3.4.33 \(\int \tan (c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [333]

3.4.33.1 Optimal result

Integrand size = 31, antiderivative size = 213 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {(a-i b)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{5/2} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 \left (a^2 A-A b^2-2 a b B\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a A-b B) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 A (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 B (a+b \tan (c+d x))^{7/2}}{7 b d} \]

output

-(a-I*b)^(5/2)*(A-I*B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d-(a+ 
I*b)^(5/2)*(A+I*B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d+2*(A*a^ 
2-A*b^2-2*B*a*b)*(a+b*tan(d*x+c))^(1/2)/d+2/3*(A*a-B*b)*(a+b*tan(d*x+c))^( 
3/2)/d+2/5*A*(a+b*tan(d*x+c))^(5/2)/d+2/7*B*(a+b*tan(d*x+c))^(7/2)/b/d
 

3.4.33.2 Mathematica [A] (verified)

Time = 1.74 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.21 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {\frac {4 B (a+b \tan (c+d x))^{7/2}}{b}-7 i (i A+B) \left (\frac {2}{5} (a+b \tan (c+d x))^{5/2}+\frac {2}{3} (a-i b) \left (-3 (a-i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+\sqrt {a+b \tan (c+d x)} (4 a-3 i b+b \tan (c+d x))\right )\right )-7 i (i A-B) \left (\frac {2}{5} (a+b \tan (c+d x))^{5/2}+\frac {2}{3} (a+i b) \left (-3 (a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+\sqrt {a+b \tan (c+d x)} (4 a+3 i b+b \tan (c+d x))\right )\right )}{14 d} \]

input

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

output

((4*B*(a + b*Tan[c + d*x])^(7/2))/b - (7*I)*(I*A + B)*((2*(a + b*Tan[c + d 
*x])^(5/2))/5 + (2*(a - I*b)*(-3*(a - I*b)^(3/2)*ArcTanh[Sqrt[a + b*Tan[c 
+ d*x]]/Sqrt[a - I*b]] + Sqrt[a + b*Tan[c + d*x]]*(4*a - (3*I)*b + b*Tan[c 
 + d*x])))/3) - (7*I)*(I*A - B)*((2*(a + b*Tan[c + d*x])^(5/2))/5 + (2*(a 
+ I*b)*(-3*(a + I*b)^(3/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] 
 + Sqrt[a + b*Tan[c + d*x]]*(4*a + (3*I)*b + b*Tan[c + d*x])))/3))/(14*d)
 

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

Time = 1.16 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.01, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 4075, 3042, 4011, 3042, 4011, 3042, 4011, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4075

\(\displaystyle \int (A \tan (c+d x)-B) (a+b \tan (c+d x))^{5/2}dx+\frac {2 B (a+b \tan (c+d x))^{7/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (A \tan (c+d x)-B) (a+b \tan (c+d x))^{5/2}dx+\frac {2 B (a+b \tan (c+d x))^{7/2}}{7 b d}\)

\(\Big \downarrow \) 4011

\(\displaystyle \int (a+b \tan (c+d x))^{3/2} (-A b-a B+(a A-b B) \tan (c+d x))dx+\frac {2 A (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 B (a+b \tan (c+d x))^{7/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (c+d x))^{3/2} (-A b-a B+(a A-b B) \tan (c+d x))dx+\frac {2 A (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 B (a+b \tan (c+d x))^{7/2}}{7 b d}\)

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

\(\displaystyle -\frac {1}{2} (-b+i a)^3 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (b+i a)^3 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2 A-2 a b B-A b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a A-b B) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 A (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 B (a+b \tan (c+d x))^{7/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {1}{2} (-b+i a)^3 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (b+i a)^3 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2 A-2 a b B-A b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a A-b B) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 A (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 B (a+b \tan (c+d x))^{7/2}}{7 b d}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {i (b+i a)^3 (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i (-b+i a)^3 (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (a^2 A-2 a b B-A b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a A-b B) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 A (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 B (a+b \tan (c+d x))^{7/2}}{7 b d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {i (b+i a)^3 (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (-b+i a)^3 (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (a^2 A-2 a b B-A b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a A-b B) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 A (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 B (a+b \tan (c+d x))^{7/2}}{7 b d}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {(-b+i a)^3 (A+i B) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(b+i a)^3 (A-i B) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {2 \left (a^2 A-2 a b B-A b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a A-b B) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 A (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 B (a+b \tan (c+d x))^{7/2}}{7 b d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 \left (a^2 A-2 a b B-A b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {(b+i a)^3 (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {(-b+i a)^3 (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}+\frac {2 (a A-b B) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 A (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 B (a+b \tan (c+d x))^{7/2}}{7 b d}\)

