6 + 6470*a^8*b^8 - 3976*a^6*b^10 + 924*a^4*b^12 - 56*a^2*b^14 
+ b^16)*d^2)/((c^4 + 2*c^2*d^2 + d^4)*f^4)) + (a^8 - 28*a^6*b^2 + 70*a^4*b 
^4 - 28*a^2*b^6 + b^8)*c + 8*(a^7*b - 7*a^5*b^3 + 7*a^3*b^5 - a*b^7)*d)/(( 
c^2 + d^2)*f^2))*log(-(8*(a^15*b - 3*a^13*b^3 - 15*a^11*b^5 - 11*a^9*b^7 + 
 11*a^7*b^9 + 15*a^5*b^11 + 3*a^3*b^13 - a*b^15)*c - (a^16 - 24*a^14*b^2 - 
 36*a^12*b^4 + 88*a^10*b^6 + 198*a^8*b^8 + 88*a^6*b^10 - 36*a^4*b^12 - 24* 
a^2*b^14 + b^16)*d)*sqrt(d*tan(f*x + e) + c) + (((a^4 - 6*a^2*b^2 + b^4)*c 
^3 + 4*(a^3*b - a*b^3)*c^2*d + (a^4 - 6*a^2*b^2 + b^4)*c*d^2 + 4*(a^3*b - 
a*b^3)*d^3)*f^3*sqrt(-(64*(a^14*b^2 - 14*a^12*b^4 + 63*a^10*b^6 - 100*a^8* 
b^8 + 63*a^6*b^10 - 14*a^4*b^12 + a^2*b^14)*c^2 - 16*(a^15*b - 35*a^13*b^3 
 + 273*a^11*b^5 - 715*a^9*b^7 + 715*a^7*b^9 - 273*a^5*b^11 + 35*a^3*b^13 - 
 a*b^15)*c*d + (a^16 - 56*a^14*b^2 + 924*a^12*b^4 - 3976*a^10*b^6 + 6470*a 
^8*b^8 - 3976*a^6*b^10 + 924*a^4*b^12 - 56*a^2*b^14 + b^16)*d^2)/((c^4 + 2 
*c^2*d^2 + d^4)*f^4)) + (32*(a^10*b^2 - 8*a^8*b^4 + 14*a^6*b^6 - 8*a^4*b^8 
 + a^2*b^10)*c^2 - 4*(3*a^11*b - 55*a^9*b^3 + 198*a^7*b^5 - 198*a^5*b^7 + 
55*a^3*b^9 - 3*a*b^11)*c*d + (a^12 - 34*a^10*b^2 + 239*a^8*b^4 - 476*a^...
 

3.13.47.6 Sympy [F]

\[ \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{4}}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \]

input

integrate((a+b*tan(f*x+e))**4/(c+d*tan(f*x+e))**(1/2),x)
 

output

Integral((a + b*tan(e + f*x))**4/sqrt(c + d*tan(e + f*x)), x)
 

3.13.47.7 Maxima [F]

\[ \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{4}}{\sqrt {d \tan \left (f x + e\right ) + c}} \,d x } \]

input

integrate((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(1/2),x, algorithm="maxima")
 

output

integrate((b*tan(f*x + e) + a)^4/sqrt(d*tan(f*x + e) + c), x)
 

3.13.47.8 Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Timed out} \]

input

integrate((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(1/2),x, algorithm="giac")
 

output

Timed out
 

3.13.47.9 Mupad [B] (verification not implemented)

Time = 21.01 (sec) , antiderivative size = 3771, normalized size of antiderivative = 15.21 \[ \int \frac {(a+b \tan (e+f x))^4}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]

input

int((a + b*tan(e + f*x))^4/(c + d*tan(e + f*x))^(1/2),x)
 

output

atan(((((32*(a^4*d^3*f^2 + b^4*d^3*f^2 - 6*a^2*b^2*d^3*f^2 + 4*a*b^3*c*d^2 
*f^2 - 4*a^3*b*c*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(a* 
b^7*8i - a^7*b*8i + a^8 + b^8 - 28*a^2*b^6 - a^3*b^5*56i + 70*a^4*b^4 + a^ 
5*b^3*56i - 28*a^6*b^2)/(4*(c*f^2 - d*f^2*1i)))^(1/2))*(-(a*b^7*8i - a^7*b 
*8i + a^8 + b^8 - 28*a^2*b^6 - a^3*b^5*56i + 70*a^4*b^4 + a^5*b^3*56i - 28 
*a^6*b^2)/(4*(c*f^2 - d*f^2*1i)))^(1/2) - (16*(c + d*tan(e + f*x))^(1/2)*( 
a^8*d^2 + b^8*d^2 - 28*a^2*b^6*d^2 + 70*a^4*b^4*d^2 - 28*a^6*b^2*d^2))/f^2 
)*(-(a*b^7*8i - a^7*b*8i + a^8 + b^8 - 28*a^2*b^6 - a^3*b^5*56i + 70*a^4*b 
^4 + a^5*b^3*56i - 28*a^6*b^2)/(4*(c*f^2 - d*f^2*1i)))^(1/2)*1i - (((32*(a 
^4*d^3*f^2 + b^4*d^3*f^2 - 6*a^2*b^2*d^3*f^2 + 4*a*b^3*c*d^2*f^2 - 4*a^3*b 
*c*d^2*f^2))/f^3 + 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(a*b^7*8i - a^7*b 
*8i + a^8 + b^8 - 28*a^2*b^6 - a^3*b^5*56i + 70*a^4*b^4 + a^5*b^3*56i - 28 
*a^6*b^2)/(4*(c*f^2 - d*f^2*1i)))^(1/2))*(-(a*b^7*8i - a^7*b*8i + a^8 + b^ 
8 - 28*a^2*b^6 - a^3*b^5*56i + 70*a^4*b^4 + a^5*b^3*56i - 28*a^6*b^2)/(4*( 
c*f^2 - d*f^2*1i)))^(1/2) + (16*(c + d*tan(e + f*x))^(1/2)*(a^8*d^2 + b^8* 
d^2 - 28*a^2*b^6*d^2 + 70*a^4*b^4*d^2 - 28*a^6*b^2*d^2))/f^2)*(-(a*b^7*8i 
- a^7*b*8i + a^8 + b^8 - 28*a^2*b^6 - a^3*b^5*56i + 70*a^4*b^4 + a^5*b^3*5 
6i - 28*a^6*b^2)/(4*(c*f^2 - d*f^2*1i)))^(1/2)*1i)/((((32*(a^4*d^3*f^2 + b 
^4*d^3*f^2 - 6*a^2*b^2*d^3*f^2 + 4*a*b^3*c*d^2*f^2 - 4*a^3*b*c*d^2*f^2))/f 
^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(a*b^7*8i - a^7*b*8i + a^8 +...