3.1 \(\int (d+e x)^4 (a+b \tan ^{-1}(c x)) \, dx\)

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

[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)

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Rubi [A]  time = 0.142807, antiderivative size = 184, normalized size of antiderivative = 1., 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 = {4862, 702, 635, 203, 260} \[ \frac{(d+e x)^5 \left (a+b \tan ^{-1}(c x)\right )}{5 e}-\frac{b e^2 x^2 \left (10 c^2 d^2-e^2\right )}{10 c^3}-\frac{b \left (-10 c^2 d^2 e^2+5 c^4 d^4+e^4\right ) \log \left (c^2 x^2+1\right )}{10 c^5}-\frac{b d e x \left (2 c^2 d^2-e^2\right )}{c^3}-\frac{b d \left (-10 c^2 d^2 e^2+c^4 d^4+5 e^4\right ) \tan ^{-1}(c x)}{5 c^4 e}-\frac{b d e^3 x^3}{3 c}-\frac{b e^4 x^4}{20 c} \]

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 4862

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]

Rule 702

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 635

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 203

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

Rule 260

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]

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*}

Mathematica [A]  time = 0.475795, size = 255, normalized size = 1.39 \[ \frac{(d+e x)^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{b \left (c^2 e^2 x \left (c^2 \left (60 d^2 e x+120 d^3+20 d e^2 x^2+3 e^3 x^3\right )-6 e^2 (10 d+e x)\right )+6 \left (-10 c^2 d^2 e^2 \left (\sqrt{-c^2} d+e\right )+c^4 d^4 \left (\sqrt{-c^2} d+5 e\right )+e^4 \left (5 \sqrt{-c^2} d+e\right )\right ) \log \left (1-\sqrt{-c^2} x\right )-6 \left (-10 c^2 d^2 e^2 \left (\sqrt{-c^2} d-e\right )+c^4 d^4 \left (\sqrt{-c^2} d-5 e\right )+e^4 \left (5 \sqrt{-c^2} d-e\right )\right ) \log \left (\sqrt{-c^2} x+1\right )\right )}{12 c^5}}{5 e} \]

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 = 0.027, size = 283, normalized size = 1.5 \begin{align*}{\frac{a{e}^{4}{x}^{5}}{5}}+a{e}^{3}{x}^{4}d+2\,a{e}^{2}{x}^{3}{d}^{2}+2\,ae{x}^{2}{d}^{3}+ax{d}^{4}+{\frac{a{d}^{5}}{5\,e}}+{\frac{b{e}^{4}\arctan \left ( cx \right ){x}^{5}}{5}}+b{e}^{3}\arctan \left ( cx \right ){x}^{4}d+2\,b{e}^{2}\arctan \left ( cx \right ){x}^{3}{d}^{2}+2\,be\arctan \left ( cx \right ){x}^{2}{d}^{3}+b\arctan \left ( cx \right ) x{d}^{4}-{\frac{b{e}^{4}{x}^{4}}{20\,c}}-{\frac{b{e}^{3}d{x}^{3}}{3\,c}}-{\frac{b{e}^{2}{x}^{2}{d}^{2}}{c}}-2\,{\frac{be{d}^{3}x}{c}}+{\frac{b{e}^{4}{x}^{2}}{10\,{c}^{3}}}+{\frac{b{e}^{3}dx}{{c}^{3}}}-{\frac{b\ln \left ({c}^{2}{x}^{2}+1 \right ){d}^{4}}{2\,c}}+{\frac{b{e}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ){d}^{2}}{{c}^{3}}}-{\frac{b{e}^{4}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{10\,{c}^{5}}}+2\,{\frac{\arctan \left ( cx \right ) be{d}^{3}}{{c}^{2}}}-{\frac{b{e}^{3}\arctan \left ( cx \right ) d}{{c}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.50502, size = 340, normalized size = 1.85 \begin{align*} \frac{1}{5} \, a e^{4} x^{5} + a d e^{3} x^{4} + 2 \, a d^{2} e^{2} x^{3} + 2 \, a d^{3} e x^{2} + 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 +{\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} + a d^{4} x + \frac{{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{4}}{2 \, c} \end{align*}

