\(\int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) [317]

Optimal result

Integrand size = 33, antiderivative size = 233 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a-i b} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {\sqrt {a+i b} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 \left (14 a A b-8 a^2 B+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}+\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d} \] Output:

(a-I*b)^(1/2)*(A-I*B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d+(a+I 
*b)^(1/2)*(A+I*B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d-2*A*(a+b 
*tan(d*x+c))^(1/2)/d-2/105*(14*A*a*b-8*B*a^2+35*B*b^2)*(a+b*tan(d*x+c))^(3 
/2)/b^3/d+2/35*(7*A*b-4*B*a)*tan(d*x+c)*(a+b*tan(d*x+c))^(3/2)/b^2/d+2/7*B 
*tan(d*x+c)^2*(a+b*tan(d*x+c))^(3/2)/b/d
 

Mathematica [A] (verified)

Time = 1.96 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.91 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a-i b} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {\sqrt {a+i b} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-14 a^2 A b-105 A b^3+8 a^3 B-35 a b^2 B-b \left (-7 a A b+4 a^2 B+35 b^2 B\right ) \tan (c+d x)+3 b^2 (7 A b+a B) \tan ^2(c+d x)+15 b^3 B \tan ^3(c+d x)\right )}{105 b^3 d} \] Input:

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

Output:

(Sqrt[a - I*b]*(A - I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/ 
d + (Sqrt[a + I*b]*(A + I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b 
]])/d + (2*Sqrt[a + b*Tan[c + d*x]]*(-14*a^2*A*b - 105*A*b^3 + 8*a^3*B - 3 
5*a*b^2*B - b*(-7*a*A*b + 4*a^2*B + 35*b^2*B)*Tan[c + d*x] + 3*b^2*(7*A*b 
+ a*B)*Tan[c + d*x]^2 + 15*b^3*B*Tan[c + d*x]^3))/(105*b^3*d)
 

Rubi [A] (warning: unable to verify)

Time = 1.47 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.12, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 4090, 27, 3042, 4130, 27, 3042, 4113, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4090

\(\displaystyle \frac {2 \int -\frac {1}{2} \tan (c+d x) \sqrt {a+b \tan (c+d x)} \left (-\left ((7 A b-4 a B) \tan ^2(c+d x)\right )+7 b B \tan (c+d x)+4 a B\right )dx}{7 b}+\frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\int \tan (c+d x) \sqrt {a+b \tan (c+d x)} \left (-\left ((7 A b-4 a B) \tan ^2(c+d x)\right )+7 b B \tan (c+d x)+4 a B\right )dx}{7 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\int \tan (c+d x) \sqrt {a+b \tan (c+d x)} \left (-\left ((7 A b-4 a B) \tan (c+d x)^2\right )+7 b B \tan (c+d x)+4 a B\right )dx}{7 b}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\frac {2 \int \frac {1}{2} \sqrt {a+b \tan (c+d x)} \left (35 A \tan (c+d x) b^2+\left (-8 B a^2+14 A b a+35 b^2 B\right ) \tan ^2(c+d x)+2 a (7 A b-4 a B)\right )dx}{5 b}-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}}{7 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\frac {\int \sqrt {a+b \tan (c+d x)} \left (35 A \tan (c+d x) b^2+\left (-8 B a^2+14 A b a+35 b^2 B\right ) \tan ^2(c+d x)+2 a (7 A b-4 a B)\right )dx}{5 b}-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}}{7 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\frac {\int \sqrt {a+b \tan (c+d x)} \left (35 A \tan (c+d x) b^2+\left (-8 B a^2+14 A b a+35 b^2 B\right ) \tan (c+d x)^2+2 a (7 A b-4 a B)\right )dx}{5 b}-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}}{7 b}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\frac {\int \sqrt {a+b \tan (c+d x)} \left (35 A b^2 \tan (c+d x)-35 b^2 B\right )dx+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}}{5 b}-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}}{7 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\frac {\int \sqrt {a+b \tan (c+d x)} \left (35 A b^2 \tan (c+d x)-35 b^2 B\right )dx+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}}{5 b}-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}}{7 b}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\frac {\int \frac {35 b^2 (a A-b B) \tan (c+d x)-35 b^2 (A b+a B)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}}{7 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {\frac {\int \frac {35 b^2 (a A-b B) \tan (c+d x)-35 b^2 (A b+a B)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}}{7 b}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {\frac {35}{2} b^2 (-b+i a) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {35}{2} b^2 (b+i a) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}}{7 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {\frac {35}{2} b^2 (-b+i a) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {35}{2} b^2 (b+i a) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}}{7 b}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {-\frac {35 i b^2 (b+i a) (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {35 i b^2 (-b+i a) (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}}{7 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {\frac {35 i b^2 (b+i a) (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {35 i b^2 (-b+i a) (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}}{7 b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {\frac {35 b (-b+i a) (A+i B) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}-\frac {35 b (b+i a) (A-i B) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}}{7 b}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 B \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}-\frac {-\frac {2 (7 A b-4 a B) \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {\frac {2 \left (-8 a^2 B+14 a A b+35 b^2 B\right ) (a+b \tan (c+d x))^{3/2}}{3 b d}-\frac {35 b^2 (b+i a) (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {35 b^2 (-b+i a) (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}+\frac {70 A b^2 \sqrt {a+b \tan (c+d x)}}{d}}{5 b}}{7 b}\)

Input:

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

Output:

(2*B*Tan[c + d*x]^2*(a + b*Tan[c + d*x])^(3/2))/(7*b*d) - ((-2*(7*A*b - 4* 
a*B)*Tan[c + d*x]*(a + b*Tan[c + d*x])^(3/2))/(5*b*d) + ((-35*b^2*(I*a + b 
)*(A - I*B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + (35*(I 
*a - b)*b^2*(A + I*B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d 
) + (70*A*b^2*Sqrt[a + b*Tan[c + d*x]])/d + (2*(14*a*A*b - 8*a^2*B + 35*b^ 
2*B)*(a + b*Tan[c + d*x])^(3/2))/(3*b*d))/(5*b))/(7*b)
 

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]
 

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])))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(936\) vs. \(2(201)=402\).

