3.14 \(\int \frac {(a+b \tan ^{-1}(c+d x))^2}{(c e+d e x)^5} \, dx\)

Optimal. Leaf size=170 \[ \frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{2 d e^5 (c+d x)}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{6 d e^5 (c+d x)^3}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}+\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d e^5}-\frac {b^2}{12 d e^5 (c+d x)^2}-\frac {2 b^2 \log (c+d x)}{3 d e^5}+\frac {b^2 \log \left ((c+d x)^2+1\right )}{3 d e^5} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {5043, 12, 4852, 4918, 266, 44, 36, 29, 31, 4884} \[ \frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{2 d e^5 (c+d x)}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{6 d e^5 (c+d x)^3}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}+\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d e^5}-\frac {b^2}{12 d e^5 (c+d x)^2}-\frac {2 b^2 \log (c+d x)}{3 d e^5}+\frac {b^2 \log \left ((c+d x)^2+1\right )}{3 d e^5} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c + d*x])^2/(c*e + d*e*x)^5,x]

[Out]

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

Rule 12

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 266

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

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 4884

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 4918

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcTan[c*x])^p)/(d + e*
x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

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]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{(c e+d e x)^5} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{e^5 x^5} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{x^5} \, dx,x,c+d x\right )}{d e^5}\\ &=-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}+\frac {b \operatorname {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{x^4 \left (1+x^2\right )} \, dx,x,c+d x\right )}{2 d e^5}\\ &=-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}+\frac {b \operatorname {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{x^4} \, dx,x,c+d x\right )}{2 d e^5}-\frac {b \operatorname {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{x^2 \left (1+x^2\right )} \, dx,x,c+d x\right )}{2 d e^5}\\ &=-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{6 d e^5 (c+d x)^3}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}-\frac {b \operatorname {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{x^2} \, dx,x,c+d x\right )}{2 d e^5}+\frac {b \operatorname {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{1+x^2} \, dx,x,c+d x\right )}{2 d e^5}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x^3 \left (1+x^2\right )} \, dx,x,c+d x\right )}{6 d e^5}\\ &=-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{6 d e^5 (c+d x)^3}+\frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{2 d e^5 (c+d x)}+\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d e^5}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x^2 (1+x)} \, dx,x,(c+d x)^2\right )}{12 d e^5}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x \left (1+x^2\right )} \, dx,x,c+d x\right )}{2 d e^5}\\ &=-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{6 d e^5 (c+d x)^3}+\frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{2 d e^5 (c+d x)}+\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d e^5}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}+\frac {b^2 \operatorname {Subst}\left (\int \left (\frac {1}{x^2}-\frac {1}{x}+\frac {1}{1+x}\right ) \, dx,x,(c+d x)^2\right )}{12 d e^5}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,(c+d x)^2\right )}{4 d e^5}\\ &=-\frac {b^2}{12 d e^5 (c+d x)^2}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{6 d e^5 (c+d x)^3}+\frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{2 d e^5 (c+d x)}+\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d e^5}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}-\frac {b^2 \log (c+d x)}{6 d e^5}+\frac {b^2 \log \left (1+(c+d x)^2\right )}{12 d e^5}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,(c+d x)^2\right )}{4 d e^5}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,(c+d x)^2\right )}{4 d e^5}\\ &=-\frac {b^2}{12 d e^5 (c+d x)^2}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{6 d e^5 (c+d x)^3}+\frac {b \left (a+b \tan ^{-1}(c+d x)\right )}{2 d e^5 (c+d x)}+\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d e^5}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^2}{4 d e^5 (c+d x)^4}-\frac {2 b^2 \log (c+d x)}{3 d e^5}+\frac {b^2 \log \left (1+(c+d x)^2\right )}{3 d e^5}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 245, normalized size = 1.44 \[ -\frac {3 a^2-2 b \tan ^{-1}(c+d x) \left (3 a \left (c^4+4 c^3 d x+6 c^2 d^2 x^2+4 c d^3 x^3+d^4 x^4-1\right )+b \left (3 c^3+9 c^2 d x+9 c d^2 x^2-c+3 d^3 x^3-d x\right )\right )-6 a b (c+d x)^3+2 a b (c+d x)-4 b^2 (c+d x)^4 \log \left (c^2+2 c d x+d^2 x^2+1\right )-3 b^2 \left (c^4+4 c^3 d x+6 c^2 d^2 x^2+4 c d^3 x^3+d^4 x^4-1\right ) \tan ^{-1}(c+d x)^2+b^2 (c+d x)^2+8 b^2 (c+d x)^4 \log (c+d x)}{12 d e^5 (c+d x)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c + d*x])^2/(c*e + d*e*x)^5,x]

