\(\int (c+a^2 c x^2)^{3/2} \arctan (a x) \, dx\) [211]

Optimal result

Integrand size = 19, antiderivative size = 298 \[ \int \left (c+a^2 c x^2\right )^{3/2} \arctan (a x) \, dx=-\frac {3 c \sqrt {c+a^2 c x^2}}{8 a}-\frac {\left (c+a^2 c x^2\right )^{3/2}}{12 a}+\frac {3}{8} c x \sqrt {c+a^2 c x^2} \arctan (a x)+\frac {1}{4} x \left (c+a^2 c x^2\right )^{3/2} \arctan (a x)-\frac {3 i c^2 \sqrt {1+a^2 x^2} \arctan (a x) \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{4 a \sqrt {c+a^2 c x^2}}+\frac {3 i c^2 \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 a \sqrt {c+a^2 c x^2}}-\frac {3 i c^2 \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 a \sqrt {c+a^2 c x^2}} \] Output:

-3/8*c*(a^2*c*x^2+c)^(1/2)/a-1/12*(a^2*c*x^2+c)^(3/2)/a+3/8*c*x*(a^2*c*x^2 
+c)^(1/2)*arctan(a*x)+1/4*x*(a^2*c*x^2+c)^(3/2)*arctan(a*x)-3/4*I*c^2*(a^2 
*x^2+1)^(1/2)*arctan(a*x)*arctan((1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))/a/(a^2*c 
*x^2+c)^(1/2)+3/8*I*c^2*(a^2*x^2+1)^(1/2)*polylog(2,-I*(1+I*a*x)^(1/2)/(1- 
I*a*x)^(1/2))/a/(a^2*c*x^2+c)^(1/2)-3/8*I*c^2*(a^2*x^2+1)^(1/2)*polylog(2, 
I*(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))/a/(a^2*c*x^2+c)^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 1.93 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.18 \[ \int \left (c+a^2 c x^2\right )^{3/2} \arctan (a x) \, dx=\frac {c \sqrt {c+a^2 c x^2} \left (2 \left (1+a^2 x^2\right )^{3/2}+96 \sqrt {1+a^2 x^2} (-1+a x \arctan (a x))+6 \left (1+a^2 x^2\right )^2 \cos (3 \arctan (a x))+96 \arctan (a x) \left (\log \left (1-i e^{i \arctan (a x)}\right )-\log \left (1+i e^{i \arctan (a x)}\right )\right )+72 i \operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )-72 i \operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )-3 \left (1+a^2 x^2\right )^2 \arctan (a x) \left (-\frac {14 a x}{\sqrt {1+a^2 x^2}}+3 \log \left (1-i e^{i \arctan (a x)}\right )+4 \cos (2 \arctan (a x)) \left (\log \left (1-i e^{i \arctan (a x)}\right )-\log \left (1+i e^{i \arctan (a x)}\right )\right )+\cos (4 \arctan (a x)) \left (\log \left (1-i e^{i \arctan (a x)}\right )-\log \left (1+i e^{i \arctan (a x)}\right )\right )-3 \log \left (1+i e^{i \arctan (a x)}\right )+2 \sin (3 \arctan (a x))\right )\right )}{192 a \sqrt {1+a^2 x^2}} \] Input:

Integrate[(c + a^2*c*x^2)^(3/2)*ArcTan[a*x],x]
 

Output:

(c*Sqrt[c + a^2*c*x^2]*(2*(1 + a^2*x^2)^(3/2) + 96*Sqrt[1 + a^2*x^2]*(-1 + 
 a*x*ArcTan[a*x]) + 6*(1 + a^2*x^2)^2*Cos[3*ArcTan[a*x]] + 96*ArcTan[a*x]* 
(Log[1 - I*E^(I*ArcTan[a*x])] - Log[1 + I*E^(I*ArcTan[a*x])]) + (72*I)*Pol 
yLog[2, (-I)*E^(I*ArcTan[a*x])] - (72*I)*PolyLog[2, I*E^(I*ArcTan[a*x])] - 
 3*(1 + a^2*x^2)^2*ArcTan[a*x]*((-14*a*x)/Sqrt[1 + a^2*x^2] + 3*Log[1 - I* 
E^(I*ArcTan[a*x])] + 4*Cos[2*ArcTan[a*x]]*(Log[1 - I*E^(I*ArcTan[a*x])] - 
Log[1 + I*E^(I*ArcTan[a*x])]) + Cos[4*ArcTan[a*x]]*(Log[1 - I*E^(I*ArcTan[ 
a*x])] - Log[1 + I*E^(I*ArcTan[a*x])]) - 3*Log[1 + I*E^(I*ArcTan[a*x])] + 
2*Sin[3*ArcTan[a*x]])))/(192*a*Sqrt[1 + a^2*x^2])
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.80, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {5413, 5413, 5425, 5421}

