3.4.18 \(\int \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) [318]

3.4.18.1 Optimal result

Integrand size = 33, antiderivative size = 186 \[ \int \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a-i b} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {\sqrt {a+i b} (i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 B \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d} \]

output

(I*A+B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))*(a-I*b)^(1/2)/d-(I*A 
-B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))*(a+I*b)^(1/2)/d-2*B*(a+b 
*tan(d*x+c))^(1/2)/d+2/15*(5*A*b-2*B*a)*(a+b*tan(d*x+c))^(3/2)/b^2/d+2/5*B 
*tan(d*x+c)*(a+b*tan(d*x+c))^(3/2)/b/d
 

3.4.18.2 Mathematica [A] (verified)

Time = 2.09 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.91 \[ \int \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {15 \sqrt {a-i b} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+15 \sqrt {a+i b} (-i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+\frac {2 \sqrt {a+b \tan (c+d x)} \left (5 a A b-2 a^2 B-15 b^2 B+b (5 A b+a B) \tan (c+d x)+3 b^2 B \tan ^2(c+d x)\right )}{b^2}}{15 d} \]

input

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

output

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

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

Time = 1.01 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.97, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 4090, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4090

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4113

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\int \frac {5 b (a A-b B)+5 b (A b+a B) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {10 b B \sqrt {a+b \tan (c+d x)}}{d}}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\int \frac {5 b (a A-b B)+5 b (A b+a B) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {10 b B \sqrt {a+b \tan (c+d x)}}{d}}{5 b}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\frac {5}{2} b (a+i b) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {5}{2} b (a-i b) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx-\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {10 b B \sqrt {a+b \tan (c+d x)}}{d}}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\frac {5}{2} b (a+i b) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {5}{2} b (a-i b) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx-\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {10 b B \sqrt {a+b \tan (c+d x)}}{d}}{5 b}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\frac {5 i b (a-i b) (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {5 i b (a+i b) (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {10 b B \sqrt {a+b \tan (c+d x)}}{d}}{5 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {-\frac {5 i b (a-i b) (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {5 i b (a+i b) (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {10 b B \sqrt {a+b \tan (c+d x)}}{d}}{5 b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\frac {5 (a+i b) (A+i B) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {5 (a-i b) (A-i B) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}-\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {10 b B \sqrt {a+b \tan (c+d x)}}{d}}{5 b}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {\frac {5 b \sqrt {a-i b} (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {5 b \sqrt {a+i b} (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}-\frac {2 (5 A b-2 a B) (a+b \tan (c+d x))^{3/2}}{3 b d}+\frac {10 b B \sqrt {a+b \tan (c+d x)}}{d}}{5 b}\)

input

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

output

(2*B*Tan[c + d*x]*(a + b*Tan[c + d*x])^(3/2))/(5*b*d) - ((5*Sqrt[a - I*b]* 
b*(A - I*B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/d + (5*Sqrt[a + I*b]*b*(A 
+ I*B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/d + (10*b*B*Sqrt[a + b*Tan[c + 
d*x]])/d - (2*(5*A*b - 2*a*B)*(a + b*Tan[c + d*x])^(3/2))/(3*b*d))/(5*b)
 

3.4.18.3.1 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]
 

3.4.18.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(872\) vs. \(2(158)=316\).

Time = 0.13 (sec) , antiderivative size = 873, normalized size of antiderivative = 4.69

method	result	size
parts	\(A \left (\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 (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}}+\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}}\right )+\frac {2 B \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 (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}}+\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 (\sqrt {a^{2}+b^{2}}-a \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}}\right )\right )}{d \,b^{2}}\)	\(873\)
derivativedivides	\(\text {Expression too large to display}\)	\(1032\)
default	\(\text {Expression too large to display}\)	\(1032\)

input

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

output

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

3.4.18.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1265 vs. \(2 (152) = 304\).

Time = 0.27 (sec) , antiderivative size = 1265, normalized size of antiderivative = 6.80 \[ \int \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

input

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

output

1/30*(15*b^2*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) + (A*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*B^2*a + (A^2*B - B^3)*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)) - 15*b^2*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) - (A*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*B^2*a + (A^2*B - B^3)*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)) - 15*b^2*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) + 
(A*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*B^2*a + (A^2*B - B^3)*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)) + 15*b^2*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 -...
 

