3.4.43 \(\int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx\) [343]

3.4.43.1 Optimal result

Integrand size = 33, antiderivative size = 213 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b} d}+\frac {(A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b} d}-\frac {2 \left (10 a A b-8 a^2 B+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}+\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d} \]

output

(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/(a+I*b)^(1/2)-2/15*(10 
*A*a*b-8*B*a^2+15*B*b^2)*(a+b*tan(d*x+c))^(1/2)/b^3/d+2/15*(5*A*b-4*B*a)*( 
a+b*tan(d*x+c))^(1/2)*tan(d*x+c)/b^2/d+2/5*B*(a+b*tan(d*x+c))^(1/2)*tan(d* 
x+c)^2/b/d
 

3.4.43.2 Mathematica [A] (verified)

Time = 4.34 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.80 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {\frac {15 (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b}}+\frac {15 (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b}}+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-10 a A b+8 a^2 B-15 b^2 B+b (5 A b-4 a B) \tan (c+d x)+3 b^2 B \tan ^2(c+d x)\right )}{b^3}}{15 d} \]

input

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

output

((15*(A - I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/Sqrt[a - I 
*b] + (15*(A + I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/Sqrt[ 
a + I*b] + (2*Sqrt[a + b*Tan[c + d*x]]*(-10*a*A*b + 8*a^2*B - 15*b^2*B + b 
*(5*A*b - 4*a*B)*Tan[c + d*x] + 3*b^2*B*Tan[c + d*x]^2))/b^3)/(15*d)
 

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

Time = 1.19 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.04, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4090, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4090

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

\(\displaystyle \frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 \int \frac {15 A \tan (c+d x) b^2+\left (-8 B a^2+10 A b a+15 b^2 B\right ) \tan ^2(c+d x)+2 a (5 A b-4 a B)}{2 \sqrt {a+b \tan (c+d x)}}dx}{3 b}-\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}}{5 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {\int \frac {15 A \tan (c+d x) b^2+\left (-8 B a^2+10 A b a+15 b^2 B\right ) \tan ^2(c+d x)+2 a (5 A b-4 a B)}{\sqrt {a+b \tan (c+d x)}}dx}{3 b}-\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {\int \frac {15 A \tan (c+d x) b^2+\left (-8 B a^2+10 A b a+15 b^2 B\right ) \tan (c+d x)^2+2 a (5 A b-4 a B)}{\sqrt {a+b \tan (c+d x)}}dx}{3 b}-\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}}{5 b}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {\int \frac {15 A b^2 \tan (c+d x)-15 b^2 B}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^2 B+10 a A b+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}-\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {\int \frac {15 A b^2 \tan (c+d x)-15 b^2 B}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^2 B+10 a A b+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}-\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}}{5 b}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {-\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}+\frac {\frac {15}{2} b^2 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {15}{2} b^2 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^2 B+10 a A b+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{5 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {-\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}+\frac {\frac {15}{2} b^2 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {15}{2} b^2 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (-8 a^2 B+10 a A b+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{5 b}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {-\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}+\frac {-\frac {15 i b^2 (B+i A) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {15 i b^2 (-B+i A) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (-8 a^2 B+10 a A b+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{5 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {-\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}+\frac {\frac {15 i b^2 (B+i A) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {15 i b^2 (-B+i A) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (-8 a^2 B+10 a A b+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{5 b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {-\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}+\frac {\frac {15 b (-B+i A) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}-\frac {15 b (B+i A) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {2 \left (-8 a^2 B+10 a A b+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{5 b}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 B \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {-\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}+\frac {\frac {2 \left (-8 a^2 B+10 a A b+15 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{b d}-\frac {15 b^2 (B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {15 b^2 (-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}}{3 b}}{5 b}\)

input

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

output

(2*B*Tan[c + d*x]^2*Sqrt[a + b*Tan[c + d*x]])/(5*b*d) - ((-2*(5*A*b - 4*a* 
B)*Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(3*b*d) + ((-15*b^2*(I*A + B)*Ar 
cTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + (15*b^2*(I*A - B)*Ar 
cTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d) + (2*(10*a*A*b - 8*a^2 
*B + 15*b^2*B)*Sqrt[a + b*Tan[c + d*x]])/(b*d))/(3*b))/(5*b)
 

3.4.43.3.1 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 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])))
 

3.4.43.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2014\) vs. \(2(183)=366\).

Time = 0.18 (sec) , antiderivative size = 2015, normalized size of antiderivative = 9.46

method	result	size
parts	\(\text {Expression too large to display}\)	\(2015\)
derivativedivides	\(\text {Expression too large to display}\)	\(4107\)
default	\(\text {Expression too large to display}\)	\(4107\)

input

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

output

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

3.4.43.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1767 vs. \(2 (177) = 354\).

Time = 0.30 (sec) , antiderivative size = 1767, normalized size of antiderivative = 8.30 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]

input

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

output

-1/30*(15*b^3*d*sqrt(((a^2 + b^2)*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)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + 2 
*A*B*b + (A^2 - B^2)*a)/((a^2 + b^2)*d^2))*log((2*(A^3*B + A*B^3)*a - (A^4 
 - B^4)*b)*sqrt(b*tan(d*x + c) + a) + ((B*a^3 - A*a^2*b + B*a*b^2 - A*b^3) 
*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)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - (2*A^2*B*a^2 - (A^3 - 3*A*B^2)*a*b 
 - (A^2*B - B^3)*b^2)*d)*sqrt(((a^2 + b^2)*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)/((a^4 + 2*a^2*b^2 + b^4)* 
d^4)) + 2*A*B*b + (A^2 - B^2)*a)/((a^2 + b^2)*d^2))) - 15*b^3*d*sqrt(((a^2 
 + b^2)*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)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + 2*A*B*b + (A^2 - B^2)*a)/(( 
a^2 + b^2)*d^2))*log((2*(A^3*B + A*B^3)*a - (A^4 - B^4)*b)*sqrt(b*tan(d*x 
+ c) + a) - ((B*a^3 - A*a^2*b + B*a*b^2 - A*b^3)*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)/((a^4 + 2*a^2*b^2 + 
 b^4)*d^4)) - (2*A^2*B*a^2 - (A^3 - 3*A*B^2)*a*b - (A^2*B - B^3)*b^2)*d)*s 
qrt(((a^2 + b^2)*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)/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + 2*A*B*b + (A^2 - B 
^2)*a)/((a^2 + b^2)*d^2))) - 15*b^3*d*sqrt(-((a^2 + b^2)*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)/((a^4 + 2*a 
^2*b^2 + b^4)*d^4)) - 2*A*B*b - (A^2 - B^2)*a)/((a^2 + b^2)*d^2))*log((...
 

3.4.43.6 Sympy [F]

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

input

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

output

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

3.4.43.7 Maxima [F]

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

input

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

output

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

3.4.43.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.4.43.9 Mupad [B] (verification not implemented)

Time = 16.40 (sec) , antiderivative size = 3054, normalized size of antiderivative = 14.34 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]

input

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

output

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