3.4.32 \(\int \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [332]

3.4.32.1 Optimal result

Integrand size = 33, antiderivative size = 252 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {(a-i b)^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{5/2} (i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 \left (2 a A b+a^2 B-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 (A b+a B) (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{63 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d} \]

output

(a-I*b)^(5/2)*(I*A+B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d-(a+I 
*b)^(5/2)*(I*A-B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d-2*(2*A*a 
*b+B*a^2-B*b^2)*(a+b*tan(d*x+c))^(1/2)/d-2/3*(A*b+B*a)*(a+b*tan(d*x+c))^(3 
/2)/d-2/5*B*(a+b*tan(d*x+c))^(5/2)/d+2/63*(9*A*b-2*B*a)*(a+b*tan(d*x+c))^( 
7/2)/b^2/d+2/9*B*tan(d*x+c)*(a+b*tan(d*x+c))^(7/2)/b/d
 

3.4.32.2 Mathematica [A] (verified)

Time = 4.97 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.17 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{b}+14 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}-\frac {63}{2} i b (A-i B) \left (\frac {2}{5} (a+b \tan (c+d x))^{5/2}+\frac {2}{3} (a-i b) \left (-3 (a-i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+\sqrt {a+b \tan (c+d x)} (4 a-3 i b+b \tan (c+d x))\right )\right )+\frac {63}{2} i b (A+i B) \left (\frac {2}{5} (a+b \tan (c+d x))^{5/2}+\frac {2}{3} (a+i b) \left (-3 (a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+\sqrt {a+b \tan (c+d x)} (4 a+3 i b+b \tan (c+d x))\right )\right )}{63 b d} \]

input

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

output

((2*(9*A*b - 2*a*B)*(a + b*Tan[c + d*x])^(7/2))/b + 14*B*Tan[c + d*x]*(a + 
 b*Tan[c + d*x])^(7/2) - ((63*I)/2)*b*(A - I*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) + ((63*I)/2)*b*(A + I*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))/(63*b* 
d)
 

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

Time = 1.61 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.98, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 4090, 27, 3042, 4113, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4090

\(\displaystyle \frac {2 \int -\frac {1}{2} (a+b \tan (c+d x))^{5/2} \left (-\left ((9 A b-2 a B) \tan ^2(c+d x)\right )+9 b B \tan (c+d x)+2 a B\right )dx}{9 b}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\int (a+b \tan (c+d x))^{5/2} \left (-\left ((9 A b-2 a B) \tan ^2(c+d x)\right )+9 b B \tan (c+d x)+2 a B\right )dx}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\int (a+b \tan (c+d x))^{5/2} \left (-\left ((9 A b-2 a B) \tan (c+d x)^2\right )+9 b B \tan (c+d x)+2 a B\right )dx}{9 b}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\int (a+b \tan (c+d x))^{5/2} (9 A b+9 B \tan (c+d x) b)dx-\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{7 b d}}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\int (a+b \tan (c+d x))^{5/2} (9 A b+9 B \tan (c+d x) b)dx-\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{7 b d}}{9 b}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\int (a+b \tan (c+d x))^{3/2} (9 b (a A-b B)+9 b (A b+a B) \tan (c+d x))dx-\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{7 b d}+\frac {18 b B (a+b \tan (c+d x))^{5/2}}{5 d}}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\int (a+b \tan (c+d x))^{3/2} (9 b (a A-b B)+9 b (A b+a B) \tan (c+d x))dx-\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{7 b d}+\frac {18 b B (a+b \tan (c+d x))^{5/2}}{5 d}}{9 b}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\int \sqrt {a+b \tan (c+d x)} \left (9 b \left (A a^2-2 b B a-A b^2\right )+9 b \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)\right )dx-\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{7 b d}+\frac {6 b (a B+A b) (a+b \tan (c+d x))^{3/2}}{d}+\frac {18 b B (a+b \tan (c+d x))^{5/2}}{5 d}}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\int \sqrt {a+b \tan (c+d x)} \left (9 b \left (A a^2-2 b B a-A b^2\right )+9 b \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)\right )dx-\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{7 b d}+\frac {6 b (a B+A b) (a+b \tan (c+d x))^{3/2}}{d}+\frac {18 b B (a+b \tan (c+d x))^{5/2}}{5 d}}{9 b}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\int \frac {9 b \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right )+9 b \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {18 b \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{7 b d}+\frac {6 b (a B+A b) (a+b \tan (c+d x))^{3/2}}{d}+\frac {18 b B (a+b \tan (c+d x))^{5/2}}{5 d}}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\int \frac {9 b \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right )+9 b \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {18 b \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{7 b d}+\frac {6 b (a B+A b) (a+b \tan (c+d x))^{3/2}}{d}+\frac {18 b B (a+b \tan (c+d x))^{5/2}}{5 d}}{9 b}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\frac {9}{2} b (a+i b)^3 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {9}{2} b (a-i b)^3 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {18 b \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{7 b d}+\frac {6 b (a B+A b) (a+b \tan (c+d x))^{3/2}}{d}+\frac {18 b B (a+b \tan (c+d x))^{5/2}}{5 d}}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\frac {9}{2} b (a+i b)^3 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {9}{2} b (a-i b)^3 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {18 b \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{7 b d}+\frac {6 b (a B+A b) (a+b \tan (c+d x))^{3/2}}{d}+\frac {18 b B (a+b \tan (c+d x))^{5/2}}{5 d}}{9 b}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\frac {9 i b (a-i b)^3 (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {9 i b (a+i b)^3 (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {18 b \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{7 b d}+\frac {6 b (a B+A b) (a+b \tan (c+d x))^{3/2}}{d}+\frac {18 b B (a+b \tan (c+d x))^{5/2}}{5 d}}{9 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {-\frac {9 i b (a-i b)^3 (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {9 i b (a+i b)^3 (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {18 b \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{7 b d}+\frac {6 b (a B+A b) (a+b \tan (c+d x))^{3/2}}{d}+\frac {18 b B (a+b \tan (c+d x))^{5/2}}{5 d}}{9 b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\frac {9 (a+i b)^3 (A+i B) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {9 (a-i b)^3 (A-i B) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {18 b \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{7 b d}+\frac {6 b (a B+A b) (a+b \tan (c+d x))^{3/2}}{d}+\frac {18 b B (a+b \tan (c+d x))^{5/2}}{5 d}}{9 b}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\frac {18 b \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {9 b (a-i b)^{5/2} (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {9 b (a+i b)^{5/2} (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}-\frac {2 (9 A b-2 a B) (a+b \tan (c+d x))^{7/2}}{7 b d}+\frac {6 b (a B+A b) (a+b \tan (c+d x))^{3/2}}{d}+\frac {18 b B (a+b \tan (c+d x))^{5/2}}{5 d}}{9 b}\)

