3.4.50 \(\int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx\) [350]

3.4.50.1 Optimal result

Integrand size = 33, antiderivative size = 264 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{3/2} d}+\frac {(A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{3/2} d}+\frac {2 a (A b-a B) \tan ^2(c+d x)}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {2 \left (6 a^2 A b+3 A b^3-8 a^3 B-5 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{3 b^3 \left (a^2+b^2\right ) d}-\frac {2 \left (3 a A b-4 a^2 B-b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b^2 \left (a^2+b^2\right ) d} \]

output

(A-I*B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(3/2)/d+(A+I 
*B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(3/2)/d+2/3*(6*A 
*a^2*b+3*A*b^3-8*B*a^3-5*B*a*b^2)*(a+b*tan(d*x+c))^(1/2)/b^3/(a^2+b^2)/d-2 
/3*(3*A*a*b-4*B*a^2-B*b^2)*(a+b*tan(d*x+c))^(1/2)*tan(d*x+c)/b^2/(a^2+b^2) 
/d+2*a*(A*b-B*a)*tan(d*x+c)^2/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(1/2)
 

3.4.50.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 3.80 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.14 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {3 i A \left (\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b}}\right )+\frac {2 \left (6 a A b-8 a^2 B+3 b^2 B\right )}{b^2 \sqrt {a+b \tan (c+d x)}}+\frac {3 i (a A+b B) \left ((a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )-(a-i b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )\right )}{\left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {2 (3 A b-4 a B) \tan (c+d x)}{b \sqrt {a+b \tan (c+d x)}}+\frac {2 B \tan ^2(c+d x)}{\sqrt {a+b \tan (c+d x)}}}{3 b d} \]

input

Integrate[(Tan[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(3/2) 
,x]
 

output

((3*I)*A*(ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]]/Sqrt[a - I*b] - 
ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]]/Sqrt[a + I*b]) + (2*(6*a*A 
*b - 8*a^2*B + 3*b^2*B))/(b^2*Sqrt[a + b*Tan[c + d*x]]) + ((3*I)*(a*A + b* 
B)*((a + I*b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a - I* 
b)] - (a - I*b)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a + 
I*b)]))/((a^2 + b^2)*Sqrt[a + b*Tan[c + d*x]]) + (2*(3*A*b - 4*a*B)*Tan[c 
+ d*x])/(b*Sqrt[a + b*Tan[c + d*x]]) + (2*B*Tan[c + d*x]^2)/Sqrt[a + b*Tan 
[c + d*x]])/(3*b*d)
 

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

Time = 1.38 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.03, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4088, 27, 3042, 4130, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4088

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

\(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 \int -\frac {-3 (a A+b B) \tan (c+d x) b^2+\left (-8 B a^3+6 A b a^2-5 b^2 B a+3 A b^3\right ) \tan ^2(c+d x)+2 a \left (-4 B a^2+3 A b a-b^2 B\right )}{2 \sqrt {a+b \tan (c+d x)}}dx}{3 b}+\frac {2 \left (-4 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 \left (-4 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\int \frac {-3 (a A+b B) \tan (c+d x) b^2+\left (-8 B a^3+6 A b a^2-5 b^2 B a+3 A b^3\right ) \tan ^2(c+d x)+2 a \left (-4 B a^2+3 A b a-b^2 B\right )}{\sqrt {a+b \tan (c+d x)}}dx}{3 b}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 \left (-4 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\int \frac {-3 (a A+b B) \tan (c+d x) b^2+\left (-8 B a^3+6 A b a^2-5 b^2 B a+3 A b^3\right ) \tan (c+d x)^2+2 a \left (-4 B a^2+3 A b a-b^2 B\right )}{\sqrt {a+b \tan (c+d x)}}dx}{3 b}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 \left (-4 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\int \frac {-3 (A b-a B) b^2-3 (a A+b B) \tan (c+d x) b^2}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^3 B+6 a^2 A b-5 a b^2 B+3 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 \left (-4 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\int \frac {-3 (A b-a B) b^2-3 (a A+b B) \tan (c+d x) b^2}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^3 B+6 a^2 A b-5 a b^2 B+3 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 \left (-4 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {-\frac {3}{2} b^2 (b+i a) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {3}{2} b^2 (a+i b) (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^3 B+6 a^2 A b-5 a b^2 B+3 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 \left (-4 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {-\frac {3}{2} b^2 (b+i a) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {3}{2} b^2 (a+i b) (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^3 B+6 a^2 A b-5 a b^2 B+3 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 \left (-4 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\frac {3 i b^2 (a+i b) (B+i A) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {3 i b^2 (b+i a) (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (-8 a^3 B+6 a^2 A b-5 a b^2 B+3 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 \left (-4 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {-\frac {3 i b^2 (a+i b) (B+i A) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {3 i b^2 (b+i a) (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (-8 a^3 B+6 a^2 A b-5 a b^2 B+3 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 \left (-4 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {-\frac {3 b (b+i a) (A+i B) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {3 b (a+i b) (B+i A) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {2 \left (-8 a^3 B+6 a^2 A b-5 a b^2 B+3 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 a (A b-a B) \tan ^2(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {2 \left (-4 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\frac {2 \left (-8 a^3 B+6 a^2 A b-5 a b^2 B+3 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{b d}+\frac {3 b^2 (a+i b) (B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {3 b^2 (b+i a) (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}}{3 b}}{b \left (a^2+b^2\right )}\)