[Out]

-1/12*(3*a^2 + 2*a*b*(c + d*x) + b^2*(c + d*x)^2 - 6*a*b*(c + d*x)^3 - 2*b*(b*(-c + 3*c^3 - d*x + 9*c^2*d*x +
9*c*d^2*x^2 + 3*d^3*x^3) + 3*a*(-1 + c^4 + 4*c^3*d*x + 6*c^2*d^2*x^2 + 4*c*d^3*x^3 + d^4*x^4))*ArcTan[c + d*x]
- 3*b^2*(-1 + c^4 + 4*c^3*d*x + 6*c^2*d^2*x^2 + 4*c*d^3*x^3 + d^4*x^4)*ArcTan[c + d*x]^2 + 8*b^2*(c + d*x)^4*
Log[c + d*x] - 4*b^2*(c + d*x)^4*Log[1 + c^2 + 2*c*d*x + d^2*x^2])/(d*e^5*(c + d*x)^4)

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fricas [B]  time = 0.47, size = 448, normalized size = 2.64 \[ \frac {6 \, a b d^{3} x^{3} + 6 \, a b c^{3} + {\left (18 \, a b c - b^{2}\right )} d^{2} x^{2} - b^{2} c^{2} - 2 \, a b c + 2 \, {\left (9 \, a b c^{2} - b^{2} c - a b\right )} d x + 3 \, {\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4} - b^{2}\right )} \arctan \left (d x + c\right )^{2} - 3 \, a^{2} + 2 \, {\left (3 \, a b d^{4} x^{4} + 3 \, {\left (4 \, a b c + b^{2}\right )} d^{3} x^{3} + 3 \, a b c^{4} + 3 \, b^{2} c^{3} + 9 \, {\left (2 \, a b c^{2} + b^{2} c\right )} d^{2} x^{2} - b^{2} c + {\left (12 \, a b c^{3} + 9 \, b^{2} c^{2} - b^{2}\right )} d x - 3 \, a b\right )} \arctan \left (d x + c\right ) + 4 \, {\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) - 8 \, {\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \left (d x + c\right )}{12 \, {\left (d^{5} e^{5} x^{4} + 4 \, c d^{4} e^{5} x^{3} + 6 \, c^{2} d^{3} e^{5} x^{2} + 4 \, c^{3} d^{2} e^{5} x + c^{4} d e^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^2/(d*e*x+c*e)^5,x, algorithm="fricas")

[Out]

1/12*(6*a*b*d^3*x^3 + 6*a*b*c^3 + (18*a*b*c - b^2)*d^2*x^2 - b^2*c^2 - 2*a*b*c + 2*(9*a*b*c^2 - b^2*c - a*b)*d
*x + 3*(b^2*d^4*x^4 + 4*b^2*c*d^3*x^3 + 6*b^2*c^2*d^2*x^2 + 4*b^2*c^3*d*x + b^2*c^4 - b^2)*arctan(d*x + c)^2 -
3*a^2 + 2*(3*a*b*d^4*x^4 + 3*(4*a*b*c + b^2)*d^3*x^3 + 3*a*b*c^4 + 3*b^2*c^3 + 9*(2*a*b*c^2 + b^2*c)*d^2*x^2
- b^2*c + (12*a*b*c^3 + 9*b^2*c^2 - b^2)*d*x - 3*a*b)*arctan(d*x + c) + 4*(b^2*d^4*x^4 + 4*b^2*c*d^3*x^3 + 6*b
^2*c^2*d^2*x^2 + 4*b^2*c^3*d*x + b^2*c^4)*log(d^2*x^2 + 2*c*d*x + c^2 + 1) - 8*(b^2*d^4*x^4 + 4*b^2*c*d^3*x^3
+ 6*b^2*c^2*d^2*x^2 + 4*b^2*c^3*d*x + b^2*c^4)*log(d*x + c))/(d^5*e^5*x^4 + 4*c*d^4*e^5*x^3 + 6*c^2*d^3*e^5*x^
2 + 4*c^3*d^2*e^5*x + c^4*d*e^5)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^2/(d*e*x+c*e)^5,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.06, size = 242, normalized size = 1.42 \[ -\frac {a^{2}}{4 d \,e^{5} \left (d x +c \right )^{4}}-\frac {b^{2} \arctan \left (d x +c \right )^{2}}{4 d \,e^{5} \left (d x +c \right )^{4}}-\frac {b^{2} \arctan \left (d x +c \right )}{6 d \,e^{5} \left (d x +c \right )^{3}}+\frac {b^{2} \arctan \left (d x +c \right )}{2 d \,e^{5} \left (d x +c \right )}+\frac {b^{2} \arctan \left (d x +c \right )^{2}}{4 d \,e^{5}}-\frac {b^{2}}{12 d \,e^{5} \left (d x +c \right )^{2}}-\frac {2 b^{2} \ln \left (d x +c \right )}{3 d \,e^{5}}+\frac {b^{2} \ln \left (1+\left (d x +c \right )^{2}\right )}{3 d \,e^{5}}-\frac {a b \arctan \left (d x +c \right )}{2 d \,e^{5} \left (d x +c \right )^{4}}-\frac {a b}{6 d \,e^{5} \left (d x +c \right )^{3}}+\frac {a b}{2 d \,e^{5} \left (d x +c \right )}+\frac {a b \arctan \left (d x +c \right )}{2 d \,e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(d*x+c))^2/(d*e*x+c*e)^5,x)

