\(\int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\) [465]

Optimal result

Integrand size = 35, antiderivative size = 244 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 \left (5 a^2 A b-A b^3-2 a^3 B+4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \] Output:

-(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I* 
a-b)^(5/2)/d+(I*A+B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c 
))^(1/2))/(I*a+b)^(5/2)/d-2/3*(A*b-B*a)*tan(d*x+c)^(1/2)/(a^2+b^2)/d/(a+b* 
tan(d*x+c))^(3/2)-2/3*(5*A*a^2*b-A*b^3-2*B*a^3+4*B*a*b^2)*tan(d*x+c)^(1/2) 
/a/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 2.39 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.31 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {\frac {2 b (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}}+\frac {6 b \left (2 a A b-a^2 B+b^2 B\right ) \tan ^{\frac {3}{2}}(c+d x)}{\left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (-\frac {\sqrt [4]{-1} a (a+i b)^2 (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {\sqrt [4]{-1} a (a-i b)^2 (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+2 \left (-2 a A b+a^2 B-b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}\right )}{a^2+b^2}}{3 a \left (a^2+b^2\right ) d} \] Input:

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

Output:

((2*b*(A*b - a*B)*Tan[c + d*x]^(3/2))/(a + b*Tan[c + d*x])^(3/2) + (6*b*(2 
*a*A*b - a^2*B + b^2*B)*Tan[c + d*x]^(3/2))/((a^2 + b^2)*Sqrt[a + b*Tan[c 
+ d*x]]) + (3*(-(((-1)^(1/4)*a*(a + I*b)^2*(A - I*B)*ArcTan[((-1)^(1/4)*Sq 
rt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] 
) + ((-1)^(1/4)*a*(a - I*b)^2*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*S 
qrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[a + I*b] + 2*(-2*a*A*b 
+ a^2*B - b^2*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]))/(a^2 + b^2) 
)/(3*a*(a^2 + b^2)*d)
 

Rubi [A] (verified)

Time = 1.48 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.22, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 4091, 27, 3042, 4132, 27, 3042, 4099, 3042, 4098, 104, 216, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4091

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4099

\(\displaystyle -\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {2 b \left (-2 a^3 B+5 a^2 A b+4 a b^2 B-A b^3\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {1}{2} a b (a-i b)^2 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a b (a+i b)^2 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {2 b \left (-2 a^3 B+5 a^2 A b+4 a b^2 B-A b^3\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {1}{2} a b (a-i b)^2 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a b (a+i b)^2 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 4098

\(\displaystyle -\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {2 b \left (-2 a^3 B+5 a^2 A b+4 a b^2 B-A b^3\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {a b (a-i b)^2 (-B+i A) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}-\frac {a b (a+i b)^2 (B+i A) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}\right )}{a \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {2 b \left (-2 a^3 B+5 a^2 A b+4 a b^2 B-A b^3\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {a b (a-i b)^2 (-B+i A) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {a b (a+i b)^2 (B+i A) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {2 b \left (-2 a^3 B+5 a^2 A b+4 a b^2 B-A b^3\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {a b (a-i b)^2 (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {a b (a+i b)^2 (B+i A) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {2 (A b-a B) \sqrt {\tan (c+d x)}}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {2 b \left (-2 a^3 B+5 a^2 A b+4 a b^2 B-A b^3\right ) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {a b (a-i b)^2 (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {a b (a+i b)^2 (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}\right )}{a \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\)

Input:

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

Output:

(-2*(A*b - a*B)*Sqrt[Tan[c + d*x]])/(3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^ 
(3/2)) + ((3*((a*(a - I*b)^2*b*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c 
+ d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d) - (a*(a + I*b)^2*b*( 
I*A + B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x 
]]])/(Sqrt[I*a + b]*d)))/(a*(a^2 + b^2)) - (2*b*(5*a^2*A*b - A*b^3 - 2*a^3 
*B + 4*a*b^2*B)*Sqrt[Tan[c + d*x]])/(a*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d* 
x]]))/(3*b*(a^2 + b^2))
 

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 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 4091

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

rule 4098

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

rule 4099

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

rule 4132

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[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + 
 f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + 
b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2))   Int[(a + b*Tan[e + 
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* 
(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d 
)*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta 
n[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] && LtQ[m, -1] && 
!(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 2.36 (sec) , antiderivative size = 2975224, normalized size of antiderivative = 12193.54

\[\text {output too large to display}\]

Input:

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

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 27593 vs. \(2 (200) = 400\).

Time = 13.58 (sec) , antiderivative size = 27593, normalized size of antiderivative = 113.09 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algo 
rithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algo 
rithm="maxima")
 

Output:

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

Giac [F]

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

integrate(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algo 
rithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a))/(tan(c + d*x)**2*b**2 + 
2*tan(c + d*x)*a*b + a**2),x)