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.

\(\displaystyle \int \arctan (a x) \left (a^2 c x^2+c\right )^{3/2} \, dx\)

\(\Big \downarrow \) 5413

\(\displaystyle \frac {3}{4} c \int \sqrt {a^2 c x^2+c} \arctan (a x)dx+\frac {1}{4} x \arctan (a x) \left (a^2 c x^2+c\right )^{3/2}-\frac {\left (a^2 c x^2+c\right )^{3/2}}{12 a}\)

\(\Big \downarrow \) 5413

\(\displaystyle \frac {3}{4} c \left (\frac {1}{2} c \int \frac {\arctan (a x)}{\sqrt {a^2 c x^2+c}}dx+\frac {1}{2} x \arctan (a x) \sqrt {a^2 c x^2+c}-\frac {\sqrt {a^2 c x^2+c}}{2 a}\right )+\frac {1}{4} x \arctan (a x) \left (a^2 c x^2+c\right )^{3/2}-\frac {\left (a^2 c x^2+c\right )^{3/2}}{12 a}\)

\(\Big \downarrow \) 5425

\(\displaystyle \frac {3}{4} c \left (\frac {c \sqrt {a^2 x^2+1} \int \frac {\arctan (a x)}{\sqrt {a^2 x^2+1}}dx}{2 \sqrt {a^2 c x^2+c}}+\frac {1}{2} x \arctan (a x) \sqrt {a^2 c x^2+c}-\frac {\sqrt {a^2 c x^2+c}}{2 a}\right )+\frac {1}{4} x \arctan (a x) \left (a^2 c x^2+c\right )^{3/2}-\frac {\left (a^2 c x^2+c\right )^{3/2}}{12 a}\)

\(\Big \downarrow \) 5421

\(\displaystyle \frac {3}{4} c \left (\frac {c \sqrt {a^2 x^2+1} \left (-\frac {2 i \arctan (a x) \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{a}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a}-\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{a}\right )}{2 \sqrt {a^2 c x^2+c}}+\frac {1}{2} x \arctan (a x) \sqrt {a^2 c x^2+c}-\frac {\sqrt {a^2 c x^2+c}}{2 a}\right )+\frac {1}{4} x \arctan (a x) \left (a^2 c x^2+c\right )^{3/2}-\frac {\left (a^2 c x^2+c\right )^{3/2}}{12 a}\)

Input:

Int[(c + a^2*c*x^2)^(3/2)*ArcTan[a*x],x]
 

Output:

-1/12*(c + a^2*c*x^2)^(3/2)/a + (x*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/4 + 
(3*c*(-1/2*Sqrt[c + a^2*c*x^2]/a + (x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/2 + 
 (c*Sqrt[1 + a^2*x^2]*(((-2*I)*ArcTan[a*x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt[1 - 
 I*a*x]])/a + (I*PolyLog[2, ((-I)*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/a - ( 
I*PolyLog[2, (I*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/a))/(2*Sqrt[c + a^2*c*x 
^2])))/4
 