[Out]

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

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maxima [B]  time = 0.51, size = 534, normalized size = 3.14 \[ \frac {1}{6} \, {\left (d {\left (\frac {3 \, d^{2} x^{2} + 6 \, c d x + 3 \, c^{2} - 1}{d^{5} e^{5} x^{3} + 3 \, c d^{4} e^{5} x^{2} + 3 \, c^{2} d^{3} e^{5} x + c^{3} d^{2} e^{5}} + \frac {3 \, \arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{2} e^{5}}\right )} - \frac {3 \, \arctan \left (d x + c\right )}{d^{5} e^{5} x^{4} + 4 \, c d^{4} e^{5} x^{3} + 6 \, c^{2} d^{3} e^{5} x^{2} + 4 \, c^{3} d^{2} e^{5} x + c^{4} d e^{5}}\right )} a b + \frac {1}{12} \, {\left (2 \, d {\left (\frac {3 \, d^{2} x^{2} + 6 \, c d x + 3 \, c^{2} - 1}{d^{5} e^{5} x^{3} + 3 \, c d^{4} e^{5} x^{2} + 3 \, c^{2} d^{3} e^{5} x + c^{3} d^{2} e^{5}} + \frac {3 \, \arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{2} e^{5}}\right )} \arctan \left (d x + c\right ) - \frac {{\left (3 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \arctan \left (d x + c\right )^{2} - 4 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) + 8 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \log \left (d x + c\right ) + 1\right )} d^{2}}{d^{5} e^{5} x^{2} + 2 \, c d^{4} e^{5} x + c^{2} d^{3} e^{5}}\right )} b^{2} - \frac {b^{2} \arctan \left (d x + c\right )^{2}}{4 \, {\left (d^{5} e^{5} x^{4} + 4 \, c d^{4} e^{5} x^{3} + 6 \, c^{2} d^{3} e^{5} x^{2} + 4 \, c^{3} d^{2} e^{5} x + c^{4} d e^{5}\right )}} - \frac {a^{2}}{4 \, {\left (d^{5} e^{5} x^{4} + 4 \, c d^{4} e^{5} x^{3} + 6 \, c^{2} d^{3} e^{5} x^{2} + 4 \, c^{3} d^{2} e^{5} x + c^{4} d e^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(d*x+c))^2/(d*e*x+c*e)^5,x, algorithm="maxima")

[Out]

1/6*(d*((3*d^2*x^2 + 6*c*d*x + 3*c^2 - 1)/(d^5*e^5*x^3 + 3*c*d^4*e^5*x^2 + 3*c^2*d^3*e^5*x + c^3*d^2*e^5) + 3*
arctan((d^2*x + c*d)/d)/(d^2*e^5)) - 3*arctan(d*x + c)/(d^5*e^5*x^4 + 4*c*d^4*e^5*x^3 + 6*c^2*d^3*e^5*x^2 + 4*
c^3*d^2*e^5*x + c^4*d*e^5))*a*b + 1/12*(2*d*((3*d^2*x^2 + 6*c*d*x + 3*c^2 - 1)/(d^5*e^5*x^3 + 3*c*d^4*e^5*x^2
+ 3*c^2*d^3*e^5*x + c^3*d^2*e^5) + 3*arctan((d^2*x + c*d)/d)/(d^2*e^5))*arctan(d*x + c) - (3*(d^2*x^2 + 2*c*d*
x + c^2)*arctan(d*x + c)^2 - 4*(d^2*x^2 + 2*c*d*x + c^2)*log(d^2*x^2 + 2*c*d*x + c^2 + 1) + 8*(d^2*x^2 + 2*c*d
*x + c^2)*log(d*x + c) + 1)*d^2/(d^5*e^5*x^2 + 2*c*d^4*e^5*x + c^2*d^3*e^5))*b^2 - 1/4*b^2*arctan(d*x + c)^2/(
d^5*e^5*x^4 + 4*c*d^4*e^5*x^3 + 6*c^2*d^3*e^5*x^2 + 4*c^3*d^2*e^5*x + c^4*d*e^5) - 1/4*a^2/(d^5*e^5*x^4 + 4*c*
d^4*e^5*x^3 + 6*c^2*d^3*e^5*x^2 + 4*c^3*d^2*e^5*x + c^4*d*e^5)

