3.4.44 \(\int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx\) [344]

3.4.44.1 Optimal result

Integrand size = 33, antiderivative size = 166 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b} d}-\frac {(i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b} d}+\frac {2 (3 A b-2 a B) \sqrt {a+b \tan (c+d x)}}{3 b^2 d}+\frac {2 B \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d} \]

output

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

3.4.44.2 Mathematica [A] (verified)

Time = 1.69 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.84 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {\frac {3 (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b}}+\frac {3 (-i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b}}+\frac {2 \sqrt {a+b \tan (c+d x)} (3 A b-2 a B+b B \tan (c+d x))}{b^2}}{3 d} \]

input

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

output

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

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

Time = 0.83 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.95, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4090, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4090

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4113

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 B \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\frac {3 i b (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {3 i b (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {2 (3 A b-2 a B) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 B \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {-\frac {3 i b (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {3 i b (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {2 (3 A b-2 a B) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}\)

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

input

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

output

(2*B*Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(3*b*d) - ((3*b*(A - I*B)*ArcT 
an[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + (3*b*(A + I*B)*ArcTan[ 
Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d) - (2*(3*A*b - 2*a*B)*Sqrt[a 
 + b*Tan[c + d*x]])/(b*d))/(3*b)
 

3.4.44.3.1 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 4090

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[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + 
 n))), x] + Simp[1/(d*(m + n))   Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Ta 
n[e + f*x])^n*Simp[a^2*A*d*(m + n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m 
 + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m - 1) - b 
*(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, 
 f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2 
, 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 1] 
&& ( !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]
 

3.4.44.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1948\) vs. \(2(140)=280\).

Time = 0.12 (sec) , antiderivative size = 1949, normalized size of antiderivative = 11.74

method	result	size
parts	\(\text {Expression too large to display}\)	\(1949\)
derivativedivides	\(\text {Expression too large to display}\)	\(4040\)
default	\(\text {Expression too large to display}\)	\(4040\)

input

int(tan(d*x+c)^2*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x,method=_RETURNV 
ERBOSE)
 

output

A*(2/d/b*(a+b*tan(d*x+c))^(1/2)-1/4/d/b/(a^2+b^2)*ln(b*tan(d*x+c)+a+(a+b*t 
an(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^ 
2)^(1/2)+2*a)^(1/2)*a^2-1/4/d*b/(a^2+b^2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c 
))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2) 
+2*a)^(1/2)+1/4/d/b/(a^2+b^2)^(3/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/ 
2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^ 
(1/2)*a^3+1/4/d*b/(a^2+b^2)^(3/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2) 
*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1 
/2)*a+1/d/b/(a^2+b^2)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*t 
an(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1 
/2))*a^2+1/d*b/(a^2+b^2)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+ 
b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a) 
^(1/2))-1/d/b/(a^2+b^2)^(3/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b 
*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^ 
(1/2))*a^4-3/d*b/(a^2+b^2)^(3/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*( 
a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2* 
a)^(1/2))*a^2-2/d*b^3/(a^2+b^2)^(3/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan 
((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/ 
2)-2*a)^(1/2))+1/4/d/b/(a^2+b^2)*ln(b*tan(d*x+c)+a-(a+b*tan(d*x+c))^(1/2)* 
(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^...
 

3.4.44.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1725 vs. \(2 (134) = 268\).

Time = 0.31 (sec) , antiderivative size = 1725, normalized size of antiderivative = 10.39 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]

