3.1.1 \(\int (d+e x)^4 (a+b \text {ArcTan}(c x)) \, dx\) [1]

Optimal. Leaf size=184 \[ -\frac {b d e \left (2 c^2 d^2-e^2\right ) x}{c^3}-\frac {b e^2 \left (10 c^2 d^2-e^2\right ) x^2}{10 c^3}-\frac {b d e^3 x^3}{3 c}-\frac {b e^4 x^4}{20 c}-\frac {b d \left (c^4 d^4-10 c^2 d^2 e^2+5 e^4\right ) \text {ArcTan}(c x)}{5 c^4 e}+\frac {(d+e x)^5 (a+b \text {ArcTan}(c x))}{5 e}-\frac {b \left (5 c^4 d^4-10 c^2 d^2 e^2+e^4\right ) \log \left (1+c^2 x^2\right )}{10 c^5} \]

[Out]

-b*d*e*(2*c^2*d^2-e^2)*x/c^3-1/10*b*e^2*(10*c^2*d^2-e^2)*x^2/c^3-1/3*b*d*e^3*x^3/c-1/20*b*e^4*x^4/c-1/5*b*d*(c
^4*d^4-10*c^2*d^2*e^2+5*e^4)*arctan(c*x)/c^4/e+1/5*(e*x+d)^5*(a+b*arctan(c*x))/e-1/10*b*(5*c^4*d^4-10*c^2*d^2*
e^2+e^4)*ln(c^2*x^2+1)/c^5

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Rubi [A]
time = 0.10, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {4972, 716, 649, 209, 266} \begin {gather*} \frac {(d+e x)^5 (a+b \text {ArcTan}(c x))}{5 e}-\frac {b d \text {ArcTan}(c x) \left (c^4 d^4-10 c^2 d^2 e^2+5 e^4\right )}{5 c^4 e}-\frac {b e^2 x^2 \left (10 c^2 d^2-e^2\right )}{10 c^3}-\frac {b d e x \left (2 c^2 d^2-e^2\right )}{c^3}-\frac {b \left (5 c^4 d^4-10 c^2 d^2 e^2+e^4\right ) \log \left (c^2 x^2+1\right )}{10 c^5}-\frac {b d e^3 x^3}{3 c}-\frac {b e^4 x^4}{20 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4*(a + b*ArcTan[c*x]),x]

[Out]

-((b*d*e*(2*c^2*d^2 - e^2)*x)/c^3) - (b*e^2*(10*c^2*d^2 - e^2)*x^2)/(10*c^3) - (b*d*e^3*x^3)/(3*c) - (b*e^4*x^
4)/(20*c) - (b*d*(c^4*d^4 - 10*c^2*d^2*e^2 + 5*e^4)*ArcTan[c*x])/(5*c^4*e) + ((d + e*x)^5*(a + b*ArcTan[c*x]))
/(5*e) - (b*(5*c^4*d^4 - 10*c^2*d^2*e^2 + e^4)*Log[1 + c^2*x^2])/(10*c^5)

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 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 716

