\(\int (a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)} \, dx\) [1281]

Optimal result

Integrand size = 29, antiderivative size = 337 \[ \int (a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)} \, dx=-\frac {i (a-i b)^{5/2} \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {i (a+i b)^{5/2} \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {\sqrt {b} \left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{4 d^{3/2} f}-\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 d f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f} \] Output:

-I*(a-I*b)^(5/2)*(c-I*d)^(1/2)*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))/f+I*(a+I*b)^(5/2)*(c+I*d)^(1/2)*ar 
ctanh((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))/f+1/4*b^(1/2)*(10*a*b*c*d+15*a^2*d^2-b^2*(c^2+8*d^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-1/4*b* 
(-9*a*d+b*c)*(a+b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/d/f+1/2*b^2*(a+ 
b*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2)/d/f
 

Mathematica [A] (verified)

Time = 4.04 (sec) , antiderivative size = 565, normalized size of antiderivative = 1.68 \[ \int (a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)} \, dx=\frac {\frac {4 b d \left (b \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right )+\sqrt {-b^2} \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )\right ) \text {arctanh}\left (\frac {\sqrt {-c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+\sqrt {-b^2}} \sqrt {-c+\frac {\sqrt {-b^2} d}{b}}}-\frac {4 b d \left (b \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right )-\sqrt {-b^2} \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )\right ) \text {arctanh}\left (\frac {\sqrt {c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+\frac {\sqrt {-b^2} d}{b}}}+b^3 (-b c+9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}+2 b^4 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}-\frac {b^{5/2} \sqrt {c-\frac {a d}{b}} \left (-10 a b c d-15 a^2 d^2+b^2 \left (c^2+8 d^2\right )\right ) \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c-\frac {a d}{b}}}\right ) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}{\sqrt {d} \sqrt {c+d \tan (e+f x)}}}{4 b^2 d f} \] Input:

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

Output:

((4*b*d*(b*(3*a^2*b*c - b^3*c + a^3*d - 3*a*b^2*d) + Sqrt[-b^2]*(a^3*c - 3 
*a*b^2*c - 3*a^2*b*d + b^3*d))*ArcTanh[(Sqrt[-c + (Sqrt[-b^2]*d)/b]*Sqrt[a 
 + b*Tan[e + f*x]])/(Sqrt[-a + Sqrt[-b^2]]*Sqrt[c + d*Tan[e + f*x]])])/(Sq 
rt[-a + Sqrt[-b^2]]*Sqrt[-c + (Sqrt[-b^2]*d)/b]) - (4*b*d*(b*(3*a^2*b*c - 
b^3*c + a^3*d - 3*a*b^2*d) - Sqrt[-b^2]*(a^3*c - 3*a*b^2*c - 3*a^2*b*d + b 
^3*d))*ArcTanh[(Sqrt[c + (Sqrt[-b^2]*d)/b]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt 
[a + Sqrt[-b^2]]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[a + Sqrt[-b^2]]*Sqrt[c 
+ (Sqrt[-b^2]*d)/b]) + b^3*(-(b*c) + 9*a*d)*Sqrt[a + b*Tan[e + f*x]]*Sqrt[ 
c + d*Tan[e + f*x]] + 2*b^4*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^ 
(3/2) - (b^(5/2)*Sqrt[c - (a*d)/b]*(-10*a*b*c*d - 15*a^2*d^2 + b^2*(c^2 + 
8*d^2))*ArcSinh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c - (a*d) 
/b])]*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])/(Sqrt[d]*Sqrt[c + d*Tan[ 
e + f*x]]))/(4*b^2*d*f)
 

Rubi [A] (verified)

Time = 1.94 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 4049, 27, 3042, 4130, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

\(\displaystyle \frac {\int -\frac {\sqrt {c+d \tan (e+f x)} \left (-4 d a^3+3 b^2 d a+b^2 (b c-9 a d) \tan ^2(e+f x)+b^3 c-4 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )}{2 \sqrt {a+b \tan (e+f x)}}dx}{2 d}+\frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac {\int \frac {\sqrt {c+d \tan (e+f x)} \left (-4 d a^3+3 b^2 d a+b^2 (b c-9 a d) \tan ^2(e+f x)+b^3 c-4 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )}{\sqrt {a+b \tan (e+f x)}}dx}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac {\int \frac {\sqrt {c+d \tan (e+f x)} \left (-4 d a^3+3 b^2 d a+b^2 (b c-9 a d) \tan (e+f x)^2+b^3 c-4 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )}{\sqrt {a+b \tan (e+f x)}}dx}{4 d}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac {\frac {\int \frac {-b^2 \left (-\left (\left (c^2+8 d^2\right ) b^2\right )+10 a c d b+15 a^2 d^2\right ) \tan ^2(e+f x)-8 b d \left (d a^3+3 b c a^2-3 b^2 d a-b^3 c\right ) \tan (e+f x)+b \left (-8 c d a^3+9 b d^2 a^2+14 b^2 c d a+b^3 c^2\right )}{2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{b}+\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac {\frac {\int \frac {-b^2 \left (-\left (\left (c^2+8 d^2\right ) b^2\right )+10 a c d b+15 a^2 d^2\right ) \tan ^2(e+f x)-8 b d \left (d a^3+3 b c a^2-3 b^2 d a-b^3 c\right ) \tan (e+f x)+b \left (-8 c d a^3+9 b d^2 a^2+14 b^2 c d a+b^3 c^2\right )}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{2 b}+\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac {\frac {\int \frac {-b^2 \left (-\left (\left (c^2+8 d^2\right ) b^2\right )+10 a c d b+15 a^2 d^2\right ) \tan (e+f x)^2-8 b d \left (d a^3+3 b c a^2-3 b^2 d a-b^3 c\right ) \tan (e+f x)+b \left (-8 c d a^3+9 b d^2 a^2+14 b^2 c d a+b^3 c^2\right )}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{2 b}+\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}}{4 d}\)

