3.4.25 \(\int \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) [325]

3.4.25.1 Optimal result

Integrand size = 33, antiderivative size = 214 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {(a-i b)^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {(a+i b)^{3/2} (i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 (A b+a B) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 B (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d} \]

output

(a-I*b)^(3/2)*(I*A+B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d-(a+I 
*b)^(3/2)*(I*A-B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d-2*(A*b+B 
*a)*(a+b*tan(d*x+c))^(1/2)/d-2/3*B*(a+b*tan(d*x+c))^(3/2)/d+2/35*(7*A*b-2* 
B*a)*(a+b*tan(d*x+c))^(5/2)/b^2/d+2/7*B*tan(d*x+c)*(a+b*tan(d*x+c))^(5/2)/ 
b/d
 

3.4.25.2 Mathematica [A] (verified)

Time = 2.80 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.18 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {\frac {2 (7 A b-2 a B) (a+b \tan (c+d x))^{5/2}}{b}+10 B \tan (c+d x) (a+b \tan (c+d x))^{5/2}+\frac {35}{3} b (A-i B) \left (3 \sqrt {a-i b} (i a+b) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )-i \sqrt {a+b \tan (c+d x)} (4 a-3 i b+b \tan (c+d x))\right )+\frac {35}{3} b (A+i B) \left (3 \sqrt {a+i b} (-i a+b) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+i \sqrt {a+b \tan (c+d x)} (4 a+3 i b+b \tan (c+d x))\right )}{35 b d} \]

input

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

output

((2*(7*A*b - 2*a*B)*(a + b*Tan[c + d*x])^(5/2))/b + 10*B*Tan[c + d*x]*(a + 
 b*Tan[c + d*x])^(5/2) + (35*b*(A - I*B)*(3*Sqrt[a - I*b]*(I*a + b)*ArcTan 
h[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] - I*Sqrt[a + b*Tan[c + d*x]]*(4* 
a - (3*I)*b + b*Tan[c + d*x])))/3 + (35*b*(A + I*B)*(3*Sqrt[a + I*b]*((-I) 
*a + b)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + I*Sqrt[a + b*Tan 
[c + d*x]]*(4*a + (3*I)*b + b*Tan[c + d*x])))/3)/(35*b*d)
 

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

Time = 1.28 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.98, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 4090, 27, 3042, 4113, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4090

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4113

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

input

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

output

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

3.4.25.3.1 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]
 

3.4.25.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1696\) vs. \(2(182)=364\).

Time = 0.15 (sec) , antiderivative size = 1697, normalized size of antiderivative = 7.93

method	result	size
parts	\(\text {Expression too large to display}\)	\(1697\)
derivativedivides	\(\text {Expression too large to display}\)	\(1717\)
default	\(\text {Expression too large to display}\)	\(1717\)

input

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

output

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

3.4.25.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3148 vs. \(2 (176) = 352\).

Time = 0.48 (sec) , antiderivative size = 3148, normalized size of antiderivative = 14.71 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

input

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

output

-1/210*(105*b^2*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 + 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)/d^4))/d^2)*log((2*(A^3*B + A*B^3)*a^5 + 3*(A^4 - B^4)*a^4*b - 4* 
(A^3*B + A*B^3)*a^3*b^2 + 2*(A^4 - B^4)*a^2*b^3 - 6*(A^3*B + A*B^3)*a*b^4 
- (A^4 - B^4)*b^5)*sqrt(b*tan(d*x + c) + a) + ((A*a - B*b)*d^3*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)/d^4) - (2*A*B^2*a^4 + 
 (5*A^2*B - 3*B^3)*a^3*b + 3*(A^3 - 3*A*B^2)*a^2*b^2 - (7*A^2*B - B^3)*a*b 
^3 - (A^3 - A*B^2)*b^4)*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 + 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)/d^4))/d^2)) - 105*b^2*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 + 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)/d^4))/d^2)*log((2*(A^3*B + A*B^3)*a^...
 

3.4.25.6 Sympy [F]

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

input

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

output

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

3.4.25.7 Maxima [F]

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

input

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

output

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

3.4.25.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.4.25.9 Mupad [B] (verification not implemented)

Time = 82.10 (sec) , antiderivative size = 2993, normalized size of antiderivative = 13.99 \[ \int \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

input

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

output

log((16*A^3*a*b^3*(a^2 + b^2)^2)/d^3 - (((16*b^2*(((-A^4*b^2*d^4*(3*a^2 - 
b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(A*b^3 + A*a^2*b 
 + a*d*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2* 
d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/d + (16*A^2*b^2*(a + b*tan(c 
+ d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^2)*(((-A^4*b^2*d^4*(3*a^2 - b^2)^ 
2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2))/2)*((6*A^4*a^2*b^4*d 
^4 - A^4*b^6*d^4 - 9*A^4*a^4*b^2*d^4)^(1/2)/(4*d^4) - (A^2*a^3)/(4*d^2) + 
(3*A^2*a*b^2)/(4*d^2))^(1/2) - log((16*A^3*a*b^3*(a^2 + b^2)^2)/d^3 - (((1 
6*b^2*(-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3*d^2 - 3*A^2*a*b^2* 
d^2)/d^4)^(1/2)*(A*b^3 + A*a^2*b - a*d*(-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^( 
1/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2 
)))/d - (16*A^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 + b^4 - 6*a^2*b^2))/d^ 
2)*(-((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2 
)/d^4)^(1/2))/2)*(-((6*A^4*a^2*b^4*d^4 - A^4*b^6*d^4 - 9*A^4*a^4*b^2*d^4)^ 
(1/2) + A^2*a^3*d^2 - 3*A^2*a*b^2*d^2)/(4*d^4))^(1/2) - log((16*A^3*a*b^3* 
(a^2 + b^2)^2)/d^3 - (((16*b^2*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^ 
2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(A*b^3 + A*a^2*b - a*d*(((-A^4*b^2 
*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*a^3*d^2 + 3*A^2*a*b^2*d^2)/d^4)^(1/2)*(a 
 + b*tan(c + d*x))^(1/2)))/d - (16*A^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^4 
 + b^4 - 6*a^2*b^2))/d^2)*(((-A^4*b^2*d^4*(3*a^2 - b^2)^2)^(1/2) - A^2*...