3.13.53 \(\int \frac {(c+d x)^2}{(a+b x)^3} \, dx\) [1253]

Optimal. Leaf size=59 \[ -\frac {(b c-a d)^2}{2 b^3 (a+b x)^2}-\frac {2 d (b c-a d)}{b^3 (a+b x)}+\frac {d^2 \log (a+b x)}{b^3} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {45} \begin {gather*} -\frac {2 d (b c-a d)}{b^3 (a+b x)}-\frac {(b c-a d)^2}{2 b^3 (a+b x)^2}+\frac {d^2 \log (a+b x)}{b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2/(a + b*x)^3,x]

[Out]

-1/2*(b*c - a*d)^2/(b^3*(a + b*x)^2) - (2*d*(b*c - a*d))/(b^3*(a + b*x)) + (d^2*Log[a + b*x])/b^3

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 49, normalized size = 0.83 \begin {gather*} \frac {-\frac {(b c-a d) (3 a d+b (c+4 d x))}{(a+b x)^2}+2 d^2 \log (a+b x)}{2 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2/(a + b*x)^3,x]

[Out]

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

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Mathics [A]
time = 2.40, size = 84, normalized size = 1.42 \begin {gather*} \frac {-a b c d+d^2 \text {Log}\left [a+b x\right ] \left (a^2+2 a b x+b^2 x^2\right )+\frac {3 a^2 d^2}{2}-\frac {b^2 c^2}{2}+2 b d x \left (a d-b c\right )}{b^3 \left (a^2+2 a b x+b^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.14, size = 69, normalized size = 1.17

method	result	size
risch	\(\frac {\frac {2 d \left (a d -b c \right ) x}{b^{2}}+\frac {3 a^{2} d^{2}-2 a b c d -b^{2} c^{2}}{2 b^{3}}}{\left (b x +a \right )^{2}}+\frac {d^{2} \ln \left (b x +a \right )}{b^{3}}\)	\(67\)
default	\(\frac {2 d \left (a d -b c \right )}{b^{3} \left (b x +a \right )}-\frac {a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{2 b^{3} \left (b x +a \right )^{2}}+\frac {d^{2} \ln \left (b x +a \right )}{b^{3}}\)	\(69\)
norman	\(\frac {\frac {3 a^{2} d^{2}-2 a b c d -b^{2} c^{2}}{2 b^{3}}+\frac {2 \left (a \,d^{2}-b d c \right ) x}{b^{2}}}{\left (b x +a \right )^{2}}+\frac {d^{2} \ln \left (b x +a \right )}{b^{3}}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2/(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 79, normalized size = 1.34 \begin {gather*} -\frac {b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2} + 4 \, {\left (b^{2} c d - a b d^{2}\right )} x}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} + \frac {d^{2} \log \left (b x + a\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.29, size = 99, normalized size = 1.68 \begin {gather*} -\frac {b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2} + 4 \, {\left (b^{2} c d - a b d^{2}\right )} x - 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.27, size = 80, normalized size = 1.36 \begin {gather*} \frac {3 a^{2} d^{2} - 2 a b c d - b^{2} c^{2} + x \left (4 a b d^{2} - 4 b^{2} c d\right )}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac {d^{2} \log {\left (a + b x \right )}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.00, size = 75, normalized size = 1.27 \begin {gather*} \frac {\frac {1}{2} \left (\left (4 d^{2} a-4 d b c\right ) x+\frac {3 d^{2} a^{2}-2 d b a c-b^{2} c^{2}}{b}\right )}{b^{2} \left (x b+a\right )^{2}}+\frac {d^{2} \ln \left |x b+a\right |}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.20, size = 77, normalized size = 1.31 \begin {gather*} \frac {d^2\,\ln \left (a+b\,x\right )}{b^3}-\frac {\frac {-3\,a^2\,d^2+2\,a\,b\,c\,d+b^2\,c^2}{2\,b^3}-\frac {2\,d\,x\,\left (a\,d-b\,c\right )}{b^2}}{a^2+2\,a\,b\,x+b^2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^2/(a + b*x)^3,x)

[Out]

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

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