3.1 \(\int (c e+d e x)^3 (a+b \tan ^{-1}(c+d x)) \, dx\)

Optimal. Leaf size=72 \[ \frac{e^3 (c+d x)^4 \left (a+b \tan ^{-1}(c+d x)\right )}{4 d}-\frac{b e^3 (c+d x)^3}{12 d}-\frac{b e^3 \tan ^{-1}(c+d x)}{4 d}+\frac{1}{4} b e^3 x \]

[Out]

(b*e^3*x)/4 - (b*e^3*(c + d*x)^3)/(12*d) - (b*e^3*ArcTan[c + d*x])/(4*d) + (e^3*(c + d*x)^4*(a + b*ArcTan[c +
d*x]))/(4*d)

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Rubi [A]  time = 0.250649, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {5043, 12, 4852, 302, 203} \[ \frac{e^3 (c+d x)^4 \left (a+b \tan ^{-1}(c+d x)\right )}{4 d}-\frac{b e^3 (c+d x)^3}{12 d}-\frac{b e^3 \tan ^{-1}(c+d x)}{4 d}+\frac{1}{4} b e^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e^3*x)/4 - (b*e^3*(c + d*x)^3)/(12*d) - (b*e^3*ArcTan[c + d*x])/(4*d) + (e^3*(c + d*x)^4*(a + b*ArcTan[c +
d*x]))/(4*d)

Rule 5043

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4852

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

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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])

Rubi steps

\begin{align*} \int (c e+d e x)^3 \left (a+b \tan ^{-1}(c+d x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int e^3 x^3 \left (a+b \tan ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int x^3 \left (a+b \tan ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 (c+d x)^4 \left (a+b \tan ^{-1}(c+d x)\right )}{4 d}-\frac{\left (b e^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{1+x^2} \, dx,x,c+d x\right )}{4 d}\\ &=\frac{e^3 (c+d x)^4 \left (a+b \tan ^{-1}(c+d x)\right )}{4 d}-\frac{\left (b e^3\right ) \operatorname{Subst}\left (\int \left (-1+x^2+\frac{1}{1+x^2}\right ) \, dx,x,c+d x\right )}{4 d}\\ &=\frac{1}{4} b e^3 x-\frac{b e^3 (c+d x)^3}{12 d}+\frac{e^3 (c+d x)^4 \left (a+b \tan ^{-1}(c+d x)\right )}{4 d}-\frac{\left (b e^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,c+d x\right )}{4 d}\\ &=\frac{1}{4} b e^3 x-\frac{b e^3 (c+d x)^3}{12 d}-\frac{b e^3 \tan ^{-1}(c+d x)}{4 d}+\frac{e^3 (c+d x)^4 \left (a+b \tan ^{-1}(c+d x)\right )}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.0595514, size = 56, normalized size = 0.78 \[ \frac{e^3 \left (\frac{1}{4} (c+d x)^4 \left (a+b \tan ^{-1}(c+d x)\right )-\frac{1}{4} b \left (\frac{1}{3} (c+d x)^3+\tan ^{-1}(c+d x)-d x\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*(-(b*(-(d*x) + (c + d*x)^3/3 + ArcTan[c + d*x]))/4 + ((c + d*x)^4*(a + b*ArcTan[c + d*x]))/4))/d

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Maple [B]  time = 0.035, size = 225, normalized size = 3.1 \begin{align*}{\frac{{d}^{3}{x}^{4}a{e}^{3}}{4}}+{d}^{2}{x}^{3}ac{e}^{3}+{\frac{3\,d{x}^{2}a{c}^{2}{e}^{3}}{2}}+xa{c}^{3}{e}^{3}+{\frac{a{c}^{4}{e}^{3}}{4\,d}}+{\frac{{d}^{3}\arctan \left ( dx+c \right ){x}^{4}b{e}^{3}}{4}}+{d}^{2}\arctan \left ( dx+c \right ){x}^{3}bc{e}^{3}+{\frac{3\,d\arctan \left ( dx+c \right ){x}^{2}b{c}^{2}{e}^{3}}{2}}+\arctan \left ( dx+c \right ) xb{c}^{3}{e}^{3}+{\frac{b\arctan \left ( dx+c \right ){c}^{4}{e}^{3}}{4\,d}}-{\frac{{d}^{2}{x}^{3}b{e}^{3}}{12}}-{\frac{d{x}^{2}bc{e}^{3}}{4}}-{\frac{xb{c}^{2}{e}^{3}}{4}}-{\frac{b{c}^{3}{e}^{3}}{12\,d}}+{\frac{b{e}^{3}x}{4}}+{\frac{bc{e}^{3}}{4\,d}}-{\frac{{e}^{3}b\arctan \left ( dx+c \right ) }{4\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*d^3*x^4*a*e^3+d^2*x^3*a*c*e^3+3/2*d*x^2*a*c^2*e^3+x*a*c^3*e^3+1/4/d*a*c^4*e^3+1/4*d^3*arctan(d*x+c)*x^4*b*
e^3+d^2*arctan(d*x+c)*x^3*b*c*e^3+3/2*d*arctan(d*x+c)*x^2*b*c^2*e^3+arctan(d*x+c)*x*b*c^3*e^3+1/4/d*arctan(d*x
+c)*b*c^4*e^3-1/12*d^2*x^3*b*e^3-1/4*d*x^2*b*c*e^3-1/4*x*b*c^2*e^3-1/12/d*b*c^3*e^3+1/4*b*e^3*x+1/4/d*b*c*e^3-
1/4*b*e^3*arctan(d*x+c)/d