input

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

output

((I*a + b)^3*(A - I*B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]* 
d) - ((I*a - b)^3*(A + I*B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + 
I*b]*d) + (2*(a^2*A - A*b^2 - 2*a*b*B)*Sqrt[a + b*Tan[c + d*x]])/d + (2*(a 
*A - b*B)*(a + b*Tan[c + d*x])^(3/2))/(3*d) + (2*A*(a + b*Tan[c + d*x])^(5 
/2))/(5*d) + (2*B*(a + b*Tan[c + d*x])^(7/2))/(7*b*d)
 

3.4.33.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 4011

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int 
[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] 
, x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 
 0] && GtQ[m, 0]
 

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 4075

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

3.4.33.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2406\) vs. \(2(181)=362\).

Time = 0.12 (sec) , antiderivative size = 2407, normalized size of antiderivative = 11.30

method	result	size
parts	\(\text {Expression too large to display}\)	\(2407\)
derivativedivides	\(\text {Expression too large to display}\)	\(2426\)
default	\(\text {Expression too large to display}\)	\(2426\)

input

int(tan(d*x+c)*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x,method=_RETURNVER 
BOSE)
 

output

2/5*A*(a+b*tan(d*x+c))^(5/2)/d+2/3/d*A*a*(a+b*tan(d*x+c))^(3/2)+2/d*(a+b*t 
an(d*x+c))^(1/2)*A*a^2-2/d*b^2*(a+b*tan(d*x+c))^(1/2)*A-3/4/d*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) 
)*A*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2+1/4/d*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))*A*(2*(a^2+b^ 
2)^(1/2)+2*a)^(1/2)+3/4*A/d*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^ 
2-1/4*A/d*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)*b^2+1/d/(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*a^3+A/d/(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^3-3/d*b^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*a-3*A/d/(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*b^2+1/2/d*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))*A*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2 
)*a-1/d/(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*(a^2+b^2)^...
 

3.4.33.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4916 vs. \(2 (175) = 350\).

Time = 0.84 (sec) , antiderivative size = 4916, normalized size of antiderivative = 23.08 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

input

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

output

1/210*(105*b*d*sqrt(-(10*A*B*a^4*b - 20*A*B*a^2*b^3 + 2*A*B*b^5 - (A^2 - B 
^2)*a^5 + 10*(A^2 - B^2)*a^3*b^2 - 5*(A^2 - B^2)*a*b^4 + d^2*sqrt(-(4*A^2* 
B^2*a^10 + 20*(A^3*B - A*B^3)*a^9*b + 5*(5*A^4 - 26*A^2*B^2 + 5*B^4)*a^8*b 
^2 - 240*(A^3*B - A*B^3)*a^7*b^3 - 20*(5*A^4 - 32*A^2*B^2 + 5*B^4)*a^6*b^4 
 + 504*(A^3*B - A*B^3)*a^5*b^5 + 10*(11*A^4 - 62*A^2*B^2 + 11*B^4)*a^4*b^6 
 - 240*(A^3*B - A*B^3)*a^3*b^7 - 20*(A^4 - 7*A^2*B^2 + B^4)*a^2*b^8 + 20*( 
A^3*B - A*B^3)*a*b^9 + (A^4 - 2*A^2*B^2 + B^4)*b^10)/d^4))/d^2)*log(-(2*(A 
^3*B + A*B^3)*a^9 + 5*(A^4 - B^4)*a^8*b - 16*(A^3*B + A*B^3)*a^7*b^2 - 28* 
(A^3*B + A*B^3)*a^5*b^4 - 14*(A^4 - B^4)*a^4*b^5 - 8*(A^4 - B^4)*a^2*b^7 + 
 10*(A^3*B + A*B^3)*a*b^8 + (A^4 - B^4)*b^9)*sqrt(b*tan(d*x + c) + a) + (( 
B*a^2 + 2*A*a*b - B*b^2)*d^3*sqrt(-(4*A^2*B^2*a^10 + 20*(A^3*B - A*B^3)*a^ 
9*b + 5*(5*A^4 - 26*A^2*B^2 + 5*B^4)*a^8*b^2 - 240*(A^3*B - A*B^3)*a^7*b^3 
 - 20*(5*A^4 - 32*A^2*B^2 + 5*B^4)*a^6*b^4 + 504*(A^3*B - A*B^3)*a^5*b^5 + 
 10*(11*A^4 - 62*A^2*B^2 + 11*B^4)*a^4*b^6 - 240*(A^3*B - A*B^3)*a^3*b^7 - 
 20*(A^4 - 7*A^2*B^2 + B^4)*a^2*b^8 + 20*(A^3*B - A*B^3)*a*b^9 + (A^4 - 2* 
A^2*B^2 + B^4)*b^10)/d^4) + (2*A^2*B*a^7 + (5*A^3 - 9*A*B^2)*a^6*b - 2*(16 
*A^2*B - 5*B^3)*a^5*b^2 - 5*(3*A^3 - 11*A*B^2)*a^4*b^3 + 10*(5*A^2*B - 2*B 
^3)*a^3*b^4 + (11*A^3 - 31*A*B^2)*a^2*b^5 - 2*(6*A^2*B - B^3)*a*b^6 - (A^3 
 - A*B^2)*b^7)*d)*sqrt(-(10*A*B*a^4*b - 20*A*B*a^2*b^3 + 2*A*B*b^5 - (A^2 
- B^2)*a^5 + 10*(A^2 - B^2)*a^3*b^2 - 5*(A^2 - B^2)*a*b^4 + d^2*sqrt(-(...
 