input

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

output

(2*B*Tan[c + d*x]*(a + b*Tan[c + d*x])^(7/2))/(9*b*d) - ((9*(a - I*b)^(5/2 
)*b*(A - I*B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/d + (9*(a + I*b)^(5/2)*b 
*(A + I*B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/d + (18*b*(2*a*A*b + a^2*B 
- b^2*B)*Sqrt[a + b*Tan[c + d*x]])/d + (6*b*(A*b + a*B)*(a + b*Tan[c + d*x 
])^(3/2))/d + (18*b*B*(a + b*Tan[c + d*x])^(5/2))/(5*d) - (2*(9*A*b - 2*a* 
B)*(a + b*Tan[c + d*x])^(7/2))/(7*b*d))/(9*b)
 

3.4.32.3.1 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 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 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]
 

3.4.32.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2437\) vs. \(2(216)=432\).

Time = 0.18 (sec) , antiderivative size = 2438, normalized size of antiderivative = 9.67

method	result	size
parts	\(\text {Expression too large to display}\)	\(2438\)
derivativedivides	\(\text {Expression too large to display}\)	\(2469\)
default	\(\text {Expression too large to display}\)	\(2469\)

input

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

output

A*(2/7/b/d*(a+b*tan(d*x+c))^(7/2)-2/3*b*(a+b*tan(d*x+c))^(3/2)/d-4*b/d*(a+ 
b*tan(d*x+c))^(1/2)*a+1/4/b/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+b^2)^(1/2)*a^2-1/4*b/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+b^2)^(1/2)-1/4/b/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^3+3/4* 
b/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*b/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^2+b^2)^(1/2)*a-3*b/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^2+b^3/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))-1/4/b/d*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+ 
2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*( 
a^2+b^2)^(1/2)*a^2+1/4*b/d*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2* 
a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^ 
2+b^2)^(1/2)+1/4/b/d*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/ 
2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3-3/...
 

3.4.32.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4963 vs. \(2 (210) = 420\).

Time = 0.86 (sec) , antiderivative size = 4963, normalized size of antiderivative = 19.69 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

input

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

output

1/630*(315*b^2*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) + ( 
(A*a^2 - 2*B*a*b - A*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*B^2*a^7 + (9*A^2*B - 5*B^3)*a^6*b + 2*(5 
*A^3 - 16*A*B^2)*a^5*b^2 - 5*(11*A^2*B - 3*B^3)*a^4*b^3 - 10*(2*A^3 - 5*A* 
B^2)*a^3*b^4 + (31*A^2*B - 11*B^3)*a^2*b^5 + 2*(A^3 - 6*A*B^2)*a*b^6 - (A^ 
2*B - B^3)*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(-(...
 

3.4.32.6 Sympy [F]

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

input

integrate(tan(d*x+c)**2*(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)**2, 
 x)
 

3.4.32.7 Maxima [F(-1)]

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

input

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

output

Timed out
 

3.4.32.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.4.32.9 Mupad [F(-1)]

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

input

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

output

\text{Hanged}