input

Int[(Tan[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(3/2),x]
 

output

(2*a*(A*b - a*B)*Tan[c + d*x]^2)/(b*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]] 
) - ((2*(3*a*A*b - 4*a^2*B - b^2*B)*Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]]) 
/(3*b*d) - ((3*(a + I*b)*b^2*(I*A + B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]]) 
/(Sqrt[a - I*b]*d) - (3*b^2*(I*a + b)*(A + I*B)*ArcTan[Tan[c + d*x]/Sqrt[a 
 + I*b]])/(Sqrt[a + I*b]*d) + (2*(6*a^2*A*b + 3*A*b^3 - 8*a^3*B - 5*a*b^2* 
B)*Sqrt[a + b*Tan[c + d*x]])/(b*d))/(3*b))/(b*(a^2 + b^2))
 

3.4.50.3.1 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 4088

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*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x 
])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(d*(n + 1)*(c^2 + d^2)) 
  Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d* 
(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1) + a*d*(n 
 + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[ 
e + f*x] - b*(d*(A*b*c + a*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(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] && GtQ[m, 1] & 
& LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

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]
 

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])))
 

3.4.50.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3753\) vs. \(2(236)=472\).

Time = 0.18 (sec) , antiderivative size = 3754, normalized size of antiderivative = 14.22

method	result	size
parts	\(\text {Expression too large to display}\)	\(3754\)
derivativedivides	\(\text {Expression too large to display}\)	\(8025\)
default	\(\text {Expression too large to display}\)	\(8025\)

input

int(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x,method=_RETURNV 
ERBOSE)
 

output

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

3.4.50.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4421 vs. \(2 (229) = 458\).

Time = 0.71 (sec) , antiderivative size = 4421, normalized size of antiderivative = 16.75 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

input

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algorith 
m="fricas")
 

output

1/6*(3*((a^2*b^4 + b^6)*d*tan(d*x + c) + (a^3*b^3 + a*b^5)*d)*sqrt((6*A*B* 
a^2*b - 2*A*B*b^3 + (A^2 - B^2)*a^3 - 3*(A^2 - B^2)*a*b^2 + (a^6 + 3*a^4*b 
^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(4*A^2*B^2*a^6 - 12*(A^3*B - A*B^3)*a^5*b 
+ 3*(3*A^4 - 14*A^2*B^2 + 3*B^4)*a^4*b^2 + 40*(A^3*B - A*B^3)*a^3*b^3 - 6* 
(A^4 - 8*A^2*B^2 + B^4)*a^2*b^4 - 12*(A^3*B - A*B^3)*a*b^5 + (A^4 - 2*A^2* 
B^2 + B^4)*b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 
 + 6*a^2*b^10 + b^12)*d^4)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))*lo 
g(-(2*(A^3*B + A*B^3)*a^3 - 3*(A^4 - B^4)*a^2*b - 6*(A^3*B + A*B^3)*a*b^2 
+ (A^4 - B^4)*b^3)*sqrt(b*tan(d*x + c) + a) + ((B*a^8 - 2*A*a^7*b + 2*B*a^ 
6*b^2 - 6*A*a^5*b^3 - 6*A*a^3*b^5 - 2*B*a^2*b^6 - 2*A*a*b^7 - B*b^8)*d^3*s 
qrt(-(4*A^2*B^2*a^6 - 12*(A^3*B - A*B^3)*a^5*b + 3*(3*A^4 - 14*A^2*B^2 + 3 
*B^4)*a^4*b^2 + 40*(A^3*B - A*B^3)*a^3*b^3 - 6*(A^4 - 8*A^2*B^2 + B^4)*a^2 
*b^4 - 12*(A^3*B - A*B^3)*a*b^5 + (A^4 - 2*A^2*B^2 + B^4)*b^6)/((a^12 + 6* 
a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) 
 - (2*A^2*B*a^5 - (3*A^3 - 7*A*B^2)*a^4*b - 2*(7*A^2*B - 3*B^3)*a^3*b^2 + 
4*(A^3 - 4*A*B^2)*a^2*b^3 + 2*(4*A^2*B - B^3)*a*b^4 - (A^3 - A*B^2)*b^5)*d 
)*sqrt((6*A*B*a^2*b - 2*A*B*b^3 + (A^2 - B^2)*a^3 - 3*(A^2 - B^2)*a*b^2 + 
(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(4*A^2*B^2*a^6 - 12*(A^3*B - 
 A*B^3)*a^5*b + 3*(3*A^4 - 14*A^2*B^2 + 3*B^4)*a^4*b^2 + 40*(A^3*B - A*B^3 
)*a^3*b^3 - 6*(A^4 - 8*A^2*B^2 + B^4)*a^2*b^4 - 12*(A^3*B - A*B^3)*a*b^...
 