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 4972

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

Rubi steps

\begin {align*} \int (d+e x)^4 \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac {(d+e x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac {(b c) \int \frac {(d+e x)^5}{1+c^2 x^2} \, dx}{5 e}\\ &=\frac {(d+e x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac {(b c) \int \left (\frac {5 d e^2 \left (2 c^2 d^2-e^2\right )}{c^4}+\frac {e^3 \left (10 c^2 d^2-e^2\right ) x}{c^4}+\frac {5 d e^4 x^2}{c^2}+\frac {e^5 x^3}{c^2}+\frac {c^4 d^5-10 c^2 d^3 e^2+5 d e^4+e \left (5 c^4 d^4-10 c^2 d^2 e^2+e^4\right ) x}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx}{5 e}\\ &=-\frac {b d e \left (2 c^2 d^2-e^2\right ) x}{c^3}-\frac {b e^2 \left (10 c^2 d^2-e^2\right ) x^2}{10 c^3}-\frac {b d e^3 x^3}{3 c}-\frac {b e^4 x^4}{20 c}+\frac {(d+e x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac {b \int \frac {c^4 d^5-10 c^2 d^3 e^2+5 d e^4+e \left (5 c^4 d^4-10 c^2 d^2 e^2+e^4\right ) x}{1+c^2 x^2} \, dx}{5 c^3 e}\\ &=-\frac {b d e \left (2 c^2 d^2-e^2\right ) x}{c^3}-\frac {b e^2 \left (10 c^2 d^2-e^2\right ) x^2}{10 c^3}-\frac {b d e^3 x^3}{3 c}-\frac {b e^4 x^4}{20 c}+\frac {(d+e x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac {\left (b \left (5 c^4 d^4-10 c^2 d^2 e^2+e^4\right )\right ) \int \frac {x}{1+c^2 x^2} \, dx}{5 c^3}-\frac {\left (b d \left (c^4 d^4-10 c^2 d^2 e^2+5 e^4\right )\right ) \int \frac {1}{1+c^2 x^2} \, dx}{5 c^3 e}\\ &=-\frac {b d e \left (2 c^2 d^2-e^2\right ) x}{c^3}-\frac {b e^2 \left (10 c^2 d^2-e^2\right ) x^2}{10 c^3}-\frac {b d e^3 x^3}{3 c}-\frac {b e^4 x^4}{20 c}-\frac {b d \left (c^4 d^4-10 c^2 d^2 e^2+5 e^4\right ) \tan ^{-1}(c x)}{5 c^4 e}+\frac {(d+e x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac {b \left (5 c^4 d^4-10 c^2 d^2 e^2+e^4\right ) \log \left (1+c^2 x^2\right )}{10 c^5}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 255, normalized size = 1.39 \begin {gather*} \frac {(d+e x)^5 (a+b \text {ArcTan}(c x))-\frac {b \left (c^2 e^2 x \left (-6 e^2 (10 d+e x)+c^2 \left (120 d^3+60 d^2 e x+20 d e^2 x^2+3 e^3 x^3\right )\right )+6 \left (-10 c^2 d^2 e^2 \left (\sqrt {-c^2} d+e\right )+e^4 \left (5 \sqrt {-c^2} d+e\right )+c^4 d^4 \left (\sqrt {-c^2} d+5 e\right )\right ) \log \left (1-\sqrt {-c^2} x\right )-6 \left (c^4 d^4 \left (\sqrt {-c^2} d-5 e\right )-10 c^2 d^2 \left (\sqrt {-c^2} d-e\right ) e^2+\left (5 \sqrt {-c^2} d-e\right ) e^4\right ) \log \left (1+\sqrt {-c^2} x\right )\right )}{12 c^5}}{5 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4*(a + b*ArcTan[c*x]),x]

[Out]

((d + e*x)^5*(a + b*ArcTan[c*x]) - (b*(c^2*e^2*x*(-6*e^2*(10*d + e*x) + c^2*(120*d^3 + 60*d^2*e*x + 20*d*e^2*x
^2 + 3*e^3*x^3)) + 6*(-10*c^2*d^2*e^2*(Sqrt[-c^2]*d + e) + e^4*(5*Sqrt[-c^2]*d + e) + c^4*d^4*(Sqrt[-c^2]*d +
5*e))*Log[1 - Sqrt[-c^2]*x] - 6*(c^4*d^4*(Sqrt[-c^2]*d - 5*e) - 10*c^2*d^2*(Sqrt[-c^2]*d - e)*e^2 + (5*Sqrt[-c
^2]*d - e)*e^4)*Log[1 + Sqrt[-c^2]*x]))/(12*c^5))/(5*e)

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Maple [A]
time = 0.25, size = 241, normalized size = 1.31