3.4.33.6 Sympy [F]

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

input

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

output

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

3.4.33.7 Maxima [F]

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

input

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

output

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

3.4.33.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.4.33.9 Mupad [B] (verification not implemented)

Time = 94.32 (sec) , antiderivative size = 3932, normalized size of antiderivative = 18.46 \[ \int \tan (c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

input

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

output

log(- ((((((-A^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + A^2*a^5*d^2 
 - 10*A^2*a^3*b^2*d^2 + 5*A^2*a*b^4*d^2)/d^4)^(1/2)*(32*A*b^6 - 32*A*a^4*b 
^2 + 32*a*b^2*d*(((-A^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + A^2* 
a^5*d^2 - 10*A^2*a^3*b^2*d^2 + 5*A^2*a*b^4*d^2)/d^4)^(1/2)*(a + b*tan(c + 
d*x))^(1/2)))/(2*d) - (16*A^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^6 - b^6 + 
15*a^2*b^4 - 15*a^4*b^2))/d^2)*(((-A^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^ 
2)^(1/2) + A^2*a^5*d^2 - 10*A^2*a^3*b^2*d^2 + 5*A^2*a*b^4*d^2)/d^4)^(1/2)) 
/2 - (8*A^3*a*b^2*(a^2 - 3*b^2)*(a^2 + b^2)^3)/d^3)*((20*A^4*a^2*b^8*d^4 - 
 A^4*b^10*d^4 - 110*A^4*a^4*b^6*d^4 + 100*A^4*a^6*b^4*d^4 - 25*A^4*a^8*b^2 
*d^4)^(1/2)/(4*d^4) + (A^2*a^5)/(4*d^2) - (5*A^2*a^3*b^2)/(2*d^2) + (5*A^2 
*a*b^4)/(4*d^2))^(1/2) - log(((((((-A^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2) 
^2)^(1/2) + A^2*a^5*d^2 - 10*A^2*a^3*b^2*d^2 + 5*A^2*a*b^4*d^2)/d^4)^(1/2) 
*(32*A*a^4*b^2 - 32*A*b^6 + 32*a*b^2*d*(((-A^4*b^2*d^4*(5*a^4 + b^4 - 10*a 
^2*b^2)^2)^(1/2) + A^2*a^5*d^2 - 10*A^2*a^3*b^2*d^2 + 5*A^2*a*b^4*d^2)/d^4 
)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/(2*d) - (16*A^2*b^2*(a + b*tan(c + d* 
x))^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2)*(((-A^4*b^2*d^4*(5*a 
^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + A^2*a^5*d^2 - 10*A^2*a^3*b^2*d^2 + 5*A^2 
*a*b^4*d^2)/d^4)^(1/2))/2 - (8*A^3*a*b^2*(a^2 - 3*b^2)*(a^2 + b^2)^3)/d^3) 
*(((20*A^4*a^2*b^8*d^4 - A^4*b^10*d^4 - 110*A^4*a^4*b^6*d^4 + 100*A^4*a^6* 
b^4*d^4 - 25*A^4*a^8*b^2*d^4)^(1/2) + A^2*a^5*d^2 - 10*A^2*a^3*b^2*d^2 ...