\(\int (c+a^2 c x^2)^{5/2} \arctan (a x) \, dx\) [219]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (warning: unable to verify)
   Maple [A] (verified)
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F(-2)]
   Mupad [F(-1)]

Optimal result

Integrand size = 19, antiderivative size = 348 \[ \int \left (c+a^2 c x^2\right )^{5/2} \arctan (a x) \, dx=-\frac {5 c^2 \sqrt {c+a^2 c x^2}}{16 a}-\frac {5 c \left (c+a^2 c x^2\right )^{3/2}}{72 a}-\frac {\left (c+a^2 c x^2\right )^{5/2}}{30 a}+\frac {5}{16} c^2 x \sqrt {c+a^2 c x^2} \arctan (a x)+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \arctan (a x)+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \arctan (a x)-\frac {5 i c^3 \sqrt {1+a^2 x^2} \arctan (a x) \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 a \sqrt {c+a^2 c x^2}}+\frac {5 i c^3 \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{16 a \sqrt {c+a^2 c x^2}}-\frac {5 i c^3 \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{16 a \sqrt {c+a^2 c x^2}} \]

[Out]

-5/72*c*(a^2*c*x^2+c)^(3/2)/a-1/30*(a^2*c*x^2+c)^(5/2)/a+5/24*c*x*(a^2*c*x^2+c)^(3/2)*arctan(a*x)+1/6*x*(a^2*c
*x^2+c)^(5/2)*arctan(a*x)-5/8*I*c^3*arctan(a*x)*arctan((1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a
^2*c*x^2+c)^(1/2)+5/16*I*c^3*polylog(2,-I*(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(
1/2)-5/16*I*c^3*polylog(2,I*(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)-5/16*c^2*
(a^2*c*x^2+c)^(1/2)/a+5/16*c^2*x*arctan(a*x)*(a^2*c*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4998, 5010, 5006} \[ \int \left (c+a^2 c x^2\right )^{5/2} \arctan (a x) \, dx=-\frac {5 i c^3 \sqrt {a^2 x^2+1} \arctan (a x) \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 a \sqrt {a^2 c x^2+c}}+\frac {5}{16} c^2 x \arctan (a x) \sqrt {a^2 c x^2+c}+\frac {5}{24} c x \arctan (a x) \left (a^2 c x^2+c\right )^{3/2}+\frac {1}{6} x \arctan (a x) \left (a^2 c x^2+c\right )^{5/2}+\frac {5 i c^3 \sqrt {a^2 x^2+1} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{16 a \sqrt {a^2 c x^2+c}}-\frac {5 i c^3 \sqrt {a^2 x^2+1} \operatorname {PolyLog}\left (2,\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{16 a \sqrt {a^2 c x^2+c}}-\frac {5 c^2 \sqrt {a^2 c x^2+c}}{16 a}-\frac {5 c \left (a^2 c x^2+c\right )^{3/2}}{72 a}-\frac {\left (a^2 c x^2+c\right )^{5/2}}{30 a} \]

[In]

Int[(c + a^2*c*x^2)^(5/2)*ArcTan[a*x],x]

[Out]

(-5*c^2*Sqrt[c + a^2*c*x^2])/(16*a) - (5*c*(c + a^2*c*x^2)^(3/2))/(72*a) - (c + a^2*c*x^2)^(5/2)/(30*a) + (5*c
^2*x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/16 + (5*c*x*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/24 + (x*(c + a^2*c*x^2)^(
5/2)*ArcTan[a*x])/6 - (((5*I)/8)*c^3*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(a
*Sqrt[c + a^2*c*x^2]) + (((5*I)/16)*c^3*Sqrt[1 + a^2*x^2]*PolyLog[2, ((-I)*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/
(a*Sqrt[c + a^2*c*x^2]) - (((5*I)/16)*c^3*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(
a*Sqrt[c + a^2*c*x^2])

Rule 4998

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

Rule 5006

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]) /; F
reeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[d, 0]