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

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Fricas [A]  time = 2.63284, size = 564, normalized size = 3.07 \begin{align*} \frac{12 \, a c^{5} e^{4} x^{5} + 3 \,{\left (20 \, a c^{5} d e^{3} - b c^{4} e^{4}\right )} x^{4} + 20 \,{\left (6 \, a c^{5} d^{2} e^{2} - b c^{4} d e^{3}\right )} x^{3} + 6 \,{\left (20 \, a c^{5} d^{3} e - 10 \, b c^{4} d^{2} e^{2} + b c^{2} e^{4}\right )} x^{2} + 60 \,{\left (a c^{5} d^{4} - 2 \, b c^{4} d^{3} e + b c^{2} d e^{3}\right )} x + 12 \,{\left (b c^{5} e^{4} x^{5} + 5 \, b c^{5} d e^{3} x^{4} + 10 \, b c^{5} d^{2} e^{2} x^{3} + 10 \, b c^{5} d^{3} e x^{2} + 5 \, b c^{5} d^{4} x + 10 \, b c^{3} d^{3} e - 5 \, b c d e^{3}\right )} \arctan \left (c x\right ) - 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{align*}

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*(12*a*c^5*e^4*x^5 + 3*(20*a*c^5*d*e^3 - b*c^4*e^4)*x^4 + 20*(6*a*c^5*d^2*e^2 - b*c^4*d*e^3)*x^3 + 6*(20*a
*c^5*d^3*e - 10*b*c^4*d^2*e^2 + b*c^2*e^4)*x^2 + 60*(a*c^5*d^4 - 2*b*c^4*d^3*e + b*c^2*d*e^3)*x + 12*(b*c^5*e^
4*x^5 + 5*b*c^5*d*e^3*x^4 + 10*b*c^5*d^2*e^2*x^3 + 10*b*c^5*d^3*e*x^2 + 5*b*c^5*d^4*x + 10*b*c^3*d^3*e - 5*b*c
*d*e^3)*arctan(c*x) - 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 [A]  time = 3.3293, size = 345, normalized size = 1.88 \begin{align*} \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{align*}

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 [A]  time = 1.24916, size = 425, normalized size = 2.31 \begin{align*} \frac{12 \, b c^{5} x^{5} \arctan \left (c x\right ) e^{4} + 60 \, b c^{5} d x^{4} \arctan \left (c x\right ) e^{3} + 120 \, b c^{5} d^{2} x^{3} \arctan \left (c x\right ) e^{2} + 120 \, b c^{5} d^{3} x^{2} \arctan \left (c x\right ) e + 60 \, b c^{5} d^{4} x \arctan \left (c x\right ) + 12 \, a c^{5} x^{5} e^{4} + 60 \, a c^{5} d x^{4} e^{3} + 120 \, a c^{5} d^{2} x^{3} e^{2} + 120 \, a c^{5} d^{3} x^{2} e + 60 \, a c^{5} d^{4} x - 120 \, \pi b c^{3} d^{3} e \mathrm{sgn}\left (c\right ) \mathrm{sgn}\left (x\right ) - 3 \, b c^{4} x^{4} e^{4} - 20 \, b c^{4} d x^{3} e^{3} - 60 \, b c^{4} d^{2} x^{2} e^{2} - 120 \, b c^{4} d^{3} x e - 30 \, b c^{4} d^{4} \log \left (c^{2} x^{2} + 1\right ) + 120 \, b c^{3} d^{3} \arctan \left (c x\right ) e + 60 \, b c^{2} d^{2} e^{2} \log \left (c^{2} x^{2} + 1\right ) + 6 \, b c^{2} x^{2} e^{4} + 60 \, b c^{2} d x e^{3} - 60 \, b c d \arctan \left (c x\right ) e^{3} - 6 \, b e^{4} \log \left (c^{2} x^{2} + 1\right )}{60 \, c^{5}} \end{align*}

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]

1/60*(12*b*c^5*x^5*arctan(c*x)*e^4 + 60*b*c^5*d*x^4*arctan(c*x)*e^3 + 120*b*c^5*d^2*x^3*arctan(c*x)*e^2 + 120*
b*c^5*d^3*x^2*arctan(c*x)*e + 60*b*c^5*d^4*x*arctan(c*x) + 12*a*c^5*x^5*e^4 + 60*a*c^5*d*x^4*e^3 + 120*a*c^5*d
^2*x^3*e^2 + 120*a*c^5*d^3*x^2*e + 60*a*c^5*d^4*x - 120*pi*b*c^3*d^3*e*sgn(c)*sgn(x) - 3*b*c^4*x^4*e^4 - 20*b*
c^4*d*x^3*e^3 - 60*b*c^4*d^2*x^2*e^2 - 120*b*c^4*d^3*x*e - 30*b*c^4*d^4*log(c^2*x^2 + 1) + 120*b*c^3*d^3*arcta
n(c*x)*e + 60*b*c^2*d^2*e^2*log(c^2*x^2 + 1) + 6*b*c^2*x^2*e^4 + 60*b*c^2*d*x*e^3 - 60*b*c*d*arctan(c*x)*e^3 -
6*b*e^4*log(c^2*x^2 + 1))/c^5