input

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

output

1/6*(3*b^2*d*sqrt(-((a^2 + b^2)*d^2*sqrt(-(4*A^2*B^2*a^2 - 4*(A^3*B - A*B^ 
3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + 2*A 
*B*b + (A^2 - B^2)*a)/((a^2 + b^2)*d^2))*log((2*(A^3*B + A*B^3)*a - (A^4 - 
 B^4)*b)*sqrt(b*tan(d*x + c) + a) + ((A*a^3 + B*a^2*b + A*a*b^2 + B*b^3)*d 
^3*sqrt(-(4*A^2*B^2*a^2 - 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)* 
b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + (2*A*B^2*a^2 - (3*A^2*B - B^3)*a*b + 
 (A^3 - A*B^2)*b^2)*d)*sqrt(-((a^2 + b^2)*d^2*sqrt(-(4*A^2*B^2*a^2 - 4*(A^ 
3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d 
^4)) + 2*A*B*b + (A^2 - B^2)*a)/((a^2 + b^2)*d^2))) - 3*b^2*d*sqrt(-((a^2 
+ b^2)*d^2*sqrt(-(4*A^2*B^2*a^2 - 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 
 + B^4)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + 2*A*B*b + (A^2 - B^2)*a)/((a 
^2 + b^2)*d^2))*log((2*(A^3*B + A*B^3)*a - (A^4 - B^4)*b)*sqrt(b*tan(d*x + 
 c) + a) - ((A*a^3 + B*a^2*b + A*a*b^2 + B*b^3)*d^3*sqrt(-(4*A^2*B^2*a^2 - 
 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/((a^4 + 2*a^2*b^2 + 
b^4)*d^4)) + (2*A*B^2*a^2 - (3*A^2*B - B^3)*a*b + (A^3 - A*B^2)*b^2)*d)*sq 
rt(-((a^2 + b^2)*d^2*sqrt(-(4*A^2*B^2*a^2 - 4*(A^3*B - A*B^3)*a*b + (A^4 - 
 2*A^2*B^2 + B^4)*b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + 2*A*B*b + (A^2 - B 
^2)*a)/((a^2 + b^2)*d^2))) - 3*b^2*d*sqrt(((a^2 + b^2)*d^2*sqrt(-(4*A^2*B^ 
2*a^2 - 4*(A^3*B - A*B^3)*a*b + (A^4 - 2*A^2*B^2 + B^4)*b^2)/((a^4 + 2*a^2 
*b^2 + b^4)*d^4)) - 2*A*B*b - (A^2 - B^2)*a)/((a^2 + b^2)*d^2))*log((2*...
 

3.4.44.6 Sympy [F]

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

input

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

output

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

3.4.44.7 Maxima [F]

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

input

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

output

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

3.4.44.8 Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.4.44.9 Mupad [B] (verification not implemented)

Time = 11.91 (sec) , antiderivative size = 2981, normalized size of antiderivative = 17.96 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]

input

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

output

2*atanh((32*A^2*b^2*((-16*A^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b^2*d^4)) - (A 
^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((16* 
A^3*a*b^3*d^3)/(a^2*d^4 + b^2*d^4) - (4*A*b^3*d^2*(-16*A^4*b^2*d^4)^(1/2)) 
/(a^2*d^5 + b^2*d^5)) + (8*a*b^2*((-16*A^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b 
^2*d^4)) - (A^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)))^(1/2)*(a + b*tan(c + d*x)) 
^(1/2)*(-16*A^4*b^2*d^4)^(1/2))/((16*A^3*a*b^5*d^5)/(a^2*d^4 + b^2*d^4) - 
(4*A*b^5*d^4*(-16*A^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5) + (16*A^3*a^3*b^ 
3*d^5)/(a^2*d^4 + b^2*d^4) - (4*A*a^2*b^3*d^4*(-16*A^4*b^2*d^4)^(1/2))/(a^ 
2*d^5 + b^2*d^5)) - (32*A^2*a^2*b^2*d^2*((-16*A^4*b^2*d^4)^(1/2)/(16*(a^2* 
d^4 + b^2*d^4)) - (A^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)))^(1/2)*(a + b*tan(c 
+ d*x))^(1/2))/((16*A^3*a*b^5*d^5)/(a^2*d^4 + b^2*d^4) - (4*A*b^5*d^4*(-16 
*A^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5) + (16*A^3*a^3*b^3*d^5)/(a^2*d^4 + 
 b^2*d^4) - (4*A*a^2*b^3*d^4*(-16*A^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5)) 
)*((-16*A^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 + b^2*d^4)) - (A^2*a*d^2)/(4*(a^2* 
d^4 + b^2*d^4)))^(1/2) - atan((a*b^2*((-16*B^4*b^2*d^4)^(1/2)/(16*(a^2*d^4 
 + b^2*d^4)) + (B^2*a*d^2)/(4*(a^2*d^4 + b^2*d^4)))^(1/2)*(a + b*tan(c + d 
*x))^(1/2)*(-16*B^4*b^2*d^4)^(1/2)*8i)/((16*B^3*a^2*b^4*d^5)/(a^2*d^4 + b^ 
2*d^4) - 16*B^3*a^2*b^2*d - 16*B^3*b^4*d + (16*B^3*a^4*b^2*d^5)/(a^2*d^4 + 
 b^2*d^4) + (4*B*a^3*b^2*d^4*(-16*B^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5) 
+ (4*B*a*b^4*d^4*(-16*B^4*b^2*d^4)^(1/2))/(a^2*d^5 + b^2*d^5)) - (B^2*b...