\(\Big \downarrow \) 4138

\(\displaystyle \frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac {\frac {\int \frac {-b^2 \left (-\left (\left (c^2+8 d^2\right ) b^2\right )+10 a c d b+15 a^2 d^2\right ) \tan ^2(e+f x)-8 b d \left (d a^3+3 b c a^2-3 b^2 d a-b^3 c\right ) \tan (e+f x)+b \left (-8 c d a^3+9 b d^2 a^2+14 b^2 c d a+b^3 c^2\right )}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left (\tan ^2(e+f x)+1\right )}d\tan (e+f x)}{2 b f}+\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}}{4 d}\)

\(\Big \downarrow \) 2348

\(\displaystyle \frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac {\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}+\frac {\int \left (\frac {\left (\left (c^2+8 d^2\right ) b^2-10 a c d b-15 a^2 d^2\right ) b^2}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {-8 c d b^4-24 a d^2 b^3+24 a^2 c d b^2+8 a^3 d^2 b+i \left (-8 d^2 b^4+24 a c d b^3+24 a^2 d^2 b^2-8 a^3 c d b\right )}{2 (i-\tan (e+f x)) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {8 c d b^4+24 a d^2 b^3-24 a^2 c d b^2-8 a^3 d^2 b+i \left (-8 d^2 b^4+24 a c d b^3+24 a^2 d^2 b^2-8 a^3 c d b\right )}{2 (\tan (e+f x)+i) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\right )d\tan (e+f x)}{2 b f}}{4 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac {\frac {b (b c-9 a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}+\frac {-\frac {2 b^{3/2} \left (15 a^2 d^2+10 a b c d-\left (b^2 \left (c^2+8 d^2\right )\right )\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d}}+8 i b d (a-i b)^{5/2} \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )-8 i b d (a+i b)^{5/2} \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{2 b f}}{4 d}\)

Input:

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

Output:

(b^2*Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2))/(2*d*f) - (((8*I 
)*(a - I*b)^(5/2)*b*Sqrt[c - I*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]])] - (8*I)*(a + I*b)^(5/ 
2)*b*Sqrt[c + I*d]*d*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqr 
t[a + I*b]*Sqrt[c + d*Tan[e + f*x]])] - (2*b^(3/2)*(10*a*b*c*d + 15*a^2*d^ 
2 - b^2*(c^2 + 8*d^2))*ArcTanh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b] 
*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[d])/(2*b*f) + (b*(b*c - 9*a*d)*Sqrt[a + 
b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/f)/(4*d)
 

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 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 4130

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] :> Simp[C*(a + b*Tan[e + f*x])^m*((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 - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C 
*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* 
m*(b*c - a*d) - b*B*d*(m + n + 1))*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[a^2 + b^2, 0] && 
 NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

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.

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

Input:

int((a+b*tan(f*x+e))^(5/2)*(c+d*tan(f*x+e))^(1/2),x)
 

Output:

int((a+b*tan(f*x+e))^(5/2)*(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. 9870 vs. \(2 (267) = 534\).

Time = 12.11 (sec) , antiderivative size = 19801, normalized size of antiderivative = 58.76 \[ \int (a+b \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(f*x+e))^(5/2)*(c+d*tan(f*x+e))^(1/2),x, algorithm="fric 
as")
                                                                                    
                                                                                    
 

Output:

Too large to include
 

Sympy [F]

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

integrate((a+b*tan(f*x+e))**(5/2)*(c+d*tan(f*x+e))**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate((a+b*tan(f*x+e))^(5/2)*(c+d*tan(f*x+e))^(1/2),x, algorithm="maxi 
ma")
 

Output:

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

Giac [F]

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

integrate((a+b*tan(f*x+e))^(5/2)*(c+d*tan(f*x+e))^(1/2),x, algorithm="giac 
")
 

Output:

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

Mupad [F(-1)]

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

int((a + b*tan(e + f*x))^(5/2)*(c + d*tan(e + f*x))^(1/2),x)
 

Output:

int((a + b*tan(e + f*x))^(5/2)*(c + d*tan(e + f*x))^(1/2), x)
 

Reduce [F]

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

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