Time = 0.45 (sec) , antiderivative size = 937, normalized size of antiderivative = 4.02

method	result	size
parts	\(\frac {2 A \left (\frac {\left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {a \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\sqrt {a +b \tan \left (d x +c \right )}\, b^{2}-b^{2} \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (-a +\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )\right )}{d \,b^{2}}+B \left (\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d \,b^{3}}-\frac {4 a \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d \,b^{3}}+\frac {2 a^{2} \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d \,b^{3}}-\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d b}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{4 d b}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{4 d b}-\frac {b \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4 d b}+\frac {b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )\)	\(937\)
derivativedivides	\(\text {Expression too large to display}\)	\(1092\)
default	\(\text {Expression too large to display}\)	\(1092\)

Input:

int(tan(d*x+c)^3*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x,method=_RETURNV 
ERBOSE)
 

Output:

2*A/d/b^2*(1/5*(a+b*tan(d*x+c))^(5/2)-1/3*a*(a+b*tan(d*x+c))^(3/2)-(a+b*ta 
n(d*x+c))^(1/2)*b^2-b^2*(1/8*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*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))+1 
/2*(-a+(a^2+b^2)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2) 
^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)) 
-1/8*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*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))+1/2*(a-(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))))+B*(2/7/d/b^3*(a+b*tan( 
d*x+c))^(7/2)-4/5/d/b^3*a*(a+b*tan(d*x+c))^(5/2)+2/3/d/b^3*a^2*(a+b*tan(d* 
x+c))^(3/2)-2/3/d/b*(a+b*tan(d*x+c))^(3/2)-1/4/d/b*(2*(a^2+b^2)^(1/2)+2*a) 
^(1/2)*a*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))+1/4/d/b*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/ 
2)*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))-1/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^ 
(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))+ 
1/4/d/b*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a*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))-1/4/d/b*(2*(a^2+b^2) 
^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*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))+1/d*b/(2*(a^2+b^2)^(1/2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1312 vs. \(2 (195) = 390\).

Time = 0.11 (sec) , antiderivative size = 1312, normalized size of antiderivative = 5.63 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorith 
m="fricas")
 

Output:

-1/210*(105*b^3*d*sqrt(-(2*A*B*b + 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)/d^4) - (A^2 - B^2)*a)/d^2)*log(-( 
2*(A^3*B + A*B^3)*a + (A^4 - B^4)*b)*sqrt(b*tan(d*x + c) + a) + (B*d^3*sqr 
t(-(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)/d 
^4) + (2*A^2*B*a + (A^3 - A*B^2)*b)*d)*sqrt(-(2*A*B*b + 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)/d^4) - (A^2 
- B^2)*a)/d^2)) - 105*b^3*d*sqrt(-(2*A*B*b + 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)/d^4) - (A^2 - B^2)*a)/d 
^2)*log(-(2*(A^3*B + A*B^3)*a + (A^4 - B^4)*b)*sqrt(b*tan(d*x + c) + a) - 
(B*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)/d^4) + (2*A^2*B*a + (A^3 - A*B^2)*b)*d)*sqrt(-(2*A*B*b + 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)/d^ 
4) - (A^2 - B^2)*a)/d^2)) - 105*b^3*d*sqrt(-(2*A*B*b - 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)/d^4) - (A^2 - 
 B^2)*a)/d^2)*log(-(2*(A^3*B + A*B^3)*a + (A^4 - B^4)*b)*sqrt(b*tan(d*x + 
c) + a) + (B*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)/d^4) - (2*A^2*B*a + (A^3 - A*B^2)*b)*d)*sqrt(-(2*A*B*b 
- 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)/d^4) - (A^2 - B^2)*a)/d^2)) + 105*b^3*d*sqrt(-(2*A*B*b - 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)/...
 

Sympy [F]

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

integrate(tan(d*x+c)**3*(a+b*tan(d*x+c))**(1/2)*(A+B*tan(d*x+c)),x)
 

Output:

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

Maxima [F]

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

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorith 
m="maxima")
 

Output:

integrate((B*tan(d*x + c) + A)*sqrt(b*tan(d*x + c) + a)*tan(d*x + c)^3, x)
 

Giac [F(-2)]

Exception generated. \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorith 
m="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[0,24,9]%%%}+%%%{10,[0,22,9]%%%}+%%%{45,[0,20,9]%%%} 
+%%%{120,
 

Mupad [B] (verification not implemented)

Time = 57.80 (sec) , antiderivative size = 1093, normalized size of antiderivative = 4.69 \[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx =\text {Too large to display} \] Input:

int(tan(c + d*x)^3*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(1/2),x)
 

Output:

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

Reduce [F]

\[ \int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\left (\int \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{4}d x \right ) b +\left (\int \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{3}d x \right ) a \] Input:

int(tan(d*x+c)^3*(a+b*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x)
 

Output:

int(sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**4,x)*b + int(sqrt(tan(c + d*x)* 
b + a)*tan(c + d*x)**3,x)*a