\(\int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx\) [436]

Optimal result

Integrand size = 35, antiderivative size = 204 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {(i a-b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {b} (2 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d} \] Output:

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

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.19 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {-\sqrt [4]{-1} (-a+i b)^{3/2} (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-(-1)^{3/4} (a+i b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+b B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+\frac {\sqrt {a} \sqrt {b} (2 A b+3 a B) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}}{d} \] Input:

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

Output:

(-((-1)^(1/4)*(-a + I*b)^(3/2)*(I*A + B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b] 
*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]) - (-1)^(3/4)*(a + I*b)^(3/ 
2)*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + 
 b*Tan[c + d*x]]] + b*B*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + (Sqr 
t[a]*Sqrt[b]*(2*A*b + 3*a*B)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]] 
*Sqrt[1 + (b*Tan[c + d*x])/a])/Sqrt[a + b*Tan[c + d*x]])/d
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3042, 4090, 27, 3042, 4138, 2035, 2257, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4090

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4138

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-((I*a - b)^(3/2)*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqr 
t[a + b*Tan[c + d*x]]]) + Sqrt[b]*(2*A*b + 3*a*B)*ArcTanh[(Sqrt[b]*Sqrt[Ta 
n[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - (I*a + b)^(3/2)*(A - I*B)*ArcTanh 
[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + (b*B*Sq 
rt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/d
 

Defintions of rubi rules used

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2257

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a 
, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

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 4138

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] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S 
imp[ff/f   Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^ 
2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f 
, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + 
 d^2, 0]
 

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 1.10 (sec) , antiderivative size = 2398415, normalized size of antiderivative = 11756.94

\[\text {output too large to display}\]

Input:

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

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 12109 vs. \(2 (164) = 328\).

Time = 4.21 (sec) , antiderivative size = 24220, normalized size of antiderivative = 118.73 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {2 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, a^{2}-2 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}}{a +\tan \left (d x +c \right ) b}d x \right ) a^{2} b d -\left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )}{a +\tan \left (d x +c \right ) b}d x \right ) a^{3} d +\left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}}{\tan \left (d x +c \right )^{2} b +\tan \left (d x +c \right ) a}d x \right ) a^{3} d +2 \left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )d x \right ) b^{2} d +4 \left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}d x \right ) a b d}{2 d} \] Input:

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

Output:

(2*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*a**2 - 2*int((sqrt(tan(c + 
d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**2)/(tan(c + d*x)*b + a),x)*a* 
*2*b*d - int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x))/(t 
an(c + d*x)*b + a),x)*a**3*d + int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b 
 + a))/(tan(c + d*x)**2*b + tan(c + d*x)*a),x)*a**3*d + 2*int(sqrt(tan(c + 
 d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x),x)*b**2*d + 4*int(sqrt(tan(c 
+ d*x))*sqrt(tan(c + d*x)*b + a),x)*a*b*d)/(2*d)