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mupad [B]  time = 3.65, size = 438, normalized size = 2.58 \[ {\mathrm {atan}\left (c+d\,x\right )}^2\,\left (\frac {b^2}{4\,d\,e^5}-\frac {b^2}{4\,d^3\,e^5\,\left (\frac {c^4}{d^2}+6\,c^2\,x^2+d^2\,x^4+\frac {4\,c^3\,x}{d}+4\,c\,d\,x^3\right )}\right )-\frac {x^2\,\left (\frac {b^2\,d}{2}-9\,a\,b\,c\,d\right )+x\,\left (b^2\,c-9\,a\,b\,c^2+a\,b\right )+\frac {3\,a^2-6\,a\,b\,c^3+2\,a\,b\,c+b^2\,c^2}{2\,d}-3\,a\,b\,d^2\,x^3}{6\,c^4\,e^5+24\,c^3\,d\,e^5\,x+36\,c^2\,d^2\,e^5\,x^2+24\,c\,d^3\,e^5\,x^3+6\,d^4\,e^5\,x^4}+\frac {\mathrm {atan}\left (c+d\,x\right )\,\left (\frac {b^2\,x^3}{2\,e^5}-\frac {a\,b}{2\,d^3\,e^5}+\frac {b^2\,c\,\left (\frac {c^2-1}{3\,d^2}+\frac {2\,c^2}{3\,d^2}\right )}{2\,d\,e^5}+\frac {b^2\,x\,\left (d\,\left (\frac {c^2-1}{3\,d^2}+\frac {2\,c^2}{3\,d^2}\right )+\frac {2\,c^2}{d}\right )}{2\,d\,e^5}+\frac {3\,b^2\,c\,x^2}{2\,d\,e^5}\right )}{\frac {c^4}{d^2}+6\,c^2\,x^2+d^2\,x^4+\frac {4\,c^3\,x}{d}+4\,c\,d\,x^3}-\frac {2\,b^2\,\ln \left (c+d\,x\right )}{3\,d\,e^5}-\frac {\ln \left (c+d\,x-\mathrm {i}\right )\,\left (-\frac {b^2}{3}+\frac {a\,b\,1{}\mathrm {i}}{4}\right )}{d\,e^5}+\frac {\ln \left (c+d\,x+1{}\mathrm {i}\right )\,\left (\frac {b^2}{3}+\frac {1{}\mathrm {i}\,a\,b}{4}\right )}{d\,e^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c + d*x))^2/(c*e + d*e*x)^5,x)

[Out]

atan(c + d*x)^2*(b^2/(4*d*e^5) - b^2/(4*d^3*e^5*(c^4/d^2 + 6*c^2*x^2 + d^2*x^4 + (4*c^3*x)/d + 4*c*d*x^3))) -
(x^2*((b^2*d)/2 - 9*a*b*c*d) + x*(a*b + b^2*c - 9*a*b*c^2) + (3*a^2 + b^2*c^2 + 2*a*b*c - 6*a*b*c^3)/(2*d) - 3
*a*b*d^2*x^3)/(6*c^4*e^5 + 6*d^4*e^5*x^4 + 24*c*d^3*e^5*x^3 + 36*c^2*d^2*e^5*x^2 + 24*c^3*d*e^5*x) + (atan(c +
d*x)*((b^2*x^3)/(2*e^5) - (a*b)/(2*d^3*e^5) + (b^2*c*((c^2 - 1)/(3*d^2) + (2*c^2)/(3*d^2)))/(2*d*e^5) + (b^2*
x*(d*((c^2 - 1)/(3*d^2) + (2*c^2)/(3*d^2)) + (2*c^2)/d))/(2*d*e^5) + (3*b^2*c*x^2)/(2*d*e^5)))/(c^4/d^2 + 6*c^
2*x^2 + d^2*x^4 + (4*c^3*x)/d + 4*c*d*x^3) - (2*b^2*log(c + d*x))/(3*d*e^5) - (log(c + d*x - 1i)*((a*b*1i)/4 -
b^2/3))/(d*e^5) + (log(c + d*x + 1i)*((a*b*1i)/4 + b^2/3))/(d*e^5)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(d*x+c))**2/(d*e*x+c*e)**5,x)

[Out]

Timed out

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