3.4.50.6 Sympy [F]

\[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{3}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

input

integrate(tan(d*x+c)**3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**(3/2),x)
 

output

Integral((A + B*tan(c + d*x))*tan(c + d*x)**3/(a + b*tan(c + d*x))**(3/2), 
 x)
 

3.4.50.7 Maxima [F(-1)]

Timed out. \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]

input

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algorith 
m="maxima")
 

output

Timed out
 

3.4.50.8 Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]

input

integrate(tan(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algorith 
m="giac")
 

output

Timed out
 

3.4.50.9 Mupad [B] (verification not implemented)

Time = 23.82 (sec) , antiderivative size = 5811, normalized size of antiderivative = 22.01 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

input

int((tan(c + d*x)^3*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x))^(3/2),x)
 

output

(log(24*A^3*a^3*b^6*d^2 - ((((((96*A^4*a^2*b^4*d^4 - 16*A^4*b^6*d^4 - 144* 
A^4*a^4*b^2*d^4)^(1/2) + 4*A^2*a^3*d^2 - 12*A^2*a*b^2*d^2)/(a^6*d^4 + b^6* 
d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2)*(((((96*A^4*a^2*b^4*d^4 - 16*A 
^4*b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) + 4*A^2*a^3*d^2 - 12*A^2*a*b^2*d^2 
)/(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2)*(a + b*tan(c 
+ d*x))^(1/2)*(64*a*b^12*d^5 + 320*a^3*b^10*d^5 + 640*a^5*b^8*d^5 + 640*a^ 
7*b^6*d^5 + 320*a^9*b^4*d^5 + 64*a^11*b^2*d^5))/4 - 32*A*b^12*d^4 - 96*A*a 
^2*b^10*d^4 - 64*A*a^4*b^8*d^4 + 64*A*a^6*b^6*d^4 + 96*A*a^8*b^4*d^4 + 32* 
A*a^10*b^2*d^4))/4 + (a + b*tan(c + d*x))^(1/2)*(16*A^2*b^10*d^3 + 32*A^2* 
a^2*b^8*d^3 - 32*A^2*a^6*b^4*d^3 - 16*A^2*a^8*b^2*d^3))*(((96*A^4*a^2*b^4* 
d^4 - 16*A^4*b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) + 4*A^2*a^3*d^2 - 12*A^2 
*a*b^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2))/4 
+ 24*A^3*a^5*b^4*d^2 + 8*A^3*a^7*b^2*d^2 + 8*A^3*a*b^8*d^2)*(((96*A^4*a^2* 
b^4*d^4 - 16*A^4*b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) + 4*A^2*a^3*d^2 - 12 
*A^2*a*b^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2) 
)/4 + (log(24*A^3*a^3*b^6*d^2 - ((((-((96*A^4*a^2*b^4*d^4 - 16*A^4*b^6*d^4 
 - 144*A^4*a^4*b^2*d^4)^(1/2) - 4*A^2*a^3*d^2 + 12*A^2*a*b^2*d^2)/(a^6*d^4 
 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2)*(((-((96*A^4*a^2*b^4*d^ 
4 - 16*A^4*b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) - 4*A^2*a^3*d^2 + 12*A^2*a 
*b^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2)*(a...