method	result	size
derivativedivides	\(\frac {\frac {\left (c e x +c d \right )^{5} a}{5 c^{4} e}+b \arctan \left (c x \right ) d^{4} c x +2 b c e \arctan \left (c x \right ) d^{3} x^{2}+2 b c \,e^{2} \arctan \left (c x \right ) d^{2} x^{3}+b c \,e^{3} \arctan \left (c x \right ) d \,x^{4}+\frac {b c \,e^{4} \arctan \left (c x \right ) x^{5}}{5}-2 b e \,d^{3} x -b \,e^{2} d^{2} x^{2}-\frac {b \,e^{3} d \,x^{3}}{3}-\frac {b \,e^{4} x^{4}}{20}+\frac {b \,e^{3} d x}{c^{2}}+\frac {b \,e^{4} x^{2}}{10 c^{2}}-\frac {b \ln \left (c^{2} x^{2}+1\right ) d^{4}}{2}+\frac {b \,e^{2} \ln \left (c^{2} x^{2}+1\right ) d^{2}}{c^{2}}-\frac {b \,e^{4} \ln \left (c^{2} x^{2}+1\right )}{10 c^{4}}+\frac {2 b e \arctan \left (c x \right ) d^{3}}{c}-\frac {b \,e^{3} \arctan \left (c x \right ) d}{c^{3}}}{c}\)	\(241\)
default	\(\frac {\frac {\left (c e x +c d \right )^{5} a}{5 c^{4} e}+b \arctan \left (c x \right ) d^{4} c x +2 b c e \arctan \left (c x \right ) d^{3} x^{2}+2 b c \,e^{2} \arctan \left (c x \right ) d^{2} x^{3}+b c \,e^{3} \arctan \left (c x \right ) d \,x^{4}+\frac {b c \,e^{4} \arctan \left (c x \right ) x^{5}}{5}-2 b e \,d^{3} x -b \,e^{2} d^{2} x^{2}-\frac {b \,e^{3} d \,x^{3}}{3}-\frac {b \,e^{4} x^{4}}{20}+\frac {b \,e^{3} d x}{c^{2}}+\frac {b \,e^{4} x^{2}}{10 c^{2}}-\frac {b \ln \left (c^{2} x^{2}+1\right ) d^{4}}{2}+\frac {b \,e^{2} \ln \left (c^{2} x^{2}+1\right ) d^{2}}{c^{2}}-\frac {b \,e^{4} \ln \left (c^{2} x^{2}+1\right )}{10 c^{4}}+\frac {2 b e \arctan \left (c x \right ) d^{3}}{c}-\frac {b \,e^{3} \arctan \left (c x \right ) d}{c^{3}}}{c}\)	\(241\)
risch	\(\frac {i e^{4} b \,x^{5} \ln \left (-i c x +1\right )}{10}+\frac {i b \,d^{5} \ln \left (c^{2} x^{2}+1\right )}{20 e}+i e b \,d^{3} x^{2} \ln \left (-i c x +1\right )+\frac {i b \,d^{4} x \ln \left (-i c x +1\right )}{2}+\frac {x^{5} e^{4} a}{5}-\frac {i \left (e x +d \right )^{5} b \ln \left (i c x +1\right )}{10 e}+x^{4} e^{3} d a +\frac {i e^{3} b d \,x^{4} \ln \left (-i c x +1\right )}{2}+2 x^{3} e^{2} d^{2} a -\frac {b \,e^{4} x^{4}}{20 c}+i e^{2} b \,d^{2} x^{3} \ln \left (-i c x +1\right )+2 x^{2} e \,d^{3} a -\frac {b d \,e^{3} x^{3}}{3 c}-\frac {b \,d^{5} \arctan \left (c x \right )}{10 e}+x a \,d^{4}-\frac {e^{2} b \,d^{2} x^{2}}{c}-\frac {b \,d^{4} \ln \left (c^{2} x^{2}+1\right )}{2 c}-\frac {2 e b \,d^{3} x}{c}+\frac {2 e b \,d^{3} \arctan \left (c x \right )}{c^{2}}+\frac {e^{4} b \,x^{2}}{10 c^{3}}+\frac {e^{2} b \,d^{2} \ln \left (c^{2} x^{2}+1\right )}{c^{3}}+\frac {e^{3} b d x}{c^{3}}-\frac {e^{3} b d \arctan \left (c x \right )}{c^{4}}-\frac {e^{4} b \ln \left (c^{2} x^{2}+1\right )}{10 c^{5}}\)	\(356\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4*(a+b*arctan(c*x)),x,method=_RETURNVERBOSE)

[Out]

1/c*(1/5*(c*e*x+c*d)^5*a/c^4/e+b*arctan(c*x)*d^4*c*x+2*b*c*e*arctan(c*x)*d^3*x^2+2*b*c*e^2*arctan(c*x)*d^2*x^3
+b*c*e^3*arctan(c*x)*d*x^4+1/5*b*c*e^4*arctan(c*x)*x^5-2*b*e*d^3*x-b*e^2*d^2*x^2-1/3*b*e^3*d*x^3-1/20*b*e^4*x^
4+b/c^2*e^3*d*x+1/10*b/c^2*e^4*x^2-1/2*b*ln(c^2*x^2+1)*d^4+b/c^2*e^2*ln(c^2*x^2+1)*d^2-1/10*b/c^4*e^4*ln(c^2*x
^2+1)+2*b/c*e*arctan(c*x)*d^3-b/c^3*e^3*arctan(c*x)*d)

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Maxima [A]
time = 0.47, size = 248, normalized size = 1.35 \begin {gather*} \frac {1}{5} \, a x^{5} e^{4} + a d x^{4} e^{3} + 2 \, a d^{2} x^{3} e^{2} + 2 \, a d^{3} x^{2} e + a d^{4} x + 2 \, {\left (x^{2} \arctan \left (c x\right ) - c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )}\right )} b d^{3} e + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{4}}{2 \, c} + {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d^{2} e^{2} + \frac {1}{3} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b d e^{3} + \frac {1}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*(a+b*arctan(c*x)),x, algorithm="maxima")