Rule 5010

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + c^2*x^2]/Sq
rt[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (c+a^2 c x^2\right )^{5/2}}{30 a}+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \arctan (a x)+\frac {1}{6} (5 c) \int \left (c+a^2 c x^2\right )^{3/2} \arctan (a x) \, dx \\ & = -\frac {5 c \left (c+a^2 c x^2\right )^{3/2}}{72 a}-\frac {\left (c+a^2 c x^2\right )^{5/2}}{30 a}+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \arctan (a x)+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \arctan (a x)+\frac {1}{8} \left (5 c^2\right ) \int \sqrt {c+a^2 c x^2} \arctan (a x) \, dx \\ & = -\frac {5 c^2 \sqrt {c+a^2 c x^2}}{16 a}-\frac {5 c \left (c+a^2 c x^2\right )^{3/2}}{72 a}-\frac {\left (c+a^2 c x^2\right )^{5/2}}{30 a}+\frac {5}{16} c^2 x \sqrt {c+a^2 c x^2} \arctan (a x)+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \arctan (a x)+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \arctan (a x)+\frac {1}{16} \left (5 c^3\right ) \int \frac {\arctan (a x)}{\sqrt {c+a^2 c x^2}} \, dx \\ & = -\frac {5 c^2 \sqrt {c+a^2 c x^2}}{16 a}-\frac {5 c \left (c+a^2 c x^2\right )^{3/2}}{72 a}-\frac {\left (c+a^2 c x^2\right )^{5/2}}{30 a}+\frac {5}{16} c^2 x \sqrt {c+a^2 c x^2} \arctan (a x)+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \arctan (a x)+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \arctan (a x)+\frac {\left (5 c^3 \sqrt {1+a^2 x^2}\right ) \int \frac {\arctan (a x)}{\sqrt {1+a^2 x^2}} \, dx}{16 \sqrt {c+a^2 c x^2}} \\ & = -\frac {5 c^2 \sqrt {c+a^2 c x^2}}{16 a}-\frac {5 c \left (c+a^2 c x^2\right )^{3/2}}{72 a}-\frac {\left (c+a^2 c x^2\right )^{5/2}}{30 a}+\frac {5}{16} c^2 x \sqrt {c+a^2 c x^2} \arctan (a x)+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \arctan (a x)+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \arctan (a x)-\frac {5 i c^3 \sqrt {1+a^2 x^2} \arctan (a x) \arctan \left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 a \sqrt {c+a^2 c x^2}}+\frac {5 i c^3 \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{16 a \sqrt {c+a^2 c x^2}}-\frac {5 i c^3 \sqrt {1+a^2 x^2} \operatorname {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{16 a \sqrt {c+a^2 c x^2}} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 5.36 (sec) , antiderivative size = 643, normalized size of antiderivative = 1.85 \[ \int \left (c+a^2 c x^2\right )^{5/2} \arctan (a x) \, dx=\frac {c^2 \sqrt {c+a^2 c x^2} \left (\frac {3}{4} \left (1+a^2 x^2\right )^{5/2}+720 \sqrt {1+a^2 x^2} (-1+a x \arctan (a x))+\frac {55}{8} \left (1+a^2 x^2\right )^3 \cos (3 \arctan (a x))-\frac {45}{8} \left (1+a^2 x^2\right )^3 \cos (5 \arctan (a x))+720 \arctan (a x) \left (\log \left (1-i e^{i \arctan (a x)}\right )-\log \left (1+i e^{i \arctan (a x)}\right )\right )+450 i \operatorname {PolyLog}\left (2,-i e^{i \arctan (a x)}\right )-450 i \operatorname {PolyLog}\left (2,i e^{i \arctan (a x)}\right )-15 \left (1+a^2 x^2\right )^2 \left (-\frac {2}{\sqrt {1+a^2 x^2}}-6 \cos (3 \arctan (a x))+3 \arctan (a x) \left (-\frac {14 a x}{\sqrt {1+a^2 x^2}}+3 \log \left (1-i e^{i \arctan (a x)}\right )+4 \cos (2 \arctan (a x)) \left (\log \left (1-i e^{i \arctan (a x)}\right )-\log \left (1+i e^{i \arctan (a x)}\right )\right )+\cos (4 \arctan (a x)) \left (\log \left (1-i e^{i \arctan (a x)}\right )-\log \left (1+i e^{i \arctan (a x)}\right )\right )-3 \log \left (1+i e^{i \arctan (a x)}\right )+2 \sin (3 \arctan (a x))\right )\right )+\frac {15}{16} \left (1+a^2 x^2\right )^3 \arctan (a x) \left (\frac {156 a x}{\sqrt {1+a^2 x^2}}+30 \log \left (1-i e^{i \arctan (a x)}\right )+3 \cos (6 \arctan (a x)) \log \left (1-i e^{i \arctan (a x)}\right )+45 \cos (2 \arctan (a x)) \left (\log \left (1-i e^{i \arctan (a x)}\right )-\log \left (1+i e^{i \arctan (a x)}\right )\right )+18 \cos (4 \arctan (a x)) \left (\log \left (1-i e^{i \arctan (a x)}\right )-\log \left (1+i e^{i \arctan (a x)}\right )\right )-30 \log \left (1+i e^{i \arctan (a x)}\right )-3 \cos (6 \arctan (a x)) \log \left (1+i e^{i \arctan (a x)}\right )-94 \sin (3 \arctan (a x))+6 \sin (5 \arctan (a x))\right )\right )}{1440 a \sqrt {1+a^2 x^2}} \]

[In]

Integrate[(c + a^2*c*x^2)^(5/2)*ArcTan[a*x],x]

