3.1.17 \(\int (d+e x) (a+b \text {ArcTan}(c x))^3 \, dx\) [17]

Optimal. Leaf size=264 \[ -\frac {3 i b e (a+b \text {ArcTan}(c x))^2}{2 c^2}-\frac {3 b e x (a+b \text {ArcTan}(c x))^2}{2 c}+\frac {i d (a+b \text {ArcTan}(c x))^3}{c}-\frac {\left (d^2-\frac {e^2}{c^2}\right ) (a+b \text {ArcTan}(c x))^3}{2 e}+\frac {(d+e x)^2 (a+b \text {ArcTan}(c x))^3}{2 e}-\frac {3 b^2 e (a+b \text {ArcTan}(c x)) \log \left (\frac {2}{1+i c x}\right )}{c^2}+\frac {3 b d (a+b \text {ArcTan}(c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c}-\frac {3 i b^3 e \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c^2}+\frac {3 i b^2 d (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c}+\frac {3 b^3 d \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c} \]

[Out]

-3/2*I*b*e*(a+b*arctan(c*x))^2/c^2-3/2*b*e*x*(a+b*arctan(c*x))^2/c+I*d*(a+b*arctan(c*x))^3/c-1/2*(d^2-e^2/c^2)
*(a+b*arctan(c*x))^3/e+1/2*(e*x+d)^2*(a+b*arctan(c*x))^3/e-3*b^2*e*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/c^2+3*b*d
*(a+b*arctan(c*x))^2*ln(2/(1+I*c*x))/c-3/2*I*b^3*e*polylog(2,1-2/(1+I*c*x))/c^2+3*I*b^2*d*(a+b*arctan(c*x))*po
lylog(2,1-2/(1+I*c*x))/c+3/2*b^3*d*polylog(3,1-2/(1+I*c*x))/c

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Rubi [A]
time = 0.38, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {4974, 4930, 5040, 4964, 2449, 2352, 5104, 5004, 5114, 6745} \begin {gather*} -\frac {3 b^2 e \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))}{c^2}+\frac {3 i b^2 d \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) (a+b \text {ArcTan}(c x))}{c}-\frac {\left (d^2-\frac {e^2}{c^2}\right ) (a+b \text {ArcTan}(c x))^3}{2 e}-\frac {3 i b e (a+b \text {ArcTan}(c x))^2}{2 c^2}+\frac {(d+e x)^2 (a+b \text {ArcTan}(c x))^3}{2 e}+\frac {i d (a+b \text {ArcTan}(c x))^3}{c}+\frac {3 b d \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^2}{c}-\frac {3 b e x (a+b \text {ArcTan}(c x))^2}{2 c}-\frac {3 i b^3 e \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{2 c^2}+\frac {3 b^3 d \text {Li}_3\left (1-\frac {2}{i c x+1}\right )}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a + b*ArcTan[c*x])^3,x]

[Out]

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

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Dist[b*c
*n*p, Int[x^n*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0
] && (EqQ[n, 1] || EqQ[p, 1])

Rule 4964

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

Rule 4974

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)*((a
+ b*ArcTan[c*x])^p/(e*(q + 1))), x] - Dist[b*c*(p/(e*(q + 1))), Int[ExpandIntegrand[(a + b*ArcTan[c*x])^(p -
1), (d + e*x)^(q + 1)/(1 + c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 1] && IntegerQ[q] && N
eQ[q, -1]

Rule 5004

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

Rule 5040

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

Rule 5104

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> I
nt[ExpandIntegrand[(a + b*ArcTan[c*x])^p/(d + e*x^2), (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& IGtQ[p, 0] && EqQ[e, c^2*d] && IGtQ[m, 0]

Rule 5114

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar
cTan[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)), x] + Dist[b*p*(I/2), Int[(a + b*ArcTan[c*x])^(p - 1)*(PolyLog[2, 1 -
u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[(1 - u)^2 - (1 - 2
*(I/(I - c*x)))^2, 0]