3.4.18.6 Sympy [F]

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

input

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

3.4.18.7 Maxima [F]

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

input

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

3.4.18.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.4.18.9 Mupad [B] (verification not implemented)

Time = 25.43 (sec) , antiderivative size = 938, normalized size of antiderivative = 5.04 \[ \int \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\mathrm {atanh}\left (\frac {d^3\,\left (\frac {16\,\left (A^2\,b^4-A^2\,a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}+\frac {16\,a\,b^2\,\left (\sqrt {-A^4\,b^2\,d^4}+A^2\,a\,d^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^4}\right )\,\sqrt {-\frac {\sqrt {-A^4\,b^2\,d^4}+A^2\,a\,d^2}{d^4}}}{16\,\left (A^3\,a^2\,b^3+A^3\,b^5\right )}\right )\,\sqrt {-\frac {\sqrt {-A^4\,b^2\,d^4}+A^2\,a\,d^2}{d^4}}-\left (\frac {2\,B\,\left (a^2+b^2\right )}{b^2\,d}-\frac {2\,B\,a^2}{b^2\,d}\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}+\mathrm {atanh}\left (\frac {d^3\,\left (\frac {16\,\left (A^2\,b^4-A^2\,a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}-\frac {16\,a\,b^2\,\left (\sqrt {-A^4\,b^2\,d^4}-A^2\,a\,d^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^4}\right )\,\sqrt {\frac {\sqrt {-A^4\,b^2\,d^4}-A^2\,a\,d^2}{d^4}}}{16\,\left (A^3\,a^2\,b^3+A^3\,b^5\right )}\right )\,\sqrt {\frac {\sqrt {-A^4\,b^2\,d^4}-A^2\,a\,d^2}{d^4}}+\frac {2\,A\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{3\,b\,d}+\frac {2\,B\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{5\,b^2\,d}-\frac {2\,B\,a\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{3\,b^2\,d}-\mathrm {atan}\left (\frac {B^2\,b^4\,\sqrt {\frac {\sqrt {-B^4\,b^2\,d^4}}{4\,d^4}+\frac {B^2\,a}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {16\,B\,b^4\,\sqrt {-B^4\,b^2\,d^4}}{d^3}+\frac {16\,B\,a^2\,b^2\,\sqrt {-B^4\,b^2\,d^4}}{d^3}}+\frac {a\,b^2\,\sqrt {\frac {\sqrt {-B^4\,b^2\,d^4}}{4\,d^4}+\frac {B^2\,a}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-B^4\,b^2\,d^4}\,32{}\mathrm {i}}{\frac {16\,B\,b^4\,\sqrt {-B^4\,b^2\,d^4}}{d}+\frac {16\,B\,a^2\,b^2\,\sqrt {-B^4\,b^2\,d^4}}{d}}\right )\,\sqrt {\frac {\sqrt {-B^4\,b^2\,d^4}+B^2\,a\,d^2}{4\,d^4}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {B^2\,b^4\,\sqrt {\frac {B^2\,a}{4\,d^2}-\frac {\sqrt {-B^4\,b^2\,d^4}}{4\,d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {16\,B\,b^4\,\sqrt {-B^4\,b^2\,d^4}}{d^3}+\frac {16\,B\,a^2\,b^2\,\sqrt {-B^4\,b^2\,d^4}}{d^3}}-\frac {a\,b^2\,\sqrt {\frac {B^2\,a}{4\,d^2}-\frac {\sqrt {-B^4\,b^2\,d^4}}{4\,d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-B^4\,b^2\,d^4}\,32{}\mathrm {i}}{\frac {16\,B\,b^4\,\sqrt {-B^4\,b^2\,d^4}}{d}+\frac {16\,B\,a^2\,b^2\,\sqrt {-B^4\,b^2\,d^4}}{d}}\right )\,\sqrt {-\frac {\sqrt {-B^4\,b^2\,d^4}-B^2\,a\,d^2}{4\,d^4}}\,2{}\mathrm {i} \]

input

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

output

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