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Maxima [B]  time = 1.51803, size = 500, normalized size = 6.94 \begin{align*} \frac{1}{4} \, a d^{3} e^{3} x^{4} + a c d^{2} e^{3} x^{3} + \frac{3}{2} \, a c^{2} d e^{3} x^{2} + \frac{3}{2} \,{\left (x^{2} \arctan \left (d x + c\right ) - d{\left (\frac{x}{d^{2}} + \frac{{\left (c^{2} - 1\right )} \arctan \left (\frac{d^{2} x + c d}{d}\right )}{d^{3}} - \frac{c \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{3}}\right )}\right )} b c^{2} d e^{3} + \frac{1}{2} \,{\left (2 \, x^{3} \arctan \left (d x + c\right ) - d{\left (\frac{d x^{2} - 4 \, c x}{d^{3}} - \frac{2 \,{\left (c^{3} - 3 \, c\right )} \arctan \left (\frac{d^{2} x + c d}{d}\right )}{d^{4}} + \frac{{\left (3 \, c^{2} - 1\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{4}}\right )}\right )} b c d^{2} e^{3} + \frac{1}{12} \,{\left (3 \, x^{4} \arctan \left (d x + c\right ) - d{\left (\frac{d^{2} x^{3} - 3 \, c d x^{2} + 3 \,{\left (3 \, c^{2} - 1\right )} x}{d^{4}} + \frac{3 \,{\left (c^{4} - 6 \, c^{2} + 1\right )} \arctan \left (\frac{d^{2} x + c d}{d}\right )}{d^{5}} - \frac{6 \,{\left (c^{3} - c\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{5}}\right )}\right )} b d^{3} e^{3} + a c^{3} e^{3} x + \frac{{\left (2 \,{\left (d x + c\right )} \arctan \left (d x + c\right ) - \log \left ({\left (d x + c\right )}^{2} + 1\right )\right )} b c^{3} e^{3}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*d^3*e^3*x^4 + a*c*d^2*e^3*x^3 + 3/2*a*c^2*d*e^3*x^2 + 3/2*(x^2*arctan(d*x + c) - d*(x/d^2 + (c^2 - 1)*ar
ctan((d^2*x + c*d)/d)/d^3 - c*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^3))*b*c^2*d*e^3 + 1/2*(2*x^3*arctan(d*x + c)
- d*((d*x^2 - 4*c*x)/d^3 - 2*(c^3 - 3*c)*arctan((d^2*x + c*d)/d)/d^4 + (3*c^2 - 1)*log(d^2*x^2 + 2*c*d*x + c^2
+ 1)/d^4))*b*c*d^2*e^3 + 1/12*(3*x^4*arctan(d*x + c) - d*((d^2*x^3 - 3*c*d*x^2 + 3*(3*c^2 - 1)*x)/d^4 + 3*(c^
4 - 6*c^2 + 1)*arctan((d^2*x + c*d)/d)/d^5 - 6*(c^3 - c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^5))*b*d^3*e^3 + a*
c^3*e^3*x + 1/2*(2*(d*x + c)*arctan(d*x + c) - log((d*x + c)^2 + 1))*b*c^3*e^3/d

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Fricas [B]  time = 1.5769, size = 315, normalized size = 4.38 \begin{align*} \frac{3 \, a d^{4} e^{3} x^{4} +{\left (12 \, a c - b\right )} d^{3} e^{3} x^{3} + 3 \,{\left (6 \, a c^{2} - b c\right )} d^{2} e^{3} x^{2} + 3 \,{\left (4 \, a c^{3} - b c^{2} + b\right )} d e^{3} x + 3 \,{\left (b d^{4} e^{3} x^{4} + 4 \, b c d^{3} e^{3} x^{3} + 6 \, b c^{2} d^{2} e^{3} x^{2} + 4 \, b c^{3} d e^{3} x +{\left (b c^{4} - b\right )} e^{3}\right )} \arctan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*d^4*e^3*x^4 + (12*a*c - b)*d^3*e^3*x^3 + 3*(6*a*c^2 - b*c)*d^2*e^3*x^2 + 3*(4*a*c^3 - b*c^2 + b)*d*e
^3*x + 3*(b*d^4*e^3*x^4 + 4*b*c*d^3*e^3*x^3 + 6*b*c^2*d^2*e^3*x^2 + 4*b*c^3*d*e^3*x + (b*c^4 - b)*e^3)*arctan(
d*x + c))/d