Rule 6745

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int (d+e x) \left (a+b \tan ^{-1}(c x)\right )^3 \, dx &=\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {(3 b c) \int \left (\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^2}+\frac {\left (c^2 d^2-e^2+2 c^2 d e x\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 \left (1+c^2 x^2\right )}\right ) \, dx}{2 e}\\ &=\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {(3 b) \int \frac {\left (c^2 d^2-e^2+2 c^2 d e x\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{2 c e}-\frac {(3 b e) \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{2 c}\\ &=-\frac {3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {(3 b) \int \left (\frac {c^2 d^2 \left (1-\frac {e^2}{c^2 d^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2}+\frac {2 c^2 d e x \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2}\right ) \, dx}{2 c e}+\left (3 b^2 e\right ) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=-\frac {3 i b e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}-\frac {3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-(3 b c d) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx-\frac {\left (3 b^2 e\right ) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx}{c}-\frac {(3 b (c d-e) (c d+e)) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{2 c e}\\ &=-\frac {3 i b e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}-\frac {3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}+\frac {i d \left (a+b \tan ^{-1}(c x)\right )^3}{c}-\frac {\left (d^2-\frac {e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {3 b^2 e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^2}+(3 b d) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{i-c x} \, dx+\frac {\left (3 b^3 e\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c}\\ &=-\frac {3 i b e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}-\frac {3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}+\frac {i d \left (a+b \tan ^{-1}(c x)\right )^3}{c}-\frac {\left (d^2-\frac {e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {3 b^2 e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^2}+\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c}-\left (6 b^2 d\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\frac {\left (3 i b^3 e\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{c^2}\\ &=-\frac {3 i b e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}-\frac {3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}+\frac {i d \left (a+b \tan ^{-1}(c x)\right )^3}{c}-\frac {\left (d^2-\frac {e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {3 b^2 e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^2}+\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c}-\frac {3 i b^3 e \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c^2}+\frac {3 i b^2 d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c}-\left (3 i b^3 d\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=-\frac {3 i b e \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2}-\frac {3 b e x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}+\frac {i d \left (a+b \tan ^{-1}(c x)\right )^3}{c}-\frac {\left (d^2-\frac {e^2}{c^2}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 e}-\frac {3 b^2 e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^2}+\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c}-\frac {3 i b^3 e \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c^2}+\frac {3 i b^2 d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c}+\frac {3 b^3 d \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 342, normalized size = 1.30 \begin {gather*} \frac {a^2 c (2 a c d-3 b e) x+a^3 c^2 e x^2+3 a^2 b e \text {ArcTan}(c x)+3 a^2 b c^2 x (2 d+e x) \text {ArcTan}(c x)-3 a^2 b c d \log \left (1+c^2 x^2\right )+3 a b^2 e \left (-2 c x \text {ArcTan}(c x)+\left (1+c^2 x^2\right ) \text {ArcTan}(c x)^2+\log \left (1+c^2 x^2\right )\right )+6 a b^2 c d \left (\text {ArcTan}(c x) \left ((-i+c x) \text {ArcTan}(c x)+2 \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )\right )-i \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )\right )+b^3 e \left (\text {ArcTan}(c x) \left ((3 i-3 c x) \text {ArcTan}(c x)+\left (1+c^2 x^2\right ) \text {ArcTan}(c x)^2-6 \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )\right )+3 i \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )\right )+b^3 c d \left (2 \text {ArcTan}(c x)^2 \left ((-i+c x) \text {ArcTan}(c x)+3 \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )\right )-6 i \text {ArcTan}(c x) \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )+3 \text {PolyLog}\left (3,-e^{2 i \text {ArcTan}(c x)}\right )\right )}{2 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)*(a + b*ArcTan[c*x])^3,x]

[Out]