[Out]

(c^2*Sqrt[c + a^2*c*x^2]*((3*(1 + a^2*x^2)^(5/2))/4 + 720*Sqrt[1 + a^2*x^2]*(-1 + a*x*ArcTan[a*x]) + (55*(1 +
a^2*x^2)^3*Cos[3*ArcTan[a*x]])/8 - (45*(1 + a^2*x^2)^3*Cos[5*ArcTan[a*x]])/8 + 720*ArcTan[a*x]*(Log[1 - I*E^(I
*ArcTan[a*x])] - Log[1 + I*E^(I*ArcTan[a*x])]) + (450*I)*PolyLog[2, (-I)*E^(I*ArcTan[a*x])] - (450*I)*PolyLog[
2, I*E^(I*ArcTan[a*x])] - 15*(1 + a^2*x^2)^2*(-2/Sqrt[1 + a^2*x^2] - 6*Cos[3*ArcTan[a*x]] + 3*ArcTan[a*x]*((-1
4*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]])) + (15*(1 + a^2*x^2)^3*ArcTan[a*x]*((156*a*x)/S
qrt[1 + a^2*x^2] + 30*Log[1 - I*E^(I*ArcTan[a*x])] + 3*Cos[6*ArcTan[a*x]]*Log[1 - I*E^(I*ArcTan[a*x])] + 45*Co
s[2*ArcTan[a*x]]*(Log[1 - I*E^(I*ArcTan[a*x])] - Log[1 + I*E^(I*ArcTan[a*x])]) + 18*Cos[4*ArcTan[a*x]]*(Log[1
- I*E^(I*ArcTan[a*x])] - Log[1 + I*E^(I*ArcTan[a*x])]) - 30*Log[1 + I*E^(I*ArcTan[a*x])] - 3*Cos[6*ArcTan[a*x]
]*Log[1 + I*E^(I*ArcTan[a*x])] - 94*Sin[3*ArcTan[a*x]] + 6*Sin[5*ArcTan[a*x]]))/16))/(1440*a*Sqrt[1 + a^2*x^2]
)

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.65

method result size
default \(\frac {c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (120 \arctan \left (a x \right ) a^{5} x^{5}-24 a^{4} x^{4}+390 \arctan \left (a x \right ) x^{3} a^{3}-98 a^{2} x^{2}+495 x \arctan \left (a x \right ) a -299\right )}{720 a}-\frac {5 c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (\arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{16 a \sqrt {a^{2} x^{2}+1}}\) \(225\)

[In]

int((a^2*c*x^2+c)^(5/2)*arctan(a*x),x,method=_RETURNVERBOSE)

[Out]

1/720*c^2/a*(c*(a*x-I)*(I+a*x))^(1/2)*(120*arctan(a*x)*a^5*x^5-24*a^4*x^4+390*arctan(a*x)*x^3*a^3-98*a^2*x^2+4
95*x*arctan(a*x)*a-299)-5/16*c^2*(c*(a*x-I)*(I+a*x))^(1/2)*(arctan(a*x)*ln(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-ar
ctan(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]

\[ \int \left (c+a^2 c x^2\right )^{5/2} \arctan (a x) \, dx=\int { {\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \arctan \left (a x\right ) \,d x } \]

[In]

integrate((a^2*c*x^2+c)^(5/2)*arctan(a*x),x, algorithm="fricas")

[Out]

integral((a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2)*sqrt(a^2*c*x^2 + c)*arctan(a*x), x)

Sympy [F]

\[ \int \left (c+a^2 c x^2\right )^{5/2} \arctan (a x) \, dx=\int \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \operatorname {atan}{\left (a x \right )}\, dx \]

[In]

integrate((a**2*c*x**2+c)**(5/2)*atan(a*x),x)

[Out]

Integral((c*(a**2*x**2 + 1))**(5/2)*atan(a*x), x)

Maxima [F]

\[ \int \left (c+a^2 c x^2\right )^{5/2} \arctan (a x) \, dx=\int { {\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \arctan \left (a x\right ) \,d x } \]

[In]

integrate((a^2*c*x^2+c)^(5/2)*arctan(a*x),x, algorithm="maxima")

[Out]

integrate((a^2*c*x^2 + c)^(5/2)*arctan(a*x), x)

Giac [F(-2)]

Exception generated. \[ \int \left (c+a^2 c x^2\right )^{5/2} \arctan (a x) \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((a^2*c*x^2+c)^(5/2)*arctan(a*x),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \left (c+a^2 c x^2\right )^{5/2} \arctan (a x) \, dx=\int \mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^{5/2} \,d x \]

[In]

int(atan(a*x)*(c + a^2*c*x^2)^(5/2),x)

[Out]

int(atan(a*x)*(c + a^2*c*x^2)^(5/2), x)