3.5.56 \(\int \frac {a+c x^2}{(d+e x)^2} \, dx\) [456]

Optimal. Leaf size=43 \[ \frac {c x}{e^2}-\frac {c d^2+a e^2}{e^3 (d+e x)}-\frac {2 c d \log (d+e x)}{e^3} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 43, 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 = {711} \begin {gather*} -\frac {a e^2+c d^2}{e^3 (d+e x)}-\frac {2 c d \log (d+e x)}{e^3}+\frac {c x}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + c*x^2)/(d + e*x)^2,x]

[Out]

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

Rule 711

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]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 39, normalized size = 0.91 \begin {gather*} \frac {c e x-\frac {c d^2+a e^2}{d+e x}-2 c d \log (d+e x)}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^2)/(d + e*x)^2,x]

[Out]

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

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Maple [A]
time = 0.45, size = 44, normalized size = 1.02

method	result	size
default	\(\frac {c x}{e^{2}}-\frac {e^{2} a +c \,d^{2}}{e^{3} \left (e x +d \right )}-\frac {2 c d \ln \left (e x +d \right )}{e^{3}}\)	\(44\)
norman	\(\frac {\frac {c \,x^{2}}{e}-\frac {e^{2} a +2 c \,d^{2}}{e^{3}}}{e x +d}-\frac {2 c d \ln \left (e x +d \right )}{e^{3}}\)	\(49\)
risch	\(\frac {c x}{e^{2}}-\frac {a}{e \left (e x +d \right )}-\frac {c \,d^{2}}{e^{3} \left (e x +d \right )}-\frac {2 c d \ln \left (e x +d \right )}{e^{3}}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 42, normalized size = 0.98 \begin {gather*} -2 \, c d e^{\left (-3\right )} \log \left (x e + d\right ) + c x e^{\left (-2\right )} - \frac {c d^{2} + a e^{2}}{x e^{4} + d e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.08, size = 58, normalized size = 1.35 \begin {gather*} \frac {c d x e - c d^{2} + {\left (c x^{2} - a\right )} e^{2} - 2 \, {\left (c d x e + c d^{2}\right )} \log \left (x e + d\right )}{x e^{4} + d e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.10, size = 42, normalized size = 0.98 \begin {gather*} - \frac {2 c d \log {\left (d + e x \right )}}{e^{3}} + \frac {c x}{e^{2}} + \frac {- a e^{2} - c d^{2}}{d e^{3} + e^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.29, size = 65, normalized size = 1.51 \begin {gather*} {\left (2 \, d e^{\left (-3\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) + {\left (x e + d\right )} e^{\left (-3\right )} - \frac {d^{2} e^{\left (-3\right )}}{x e + d}\right )} c - \frac {a e^{\left (-1\right )}}{x e + d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.06, size = 49, normalized size = 1.14 \begin {gather*} \frac {c\,x}{e^2}-\frac {c\,d^2+a\,e^2}{e\,\left (x\,e^3+d\,e^2\right )}-\frac {2\,c\,d\,\ln \left (d+e\,x\right )}{e^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)/(d + e*x)^2,x)

[Out]

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

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