(a^2*c*(2*a*c*d - 3*b*e)*x + a^3*c^2*e*x^2 + 3*a^2*b*e*ArcTan[c*x] + 3*a^2*b*c^2*x*(2*d + e*x)*ArcTan[c*x] - 3
*a^2*b*c*d*Log[1 + c^2*x^2] + 3*a*b^2*e*(-2*c*x*ArcTan[c*x] + (1 + c^2*x^2)*ArcTan[c*x]^2 + Log[1 + c^2*x^2])
+ 6*a*b^2*c*d*(ArcTan[c*x]*((-I + c*x)*ArcTan[c*x] + 2*Log[1 + E^((2*I)*ArcTan[c*x])]) - I*PolyLog[2, -E^((2*I
)*ArcTan[c*x])]) + b^3*e*(ArcTan[c*x]*((3*I - 3*c*x)*ArcTan[c*x] + (1 + c^2*x^2)*ArcTan[c*x]^2 - 6*Log[1 + E^(
(2*I)*ArcTan[c*x])]) + (3*I)*PolyLog[2, -E^((2*I)*ArcTan[c*x])]) + b^3*c*d*(2*ArcTan[c*x]^2*((-I + c*x)*ArcTan
[c*x] + 3*Log[1 + E^((2*I)*ArcTan[c*x])]) - (6*I)*ArcTan[c*x]*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + 3*PolyLog[3
, -E^((2*I)*ArcTan[c*x])]))/(2*c^2)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.45, size = 7195, normalized size = 27.25

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(7195\)
default	\(\text {Expression too large to display}\)	\(7195\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(a+b*arctan(c*x))^3,x,method=_RETURNVERBOSE)

[Out]

result too large to display

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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((e*x+d)*(a+b*arctan(c*x))^3,x, algorithm="maxima")

[Out]

7/32*b^3*d*arctan(c*x)^4/c + 56*b^3*c^2*e*integrate(1/64*x^3*arctan(c*x)^3/(c^2*x^2 + 1), x) + 6*b^3*c^2*e*int
egrate(1/64*x^3*arctan(c*x)*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 192*a*b^2*c^2*e*integrate(1/64*x^3*arctan(c
*x)^2/(c^2*x^2 + 1), x) + 56*b^3*c^2*d*integrate(1/64*x^2*arctan(c*x)^3/(c^2*x^2 + 1), x) + 12*b^3*c^2*e*integ
rate(1/64*x^3*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 6*b^3*c^2*d*integrate(1/64*x^2*arctan(c*x)*log(
c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 192*a*b^2*c^2*d*integrate(1/64*x^2*arctan(c*x)^2/(c^2*x^2 + 1), x) + 24*b^3
*c^2*d*integrate(1/64*x^2*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + a*b^2*d*arctan(c*x)^3/c + 1/2*a^3*x
^2*e - 12*b^3*c*e*integrate(1/64*x^2*arctan(c*x)^2/(c^2*x^2 + 1), x) + 3*b^3*c*e*integrate(1/64*x^2*log(c^2*x^
2 + 1)^2/(c^2*x^2 + 1), x) - 24*b^3*c*d*integrate(1/64*x*arctan(c*x)^2/(c^2*x^2 + 1), x) + 6*b^3*c*d*integrate
(1/64*x*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + a^3*d*x + 3/2*(x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*a
^2*b*e + 56*b^3*e*integrate(1/64*x*arctan(c*x)^3/(c^2*x^2 + 1), x) + 6*b^3*e*integrate(1/64*x*arctan(c*x)*log(
c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 192*a*b^2*e*integrate(1/64*x*arctan(c*x)^2/(c^2*x^2 + 1), x) + 6*b^3*d*inte
grate(1/64*arctan(c*x)*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 3/2*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*a^2*b
*d/c + 1/16*(b^3*x^2*e + 2*b^3*d*x)*arctan(c*x)^3 - 3/64*(b^3*x^2*e + 2*b^3*d*x)*arctan(c*x)*log(c^2*x^2 + 1)^
2

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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((e*x+d)*(a+b*arctan(c*x))^3,x, algorithm="fricas")

[Out]

integral(a^3*x*e + a^3*d + (b^3*x*e + b^3*d)*arctan(c*x)^3 + 3*(a*b^2*x*e + a*b^2*d)*arctan(c*x)^2 + 3*(a^2*b*
x*e + a^2*b*d)*arctan(c*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(a+b*atan(c*x))**3,x)

[Out]

Integral((a + b*atan(c*x))**3*(d + e*x), x)

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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((e*x+d)*(a+b*arctan(c*x))^3,x, algorithm="giac")

[Out]

sage0*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))^3*(d + e*x),x)

[Out]

int((a + b*atan(c*x))^3*(d + e*x), x)

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