3.1.23 \(\int \frac {a+b \text {ArcTan}(c x^2)}{d+e x} \, dx\) [23]

Optimal. Leaf size=501 \[ \frac {\left (a+b \text {ArcTan}\left (c x^2\right )\right ) \log (d+e x)}{e}+\frac {b c \log \left (\frac {e \left (1-\sqrt [4]{-c^2} x\right )}{\sqrt [4]{-c^2} d+e}\right ) \log (d+e x)}{2 \sqrt {-c^2} e}+\frac {b c \log \left (-\frac {e \left (1+\sqrt [4]{-c^2} x\right )}{\sqrt [4]{-c^2} d-e}\right ) \log (d+e x)}{2 \sqrt {-c^2} e}-\frac {b c \log \left (\frac {e \left (1-\sqrt {-\sqrt {-c^2}} x\right )}{\sqrt {-\sqrt {-c^2}} d+e}\right ) \log (d+e x)}{2 \sqrt {-c^2} e}-\frac {b c \log \left (-\frac {e \left (1+\sqrt {-\sqrt {-c^2}} x\right )}{\sqrt {-\sqrt {-c^2}} d-e}\right ) \log (d+e x)}{2 \sqrt {-c^2} e}+\frac {b c \text {PolyLog}\left (2,\frac {\sqrt [4]{-c^2} (d+e x)}{\sqrt [4]{-c^2} d-e}\right )}{2 \sqrt {-c^2} e}-\frac {b c \text {PolyLog}\left (2,\frac {\sqrt {-\sqrt {-c^2}} (d+e x)}{\sqrt {-\sqrt {-c^2}} d-e}\right )}{2 \sqrt {-c^2} e}+\frac {b c \text {PolyLog}\left (2,\frac {\sqrt [4]{-c^2} (d+e x)}{\sqrt [4]{-c^2} d+e}\right )}{2 \sqrt {-c^2} e}-\frac {b c \text {PolyLog}\left (2,\frac {\sqrt {-\sqrt {-c^2}} (d+e x)}{\sqrt {-\sqrt {-c^2}} d+e}\right )}{2 \sqrt {-c^2} e} \]

[Out]

(a+b*arctan(c*x^2))*ln(e*x+d)/e+1/2*b*c*ln(e*(1-(-c^2)^(1/4)*x)/((-c^2)^(1/4)*d+e))*ln(e*x+d)/e/(-c^2)^(1/2)+1
/2*b*c*ln(-e*(1+(-c^2)^(1/4)*x)/((-c^2)^(1/4)*d-e))*ln(e*x+d)/e/(-c^2)^(1/2)-1/2*b*c*ln(e*x+d)*ln(e*(1-x*(-(-c
^2)^(1/2))^(1/2))/(e+d*(-(-c^2)^(1/2))^(1/2)))/e/(-c^2)^(1/2)-1/2*b*c*ln(e*x+d)*ln(-e*(1+x*(-(-c^2)^(1/2))^(1/
2))/(-e+d*(-(-c^2)^(1/2))^(1/2)))/e/(-c^2)^(1/2)+1/2*b*c*polylog(2,(-c^2)^(1/4)*(e*x+d)/((-c^2)^(1/4)*d-e))/e/
(-c^2)^(1/2)+1/2*b*c*polylog(2,(-c^2)^(1/4)*(e*x+d)/((-c^2)^(1/4)*d+e))/e/(-c^2)^(1/2)-1/2*b*c*polylog(2,(e*x+
d)*(-(-c^2)^(1/2))^(1/2)/(-e+d*(-(-c^2)^(1/2))^(1/2)))/e/(-c^2)^(1/2)-1/2*b*c*polylog(2,(e*x+d)*(-(-c^2)^(1/2)
)^(1/2)/(e+d*(-(-c^2)^(1/2))^(1/2)))/e/(-c^2)^(1/2)

