3.14.56 \(\int \frac {(a+b x)^2}{(c+d x)^3} \, dx\) [1356]

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

[Out]

-1/2*(-a*d+b*c)^2/d^3/(d*x+c)^2+2*b*(-a*d+b*c)/d^3/(d*x+c)+b^2*ln(d*x+c)/d^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 b (b c-a d)}{d^3 (c+d x)}-\frac {(b c-a d)^2}{2 d^3 (c+d x)^2}+\frac {b^2 \log (c+d x)}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/(c + d*x)^3,x]

[Out]

-1/2*(b*c - a*d)^2/(d^3*(c + d*x)^2) + (2*b*(b*c - a*d))/(d^3*(c + d*x)) + (b^2*Log[c + d*x])/d^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 {(a+b x)^2}{(c+d x)^3} \, dx &=\int \left (\frac {(-b c+a d)^2}{d^2 (c+d x)^3}-\frac {2 b (b c-a d)}{d^2 (c+d x)^2}+\frac {b^2}{d^2 (c+d x)}\right ) \, dx\\ &=-\frac {(b c-a d)^2}{2 d^3 (c+d x)^2}+\frac {2 b (b c-a d)}{d^3 (c+d x)}+\frac {b^2 \log (c+d x)}{d^3}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 48, normalized size = 0.81 \begin {gather*} \frac {\frac {(b c-a d) (3 b c+a d+4 b d x)}{(c+d x)^2}+2 b^2 \log (c+d x)}{2 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/(c + d*x)^3,x]

[Out]

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

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Mathics [A]
time = 2.58, size = 84, normalized size = 1.42 \begin {gather*} \frac {-\frac {a^2 d^2}{2}-a b c d+b^2 \text {Log}\left [c+d x\right ] \left (c^2+2 c d x+d^2 x^2\right )+\frac {3 b^2 c^2}{2}-2 b d x \left (a d-b c\right )}{d^3 \left (c^2+2 c d x+d^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.13, size = 69, normalized size = 1.17

method	result	size
risch	\(\frac {-\frac {2 b \left (a d -b c \right ) x}{d^{2}}-\frac {a^{2} d^{2}+2 a b c d -3 b^{2} c^{2}}{2 d^{3}}}{\left (d x +c \right )^{2}}+\frac {b^{2} \ln \left (d x +c \right )}{d^{3}}\)	\(66\)
norman	\(\frac {-\frac {a^{2} d^{2}+2 a b c d -3 b^{2} c^{2}}{2 d^{3}}-\frac {2 \left (a b d -b^{2} c \right ) x}{d^{2}}}{\left (d x +c \right )^{2}}+\frac {b^{2} \ln \left (d x +c \right )}{d^{3}}\)	\(68\)
default	\(-\frac {2 b \left (a d -b c \right )}{d^{3} \left (d x +c \right )}+\frac {b^{2} \ln \left (d x +c \right )}{d^{3}}-\frac {a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{2 d^{3} \left (d x +c \right )^{2}}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.29, size = 100, normalized size = 1.69 \begin {gather*} \frac {3 \, b^{2} c^{2} - 2 \, a b c d - 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 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (d x + c\right )}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.23, size = 77, normalized size = 1.31 \begin {gather*} \frac {b^2\,\ln \left (c+d\,x\right )}{d^3}-\frac {\frac {a^2\,d^2+2\,a\,b\,c\,d-3\,b^2\,c^2}{2\,d^3}+\frac {2\,b\,x\,\left (a\,d-b\,c\right )}{d^2}}{c^2+2\,c\,d\,x+d^2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^2/(c + d*x)^3,x)

[Out]

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

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