[Out]

1/5*a*x^5*e^4 + a*d*x^4*e^3 + 2*a*d^2*x^3*e^2 + 2*a*d^3*x^2*e + a*d^4*x + 2*(x^2*arctan(c*x) - c*(x/c^2 - arct
an(c*x)/c^3))*b*d^3*e + 1/2*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*d^4/c + (2*x^3*arctan(c*x) - c*(x^2/c^2 -
log(c^2*x^2 + 1)/c^4))*b*d^2*e^2 + 1/3*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*b*d*
e^3 + 1/20*(4*x^5*arctan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*b*e^4

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Fricas [A]
time = 1.38, size = 257, normalized size = 1.40 \begin {gather*} \frac {60 \, a c^{5} d^{4} x + 12 \, {\left (b c^{5} x^{5} e^{4} + 10 \, b c^{5} d^{2} x^{3} e^{2} + 5 \, b c^{5} d^{4} x + 5 \, {\left (b c^{5} d x^{4} - b c d\right )} e^{3} + 10 \, {\left (b c^{5} d^{3} x^{2} + b c^{3} d^{3}\right )} e\right )} \arctan \left (c x\right ) + 3 \, {\left (4 \, a c^{5} x^{5} - b c^{4} x^{4} + 2 \, b c^{2} x^{2}\right )} e^{4} + 20 \, {\left (3 \, a c^{5} d x^{4} - b c^{4} d x^{3} + 3 \, b c^{2} d x\right )} e^{3} + 60 \, {\left (2 \, a c^{5} d^{2} x^{3} - b c^{4} d^{2} x^{2}\right )} e^{2} + 120 \, {\left (a c^{5} d^{3} x^{2} - b c^{4} d^{3} x\right )} e - 6 \, {\left (5 \, b c^{4} d^{4} - 10 \, b c^{2} d^{2} e^{2} + b e^{4}\right )} \log \left (c^{2} x^{2} + 1\right )}{60 \, c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*(a+b*arctan(c*x)),x, algorithm="fricas")

[Out]

1/60*(60*a*c^5*d^4*x + 12*(b*c^5*x^5*e^4 + 10*b*c^5*d^2*x^3*e^2 + 5*b*c^5*d^4*x + 5*(b*c^5*d*x^4 - b*c*d)*e^3
+ 10*(b*c^5*d^3*x^2 + b*c^3*d^3)*e)*arctan(c*x) + 3*(4*a*c^5*x^5 - b*c^4*x^4 + 2*b*c^2*x^2)*e^4 + 20*(3*a*c^5*
d*x^4 - b*c^4*d*x^3 + 3*b*c^2*d*x)*e^3 + 60*(2*a*c^5*d^2*x^3 - b*c^4*d^2*x^2)*e^2 + 120*(a*c^5*d^3*x^2 - b*c^4
*d^3*x)*e - 6*(5*b*c^4*d^4 - 10*b*c^2*d^2*e^2 + b*e^4)*log(c^2*x^2 + 1))/c^5

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (170) = 340\).
time = 0.44, size = 345, normalized size = 1.88 \begin {gather*} \begin {cases} a d^{4} x + 2 a d^{3} e x^{2} + 2 a d^{2} e^{2} x^{3} + a d e^{3} x^{4} + \frac {a e^{4} x^{5}}{5} + b d^{4} x \operatorname {atan}{\left (c x \right )} + 2 b d^{3} e x^{2} \operatorname {atan}{\left (c x \right )} + 2 b d^{2} e^{2} x^{3} \operatorname {atan}{\left (c x \right )} + b d e^{3} x^{4} \operatorname {atan}{\left (c x \right )} + \frac {b e^{4} x^{5} \operatorname {atan}{\left (c x \right )}}{5} - \frac {b d^{4} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c} - \frac {2 b d^{3} e x}{c} - \frac {b d^{2} e^{2} x^{2}}{c} - \frac {b d e^{3} x^{3}}{3 c} - \frac {b e^{4} x^{4}}{20 c} + \frac {2 b d^{3} e \operatorname {atan}{\left (c x \right )}}{c^{2}} + \frac {b d^{2} e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{c^{3}} + \frac {b d e^{3} x}{c^{3}} + \frac {b e^{4} x^{2}}{10 c^{3}} - \frac {b d e^{3} \operatorname {atan}{\left (c x \right )}}{c^{4}} - \frac {b e^{4} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10 c^{5}} & \text {for}\: c \neq 0 \\a \left (d^{4} x + 2 d^{3} e x^{2} + 2 d^{2} e^{2} x^{3} + d e^{3} x^{4} + \frac {e^{4} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4*(a+b*atan(c*x)),x)