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Rubi [A]
time = 0.65, antiderivative size = 501, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {4976, 281, 209, 2463, 266, 2441, 2440, 2438} \begin {gather*} \frac {\log (d+e x) \left (a+b \text {ArcTan}\left (c x^2\right )\right )}{e}+\frac {b c \text {Li}_2\left (\frac {\sqrt [4]{-c^2} (d+e x)}{\sqrt [4]{-c^2} d-e}\right )}{2 \sqrt {-c^2} e}-\frac {b c \text {Li}_2\left (\frac {\sqrt {-\sqrt {-c^2}} (d+e x)}{\sqrt {-\sqrt {-c^2}} d-e}\right )}{2 \sqrt {-c^2} e}+\frac {b c \text {Li}_2\left (\frac {\sqrt [4]{-c^2} (d+e x)}{\sqrt [4]{-c^2} d+e}\right )}{2 \sqrt {-c^2} e}-\frac {b c \text {Li}_2\left (\frac {\sqrt {-\sqrt {-c^2}} (d+e x)}{\sqrt {-\sqrt {-c^2}} d+e}\right )}{2 \sqrt {-c^2} e}+\frac {b c \log (d+e x) \log \left (\frac {e \left (1-\sqrt [4]{-c^2} x\right )}{\sqrt [4]{-c^2} d+e}\right )}{2 \sqrt {-c^2} e}+\frac {b c \log (d+e x) \log \left (-\frac {e \left (\sqrt [4]{-c^2} x+1\right )}{\sqrt [4]{-c^2} d-e}\right )}{2 \sqrt {-c^2} e}-\frac {b c \log (d+e x) \log \left (\frac {e \left (1-\sqrt {-\sqrt {-c^2}} x\right )}{\sqrt {-\sqrt {-c^2}} d+e}\right )}{2 \sqrt {-c^2} e}-\frac {b c \log (d+e x) \log \left (-\frac {e \left (\sqrt {-\sqrt {-c^2}} x+1\right )}{\sqrt {-\sqrt {-c^2}} d-e}\right )}{2 \sqrt {-c^2} e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x^2])/(d + e*x),x]

[Out]

((a + b*ArcTan[c*x^2])*Log[d + e*x])/e + (b*c*Log[(e*(1 - (-c^2)^(1/4)*x))/((-c^2)^(1/4)*d + e)]*Log[d + e*x])
/(2*Sqrt[-c^2]*e) + (b*c*Log[-((e*(1 + (-c^2)^(1/4)*x))/((-c^2)^(1/4)*d - e))]*Log[d + e*x])/(2*Sqrt[-c^2]*e)
- (b*c*Log[(e*(1 - Sqrt[-Sqrt[-c^2]]*x))/(Sqrt[-Sqrt[-c^2]]*d + e)]*Log[d + e*x])/(2*Sqrt[-c^2]*e) - (b*c*Log[
-((e*(1 + Sqrt[-Sqrt[-c^2]]*x))/(Sqrt[-Sqrt[-c^2]]*d - e))]*Log[d + e*x])/(2*Sqrt[-c^2]*e) + (b*c*PolyLog[2, (
(-c^2)^(1/4)*(d + e*x))/((-c^2)^(1/4)*d - e)])/(2*Sqrt[-c^2]*e) - (b*c*PolyLog[2, (Sqrt[-Sqrt[-c^2]]*(d + e*x)
)/(Sqrt[-Sqrt[-c^2]]*d - e)])/(2*Sqrt[-c^2]*e) + (b*c*PolyLog[2, ((-c^2)^(1/4)*(d + e*x))/((-c^2)^(1/4)*d + e)
])/(2*Sqrt[-c^2]*e) - (b*c*PolyLog[2, (Sqrt[-Sqrt[-c^2]]*(d + e*x))/(Sqrt[-Sqrt[-c^2]]*d + e)])/(2*Sqrt[-c^2]*
e)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2440

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + c*e*(x/g)])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2441

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[Log[e*((f + g
*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x)^n])/g), x] - Dist[b*e*(n/g), Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 4976