Defintions of rubi rules used

rule 5413

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbo 
l] :> Simp[(-b)*((d + e*x^2)^q/(2*c*q*(2*q + 1))), x] + (Simp[x*(d + e*x^2) 
^q*((a + b*ArcTan[c*x])/(2*q + 1)), x] + Simp[2*d*(q/(2*q + 1))   Int[(d + 
e*x^2)^(q - 1)*(a + b*ArcTan[c*x]), x], x]) /; FreeQ[{a, b, c, d, e}, x] && 
 EqQ[e, c^2*d] && GtQ[q, 0]
 

rule 5421

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] 
 :> Simp[-2*I*(a + b*ArcTan[c*x])*(ArcTan[Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]]/ 
(c*Sqrt[d])), x] + (Simp[I*b*(PolyLog[2, (-I)*(Sqrt[1 + I*c*x]/Sqrt[1 - I*c 
*x])]/(c*Sqrt[d])), x] - Simp[I*b*(PolyLog[2, I*(Sqrt[1 + I*c*x]/Sqrt[1 - I 
*c*x])]/(c*Sqrt[d])), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && 
GtQ[d, 0]
 

rule 5425

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S 
ymbol] :> Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]   Int[(a + b*ArcTan[c*x])^ 
p/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] & 
& IGtQ[p, 0] &&  !GtQ[d, 0]
 

Maple [A] (verified)

Time = 1.36 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.67

method	result	size
default	\(\frac {c \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (6 \arctan \left (a x \right ) x^{3} a^{3}-2 a^{2} x^{2}+15 \arctan \left (a x \right ) a x -11\right )}{24 a}-\frac {3 c \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (\arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{8 a \sqrt {a^{2} x^{2}+1}}\)	\(201\)

Input:

int((a^2*c*x^2+c)^(3/2)*arctan(a*x),x,method=_RETURNVERBOSE)
 

Output:

1/24*c/a*(c*(a*x-I)*(a*x+I))^(1/2)*(6*arctan(a*x)*x^3*a^3-2*a^2*x^2+15*arc 
tan(a*x)*a*x-11)-3/8*c*(c*(a*x-I)*(a*x+I))^(1/2)*(arctan(a*x)*ln(1+I*(1+I* 
a*x)/(a^2*x^2+1)^(1/2))-arctan(a*x)*ln(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+I* 
dilog(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-I*dilog(1+I*(1+I*a*x)/(a^2*x^2+1)^( 
1/2)))/a/(a^2*x^2+1)^(1/2)
 

Fricas [F]

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

integrate((a^2*c*x^2+c)^(3/2)*arctan(a*x),x, algorithm="fricas")
 

Output:

integral((a^2*c*x^2 + c)^(3/2)*arctan(a*x), x)
 

Sympy [F]

\[ \int \left (c+a^2 c x^2\right )^{3/2} \arctan (a x) \, dx=\int \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \operatorname {atan}{\left (a x \right )}\, dx \] Input:

integrate((a**2*c*x**2+c)**(3/2)*atan(a*x),x)
 

Output:

Integral((c*(a**2*x**2 + 1))**(3/2)*atan(a*x), x)
 

Maxima [F]

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

integrate((a^2*c*x^2+c)^(3/2)*arctan(a*x),x, algorithm="maxima")
 

Output:

integrate((a^2*c*x^2 + c)^(3/2)*arctan(a*x), x)
 

Giac [F(-2)]

Exception generated. \[ \int \left (c+a^2 c x^2\right )^{3/2} \arctan (a x) \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a^2*c*x^2+c)^(3/2)*arctan(a*x),x, algorithm="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. \[ \int \left (c+a^2 c x^2\right )^{3/2} \arctan (a x) \, dx=\int \mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^{3/2} \,d x \] Input:

int(atan(a*x)*(c + a^2*c*x^2)^(3/2),x)
 

Output:

int(atan(a*x)*(c + a^2*c*x^2)^(3/2), x)
 

Reduce [F]

\[ \int \left (c+a^2 c x^2\right )^{3/2} \arctan (a x) \, dx=\sqrt {c}\, c \left (\left (\int \sqrt {a^{2} x^{2}+1}\, \mathit {atan} \left (a x \right ) x^{2}d x \right ) a^{2}+\int \sqrt {a^{2} x^{2}+1}\, \mathit {atan} \left (a x \right )d x \right ) \] Input:

int((a^2*c*x^2+c)^(3/2)*atan(a*x),x)
 

Output:

sqrt(c)*c*(int(sqrt(a**2*x**2 + 1)*atan(a*x)*x**2,x)*a**2 + int(sqrt(a**2* 
x**2 + 1)*atan(a*x),x))