[Out]

Piecewise((a*d**4*x + 2*a*d**3*e*x**2 + 2*a*d**2*e**2*x**3 + a*d*e**3*x**4 + a*e**4*x**5/5 + b*d**4*x*atan(c*x
) + 2*b*d**3*e*x**2*atan(c*x) + 2*b*d**2*e**2*x**3*atan(c*x) + b*d*e**3*x**4*atan(c*x) + b*e**4*x**5*atan(c*x)
/5 - b*d**4*log(x**2 + c**(-2))/(2*c) - 2*b*d**3*e*x/c - b*d**2*e**2*x**2/c - b*d*e**3*x**3/(3*c) - b*e**4*x**
4/(20*c) + 2*b*d**3*e*atan(c*x)/c**2 + b*d**2*e**2*log(x**2 + c**(-2))/c**3 + b*d*e**3*x/c**3 + b*e**4*x**2/(1
0*c**3) - b*d*e**3*atan(c*x)/c**4 - b*e**4*log(x**2 + c**(-2))/(10*c**5), Ne(c, 0)), (a*(d**4*x + 2*d**3*e*x**
2 + 2*d**2*e**2*x**3 + d*e**3*x**4 + e**4*x**5/5), True))

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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)^4*(a+b*arctan(c*x)),x, algorithm="giac")

[Out]

sage0*x

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Mupad [B]
time = 1.05, size = 273, normalized size = 1.48 \begin {gather*} \frac {a\,e^4\,x^5}{5}+a\,d^4\,x-\frac {b\,d^4\,\ln \left (c^2\,x^2+1\right )}{2\,c}-\frac {b\,e^4\,\ln \left (c^2\,x^2+1\right )}{10\,c^5}+2\,a\,d^2\,e^2\,x^3-\frac {b\,e^4\,x^4}{20\,c}+\frac {b\,e^4\,x^2}{10\,c^3}+b\,d^4\,x\,\mathrm {atan}\left (c\,x\right )+2\,a\,d^3\,e\,x^2+a\,d\,e^3\,x^4+\frac {b\,e^4\,x^5\,\mathrm {atan}\left (c\,x\right )}{5}-\frac {2\,b\,d^3\,e\,x}{c}+\frac {b\,d\,e^3\,x}{c^3}+\frac {2\,b\,d^3\,e\,\mathrm {atan}\left (c\,x\right )}{c^2}-\frac {b\,d\,e^3\,\mathrm {atan}\left (c\,x\right )}{c^4}+2\,b\,d^3\,e\,x^2\,\mathrm {atan}\left (c\,x\right )+b\,d\,e^3\,x^4\,\mathrm {atan}\left (c\,x\right )-\frac {b\,d\,e^3\,x^3}{3\,c}+2\,b\,d^2\,e^2\,x^3\,\mathrm {atan}\left (c\,x\right )+\frac {b\,d^2\,e^2\,\ln \left (c^2\,x^2+1\right )}{c^3}-\frac {b\,d^2\,e^2\,x^2}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))*(d + e*x)^4,x)

[Out]

(a*e^4*x^5)/5 + a*d^4*x - (b*d^4*log(c^2*x^2 + 1))/(2*c) - (b*e^4*log(c^2*x^2 + 1))/(10*c^5) + 2*a*d^2*e^2*x^3
- (b*e^4*x^4)/(20*c) + (b*e^4*x^2)/(10*c^3) + b*d^4*x*atan(c*x) + 2*a*d^3*e*x^2 + a*d*e^3*x^4 + (b*e^4*x^5*at
an(c*x))/5 - (2*b*d^3*e*x)/c + (b*d*e^3*x)/c^3 + (2*b*d^3*e*atan(c*x))/c^2 - (b*d*e^3*atan(c*x))/c^4 + 2*b*d^3
*e*x^2*atan(c*x) + b*d*e^3*x^4*atan(c*x) - (b*d*e^3*x^3)/(3*c) + 2*b*d^2*e^2*x^3*atan(c*x) + (b*d^2*e^2*log(c^
2*x^2 + 1))/c^3 - (b*d^2*e^2*x^2)/c

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