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

Rubi steps

\begin {align*} \int \frac {a+b \tan ^{-1}\left (c x^2\right )}{d+e x} \, dx &=\int \left (\frac {a}{d+e x}+\frac {b \tan ^{-1}\left (c x^2\right )}{d+e x}\right ) \, dx\\ &=\frac {a \log (d+e x)}{e}+b \int \frac {\tan ^{-1}\left (c x^2\right )}{d+e x} \, dx\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 22.38, size = 326, normalized size = 0.65 \begin {gather*} \frac {a \log (d+e x)}{e}+\frac {b \left (2 \text {ArcTan}\left (c x^2\right ) \log (d+e x)+i \left (\log (d+e x) \log \left (1-\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt [4]{-1} e}\right )+\log (d+e x) \log \left (1-\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt [4]{-1} e}\right )-\log (d+e x) \log \left (1-\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-(-1)^{3/4} e}\right )-\log (d+e x) \log \left (1-\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+(-1)^{3/4} e}\right )+\text {PolyLog}\left (2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt [4]{-1} e}\right )+\text {PolyLog}\left (2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt [4]{-1} e}\right )-\text {PolyLog}\left (2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-(-1)^{3/4} e}\right )-\text {PolyLog}\left (2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+(-1)^{3/4} e}\right )\right )\right )}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x^2])/(d + e*x),x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 138, normalized size = 0.28

method	result	size
default	\(\frac {a \ln \left (e x +d \right )}{e}+\frac {b \ln \left (e x +d \right ) \arctan \left (c \,x^{2}\right )}{e}-\frac {b e \left (\munderset {\textit {\_R1} =\RootOf \left (c^{2} \textit {\_Z}^{4}-4 c^{2} d \,\textit {\_Z}^{3}+6 c^{2} d^{2} \textit {\_Z}^{2}-4 c^{2} d^{3} \textit {\_Z} +c^{2} d^{4}+e^{4}\right )}{\sum }\frac {\ln \left (e x +d \right ) \ln \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )+\dilog \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )}{\textit {\_R1}^{2}-2 \textit {\_R1} d +d^{2}}\right )}{2 c}\)	\(138\)
risch	\(\frac {i b \ln \left (e x +d \right ) \ln \left (-i c \,x^{2}+1\right )}{2 e}-\frac {i b \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-i c}-\left (e x +d \right ) c +c d}{e \sqrt {-i c}+c d}\right )}{2 e}-\frac {i b \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-i c}+\left (e x +d \right ) c -c d}{e \sqrt {-i c}-c d}\right )}{2 e}-\frac {i b \dilog \left (\frac {e \sqrt {-i c}-\left (e x +d \right ) c +c d}{e \sqrt {-i c}+c d}\right )}{2 e}-\frac {i b \dilog \left (\frac {e \sqrt {-i c}+\left (e x +d \right ) c -c d}{e \sqrt {-i c}-c d}\right )}{2 e}+\frac {a \ln \left (e x +d \right )}{e}-\frac {i b \ln \left (e x +d \right ) \ln \left (i c \,x^{2}+1\right )}{2 e}+\frac {i b \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {i c}-\left (e x +d \right ) c +c d}{e \sqrt {i c}+c d}\right )}{2 e}+\frac {i b \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {i c}+\left (e x +d \right ) c -c d}{e \sqrt {i c}-c d}\right )}{2 e}+\frac {i b \dilog \left (\frac {e \sqrt {i c}-\left (e x +d \right ) c +c d}{e \sqrt {i c}+c d}\right )}{2 e}+\frac {i b \dilog \left (\frac {e \sqrt {i c}+\left (e x +d \right ) c -c d}{e \sqrt {i c}-c d}\right )}{2 e}\)	\(431\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*ln(e*x+d)/e+b*ln(e*x+d)/e*arctan(c*x^2)-1/2*b*e/c*sum(1/(_R1^2-2*_R1*d+d^2)*(ln(e*x+d)*ln((-e*x+_R1-d)/_R1)+
dilog((-e*x+_R1-d)/_R1)),_R1=RootOf(_Z^4*c^2-4*_Z^3*c^2*d+6*_Z^2*c^2*d^2-4*_Z*c^2*d^3+c^2*d^4+e^4))

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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+b*arctan(c*x^2))/(e*x+d),x, algorithm="maxima")

[Out]

a*e^(-1)*log(x*e + d) + 2*b*integrate(1/2*arctan(c*x^2)/(x*e + d), 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+b*arctan(c*x^2))/(e*x+d),x, algorithm="fricas")

[Out]

integral((b*arctan(c*x^2) + a)/(x*e + d), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x**2))/(e*x+d),x)

[Out]

Timed out

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Giac [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+b*arctan(c*x^2))/(e*x+d),x, algorithm="giac")

[Out]

integrate((b*arctan(c*x^2) + a)/(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^2))/(d + e*x),x)

[Out]

int((a + b*atan(c*x^2))/(d + e*x), x)

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