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

Optimal. Leaf size=298 \[ -\frac {3 c \sqrt {c+a^2 c x^2}}{8 a}-\frac {\left (c+a^2 c x^2\right )^{3/2}}{12 a}+\frac {3}{8} c x \sqrt {c+a^2 c x^2} \text {ArcTan}(a x)+\frac {1}{4} x \left (c+a^2 c x^2\right )^{3/2} \text {ArcTan}(a x)-\frac {3 i c^2 \sqrt {1+a^2 x^2} \text {ArcTan}(a x) \text {ArcTan}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{4 a \sqrt {c+a^2 c x^2}}+\frac {3 i c^2 \sqrt {1+a^2 x^2} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 a \sqrt {c+a^2 c x^2}}-\frac {3 i c^2 \sqrt {1+a^2 x^2} \text {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 a \sqrt {c+a^2 c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4998, 5010, 5006} \begin {gather*} -\frac {3 i c^2 \sqrt {a^2 x^2+1} \text {ArcTan}(a x) \text {ArcTan}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{4 a \sqrt {a^2 c x^2+c}}+\frac {3}{8} c x \text {ArcTan}(a x) \sqrt {a^2 c x^2+c}+\frac {1}{4} x \text {ArcTan}(a x) \left (a^2 c x^2+c\right )^{3/2}+\frac {3 i c^2 \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{8 a \sqrt {a^2 c x^2+c}}-\frac {3 i c^2 \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{8 a \sqrt {a^2 c x^2+c}}-\frac {3 c \sqrt {a^2 c x^2+c}}{8 a}-\frac {\left (a^2 c x^2+c\right )^{3/2}}{12 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*c*Sqrt[c + a^2*c*x^2])/(8*a) - (c + a^2*c*x^2)^(3/2)/(12*a) + (3*c*x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/8 +
(x*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/4 - (((3*I)/4)*c^2*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]) + (((3*I)/8)*c^2*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]) - (((3*I)/8)*c^2*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqrt[1 + I*a*x])/S
qrt[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*} \int \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x) \, dx &=-\frac {\left (c+a^2 c x^2\right )^{3/2}}{12 a}+\frac {1}{4} x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)+\frac {1}{4} (3 c) \int \sqrt {c+a^2 c x^2} \tan ^{-1}(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} \tan ^{-1}(a x)+\frac {1}{4} x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)+\frac {1}{8} \left (3 c^2\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, 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} \tan ^{-1}(a x)+\frac {1}{4} x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)+\frac {\left (3 c^2 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{8 \sqrt {c+a^2 c x^2}}\\ &=-\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} \tan ^{-1}(a x)+\frac {1}{4} x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)-\frac {3 i c^2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\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} \text {Li}_2\left (-\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} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 a \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]
time = 1.87, size = 351, normalized size = 1.18 \begin {gather*} \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 \text {ArcTan}(a x))+6 \left (1+a^2 x^2\right )^2 \cos (3 \text {ArcTan}(a x))+96 \text {ArcTan}(a x) \left (\log \left (1-i e^{i \text {ArcTan}(a x)}\right )-\log \left (1+i e^{i \text {ArcTan}(a x)}\right )\right )+72 i \text {PolyLog}\left (2,-i e^{i \text {ArcTan}(a x)}\right )-72 i \text {PolyLog}\left (2,i e^{i \text {ArcTan}(a x)}\right )-3 \left (1+a^2 x^2\right )^2 \text {ArcTan}(a x) \left (-\frac {14 a x}{\sqrt {1+a^2 x^2}}+3 \log \left (1-i e^{i \text {ArcTan}(a x)}\right )+4 \cos (2 \text {ArcTan}(a x)) \left (\log \left (1-i e^{i \text {ArcTan}(a x)}\right )-\log \left (1+i e^{i \text {ArcTan}(a x)}\right )\right )+\cos (4 \text {ArcTan}(a x)) \left (\log \left (1-i e^{i \text {ArcTan}(a x)}\right )-\log \left (1+i e^{i \text {ArcTan}(a x)}\right )\right )-3 \log \left (1+i e^{i \text {ArcTan}(a x)}\right )+2 \sin (3 \text {ArcTan}(a x))\right )\right )}{192 a \sqrt {1+a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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)*P
olyLog[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])

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Maple [A]
time = 0.24, size = 201, normalized size = 0.67

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \operatorname {atan}{\left (a x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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