3.159 \(\int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{(c+d x)^2} \, dx\)

Optimal. Leaf size=94 \[ \frac {\left (a d^2+b c^2\right )^2 \log (c+d x)}{d^5}-\frac {b c x \left (2 a d^2+b c^2\right )}{d^4}+\frac {b x^2 \left (2 a d^2+b c^2\right )}{2 d^3}-\frac {b^2 c x^3}{3 d^2}+\frac {b^2 x^4}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {1586, 28, 697} \[ \frac {b x^2 \left (2 a d^2+b c^2\right )}{2 d^3}-\frac {b c x \left (2 a d^2+b c^2\right )}{d^4}+\frac {\left (a d^2+b c^2\right )^2 \log (c+d x)}{d^5}-\frac {b^2 c x^3}{3 d^2}+\frac {b^2 x^4}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a^2*c + a^2*d*x + 2*a*b*c*x^2 + 2*a*b*d*x^3 + b^2*c*x^4 + b^2*d*x^5)/(c + d*x)^2,x]

[Out]

-((b*c*(b*c^2 + 2*a*d^2)*x)/d^4) + (b*(b*c^2 + 2*a*d^2)*x^2)/(2*d^3) - (b^2*c*x^3)/(3*d^2) + (b^2*x^4)/(4*d) +
((b*c^2 + a*d^2)^2*Log[c + d*x])/d^5

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 697

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

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 79, normalized size = 0.84 \[ \frac {12 \left (a d^2+b c^2\right )^2 \log (c+d x)+b d x \left (12 a d^2 (d x-2 c)+b \left (-12 c^3+6 c^2 d x-4 c d^2 x^2+3 d^3 x^3\right )\right )}{12 d^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2*c + a^2*d*x + 2*a*b*c*x^2 + 2*a*b*d*x^3 + b^2*c*x^4 + b^2*d*x^5)/(c + d*x)^2,x]

[Out]

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

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fricas [A]  time = 0.96, size = 105, normalized size = 1.12 \[ \frac {3 \, b^{2} d^{4} x^{4} - 4 \, b^{2} c d^{3} x^{3} + 6 \, {\left (b^{2} c^{2} d^{2} + 2 \, a b d^{4}\right )} x^{2} - 12 \, {\left (b^{2} c^{3} d + 2 \, a b c d^{3}\right )} x + 12 \, {\left (b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}\right )} \log \left (d x + c\right )}{12 \, d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*d*x^5+b^2*c*x^4+2*a*b*d*x^3+2*a*b*c*x^2+a^2*d*x+a^2*c)/(d*x+c)^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.26, size = 365, normalized size = 3.88 \[ -\frac {1}{12} \, b^{2} d {\left (\frac {{\left (d x + c\right )}^{4} {\left (\frac {20 \, c}{d x + c} - \frac {60 \, c^{2}}{{\left (d x + c\right )}^{2}} + \frac {120 \, c^{3}}{{\left (d x + c\right )}^{3}} - 3\right )}}{d^{6}} + \frac {60 \, c^{4} \log \left (\frac {{\left | d x + c \right |}}{{\left (d x + c\right )}^{2} {\left | d \right |}}\right )}{d^{6}} - \frac {12 \, c^{5}}{{\left (d x + c\right )} d^{6}}\right )} - \frac {1}{3} \, b^{2} c {\left (\frac {{\left (d x + c\right )}^{3} {\left (\frac {6 \, c}{d x + c} - \frac {18 \, c^{2}}{{\left (d x + c\right )}^{2}} - 1\right )}}{d^{5}} - \frac {12 \, c^{3} \log \left (\frac {{\left | d x + c \right |}}{{\left (d x + c\right )}^{2} {\left | d \right |}}\right )}{d^{5}} + \frac {3 \, c^{4}}{{\left (d x + c\right )} d^{5}}\right )} - a b d {\left (\frac {{\left (d x + c\right )}^{2} {\left (\frac {6 \, c}{d x + c} - 1\right )}}{d^{4}} + \frac {6 \, c^{2} \log \left (\frac {{\left | d x + c \right |}}{{\left (d x + c\right )}^{2} {\left | d \right |}}\right )}{d^{4}} - \frac {2 \, c^{3}}{{\left (d x + c\right )} d^{4}}\right )} + 2 \, a b c {\left (\frac {2 \, c \log \left (\frac {{\left | d x + c \right |}}{{\left (d x + c\right )}^{2} {\left | d \right |}}\right )}{d^{3}} + \frac {d x + c}{d^{3}} - \frac {c^{2}}{{\left (d x + c\right )} d^{3}}\right )} - a^{2} {\left (\frac {\log \left (\frac {{\left | d x + c \right |}}{{\left (d x + c\right )}^{2} {\left | d \right |}}\right )}{d} - \frac {c}{{\left (d x + c\right )} d}\right )} - \frac {a^{2} c}{{\left (d x + c\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*d*x^5+b^2*c*x^4+2*a*b*d*x^3+2*a*b*c*x^2+a^2*d*x+a^2*c)/(d*x+c)^2,x, algorithm="giac")