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Sympy [A]  time = 4.43974, size = 231, normalized size = 3.21 \begin{align*} \begin{cases} a c^{3} e^{3} x + \frac{3 a c^{2} d e^{3} x^{2}}{2} + a c d^{2} e^{3} x^{3} + \frac{a d^{3} e^{3} x^{4}}{4} + \frac{b c^{4} e^{3} \operatorname{atan}{\left (c + d x \right )}}{4 d} + b c^{3} e^{3} x \operatorname{atan}{\left (c + d x \right )} + \frac{3 b c^{2} d e^{3} x^{2} \operatorname{atan}{\left (c + d x \right )}}{2} - \frac{b c^{2} e^{3} x}{4} + b c d^{2} e^{3} x^{3} \operatorname{atan}{\left (c + d x \right )} - \frac{b c d e^{3} x^{2}}{4} + \frac{b d^{3} e^{3} x^{4} \operatorname{atan}{\left (c + d x \right )}}{4} - \frac{b d^{2} e^{3} x^{3}}{12} + \frac{b e^{3} x}{4} - \frac{b e^{3} \operatorname{atan}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\c^{3} e^{3} x \left (a + b \operatorname{atan}{\left (c \right )}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**3*e**3*x + 3*a*c**2*d*e**3*x**2/2 + a*c*d**2*e**3*x**3 + a*d**3*e**3*x**4/4 + b*c**4*e**3*atan
(c + d*x)/(4*d) + b*c**3*e**3*x*atan(c + d*x) + 3*b*c**2*d*e**3*x**2*atan(c + d*x)/2 - b*c**2*e**3*x/4 + b*c*d
**2*e**3*x**3*atan(c + d*x) - b*c*d*e**3*x**2/4 + b*d**3*e**3*x**4*atan(c + d*x)/4 - b*d**2*e**3*x**3/12 + b*e
**3*x/4 - b*e**3*atan(c + d*x)/(4*d), Ne(d, 0)), (c**3*e**3*x*(a + b*atan(c)), True))

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Giac [B]  time = 1.15081, size = 317, normalized size = 4.4 \begin{align*} \frac{6 \, b d^{4} x^{4} \arctan \left (d x + c\right ) e^{3} + 6 \, a d^{4} x^{4} e^{3} + 24 \, b c d^{3} x^{3} \arctan \left (d x + c\right ) e^{3} + 24 \, a c d^{3} x^{3} e^{3} + 36 \, b c^{2} d^{2} x^{2} \arctan \left (d x + c\right ) e^{3} + 36 \, a c^{2} d^{2} x^{2} e^{3} - 2 \, b d^{3} x^{3} e^{3} + 24 \, b c^{3} d x \arctan \left (d x + c\right ) e^{3} + 3 \, \pi b c^{4} e^{3} \mathrm{sgn}\left (d x + c\right ) - 3 \, \pi b c^{4} e^{3} + 24 \, a c^{3} d x e^{3} - 6 \, b c d^{2} x^{2} e^{3} - 6 \, b c^{4} \arctan \left (\frac{1}{d x + c}\right ) e^{3} - 6 \, b c^{2} d x e^{3} + 6 \, b d x e^{3} - 3 \, \pi b e^{3} \mathrm{sgn}\left (d x + c\right ) + 3 \, \pi b e^{3} + 6 \, b \arctan \left (\frac{1}{d x + c}\right ) e^{3}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*b*d^4*x^4*arctan(d*x + c)*e^3 + 6*a*d^4*x^4*e^3 + 24*b*c*d^3*x^3*arctan(d*x + c)*e^3 + 24*a*c*d^3*x^3*
e^3 + 36*b*c^2*d^2*x^2*arctan(d*x + c)*e^3 + 36*a*c^2*d^2*x^2*e^3 - 2*b*d^3*x^3*e^3 + 24*b*c^3*d*x*arctan(d*x
+ c)*e^3 + 3*pi*b*c^4*e^3*sgn(d*x + c) - 3*pi*b*c^4*e^3 + 24*a*c^3*d*x*e^3 - 6*b*c*d^2*x^2*e^3 - 6*b*c^4*arcta
n(1/(d*x + c))*e^3 - 6*b*c^2*d*x*e^3 + 6*b*d*x*e^3 - 3*pi*b*e^3*sgn(d*x + c) + 3*pi*b*e^3 + 6*b*arctan(1/(d*x
+ c))*e^3)/d