[Out]

-1/12*b^2*d*((d*x + c)^4*(20*c/(d*x + c) - 60*c^2/(d*x + c)^2 + 120*c^3/(d*x + c)^3 - 3)/d^6 + 60*c^4*log(abs(
d*x + c)/((d*x + c)^2*abs(d)))/d^6 - 12*c^5/((d*x + c)*d^6)) - 1/3*b^2*c*((d*x + c)^3*(6*c/(d*x + c) - 18*c^2/
(d*x + c)^2 - 1)/d^5 - 12*c^3*log(abs(d*x + c)/((d*x + c)^2*abs(d)))/d^5 + 3*c^4/((d*x + c)*d^5)) - a*b*d*((d*
x + c)^2*(6*c/(d*x + c) - 1)/d^4 + 6*c^2*log(abs(d*x + c)/((d*x + c)^2*abs(d)))/d^4 - 2*c^3/((d*x + c)*d^4)) +
2*a*b*c*(2*c*log(abs(d*x + c)/((d*x + c)^2*abs(d)))/d^3 + (d*x + c)/d^3 - c^2/((d*x + c)*d^3)) - a^2*(log(abs
(d*x + c)/((d*x + c)^2*abs(d)))/d - c/((d*x + c)*d)) - a^2*c/((d*x + c)*d)

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maple [A]  time = 0.00, size = 114, normalized size = 1.21 \[ \frac {b^{2} x^{4}}{4 d}-\frac {b^{2} c \,x^{3}}{3 d^{2}}+\frac {a b \,x^{2}}{d}+\frac {b^{2} c^{2} x^{2}}{2 d^{3}}+\frac {a^{2} \ln \left (d x +c \right )}{d}+\frac {2 a b \,c^{2} \ln \left (d x +c \right )}{d^{3}}-\frac {2 a b c x}{d^{2}}+\frac {b^{2} c^{4} \ln \left (d x +c \right )}{d^{5}}-\frac {b^{2} c^{3} x}{d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*d*x^5+b^2*c*x^4+2*a*b*d*x^3+2*a*b*c*x^2+a^2*d*x+a^2*c)/(d*x+c)^2,x)

[Out]

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

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maxima [A]  time = 0.44, size = 105, normalized size = 1.12 \[ \frac {3 \, b^{2} d^{3} x^{4} - 4 \, b^{2} c d^{2} x^{3} + 6 \, {\left (b^{2} c^{2} d + 2 \, a b d^{3}\right )} x^{2} - 12 \, {\left (b^{2} c^{3} + 2 \, a b c d^{2}\right )} x}{12 \, d^{4}} + \frac {{\left (b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}\right )} \log \left (d x + c\right )}{d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*d*x^5+b^2*c*x^4+2*a*b*d*x^3+2*a*b*c*x^2+a^2*d*x+a^2*c)/(d*x+c)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.06, size = 106, normalized size = 1.13 \[ x^2\,\left (\frac {b^2\,c^2}{2\,d^3}+\frac {a\,b}{d}\right )+\frac {\ln \left (c+d\,x\right )\,\left (a^2\,d^4+2\,a\,b\,c^2\,d^2+b^2\,c^4\right )}{d^5}+\frac {b^2\,x^4}{4\,d}-\frac {b^2\,c\,x^3}{3\,d^2}-\frac {c\,x\,\left (\frac {b^2\,c^2}{d^3}+\frac {2\,a\,b}{d}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c + b^2*c*x^4 + b^2*d*x^5 + a^2*d*x + 2*a*b*c*x^2 + 2*a*b*d*x^3)/(c + d*x)^2,x)

[Out]

x^2*((b^2*c^2)/(2*d^3) + (a*b)/d) + (log(c + d*x)*(a^2*d^4 + b^2*c^4 + 2*a*b*c^2*d^2))/d^5 + (b^2*x^4)/(4*d) -
(b^2*c*x^3)/(3*d^2) - (c*x*((b^2*c^2)/d^3 + (2*a*b)/d))/d

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sympy [A]  time = 0.31, size = 88, normalized size = 0.94 \[ - \frac {b^{2} c x^{3}}{3 d^{2}} + \frac {b^{2} x^{4}}{4 d} + x^{2} \left (\frac {a b}{d} + \frac {b^{2} c^{2}}{2 d^{3}}\right ) + x \left (- \frac {2 a b c}{d^{2}} - \frac {b^{2} c^{3}}{d^{4}}\right ) + \frac {\left (a d^{2} + b c^{2}\right )^{2} \log {\left (c + d x \right )}}{d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*d*x**5+b**2*c*x**4+2*a*b*d*x**3+2*a*b*c*x**2+a**2*d*x+a**2*c)/(d*x+c)**2,x)

[Out]

-b**2*c*x**3/(3*d**2) + b**2*x**4/(4*d) + x**2*(a*b/d + b**2*c**2/(2*d**3)) + x*(-2*a*b*c/d**2 - b**2*c**3/d**
4) + (a*d**2 + b*c**2)**2*log(c + d*x)/d**5

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