3.6.51 \(\int \frac {(d+e x) (f+g x)^2}{d^2-e^2 x^2} \, dx\) [551]

Optimal. Leaf size=50 \[ -\frac {g (e f+d g) x}{e^2}-\frac {(f+g x)^2}{2 e}-\frac {(e f+d g)^2 \log (d-e x)}{e^3} \]

[Out]

-g*(d*g+e*f)*x/e^2-1/2*(g*x+f)^2/e-(d*g+e*f)^2*ln(-e*x+d)/e^3

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Rubi [A]
time = 0.02, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {813, 45} \begin {gather*} -\frac {(d g+e f)^2 \log (d-e x)}{e^3}-\frac {g x (d g+e f)}{e^2}-\frac {(f+g x)^2}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x)*(f + g*x)^2)/(d^2 - e^2*x^2),x]

[Out]

-((g*(e*f + d*g)*x)/e^2) - (f + g*x)^2/(2*e) - ((e*f + d*g)^2*Log[d - e*x])/e^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])

Rule 813

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^m*(
f + g*x)^(p + 1)*(a/f + (c/g)*x)^p, x] /; FreeQ[{a, c, d, e, f, g, m}, x] && EqQ[c*f^2 + a*g^2, 0] && (Integer
Q[p] || (GtQ[a, 0] && GtQ[f, 0] && EqQ[p, -1]))

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 43, normalized size = 0.86 \begin {gather*} -\frac {e g x (4 e f+2 d g+e g x)+2 (e f+d g)^2 \log (d-e x)}{2 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)*(f + g*x)^2)/(d^2 - e^2*x^2),x]

[Out]

-1/2*(e*g*x*(4*e*f + 2*d*g + e*g*x) + 2*(e*f + d*g)^2*Log[d - e*x])/e^3

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Maple [A]
time = 0.08, size = 59, normalized size = 1.18

method	result	size
default	\(-\frac {g \left (\frac {1}{2} e g \,x^{2}+d g x +2 e f x \right )}{e^{2}}+\frac {\left (-d^{2} g^{2}-2 d e f g -e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(59\)
norman	\(-\frac {g^{2} x^{2}}{2 e}-\frac {g \left (d g +2 e f \right ) x}{e^{2}}-\frac {\left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\)	\(61\)
risch	\(-\frac {g^{2} x^{2}}{2 e}-\frac {g^{2} d x}{e^{2}}-\frac {2 g f x}{e}-\frac {\ln \left (-e x +d \right ) d^{2} g^{2}}{e^{3}}-\frac {2 \ln \left (-e x +d \right ) d f g}{e^{2}}-\frac {\ln \left (-e x +d \right ) f^{2}}{e}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(g*x+f)^2/(-e^2*x^2+d^2),x,method=_RETURNVERBOSE)

[Out]

-g/e^2*(1/2*e*g*x^2+d*g*x+2*e*f*x)+(-d^2*g^2-2*d*e*f*g-e^2*f^2)/e^3*ln(-e*x+d)

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Maxima [A]
time = 0.28, size = 64, normalized size = 1.28 \begin {gather*} -{\left (d^{2} g^{2} + 2 \, d f g e + f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (x e - d\right ) - \frac {1}{2} \, {\left (g^{2} x^{2} e + 2 \, {\left (d g^{2} + 2 \, f g e\right )} x\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(g*x+f)^2/(-e^2*x^2+d^2),x, algorithm="maxima")

[Out]

-(d^2*g^2 + 2*d*f*g*e + f^2*e^2)*e^(-3)*log(x*e - d) - 1/2*(g^2*x^2*e + 2*(d*g^2 + 2*f*g*e)*x)*e^(-2)

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Fricas [A]
time = 2.06, size = 62, normalized size = 1.24 \begin {gather*} -\frac {1}{2} \, {\left (2 \, d g^{2} x e + {\left (g^{2} x^{2} + 4 \, f g x\right )} e^{2} + 2 \, {\left (d^{2} g^{2} + 2 \, d f g e + f^{2} e^{2}\right )} \log \left (x e - d\right )\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(g*x+f)^2/(-e^2*x^2+d^2),x, algorithm="fricas")

[Out]

-1/2*(2*d*g^2*x*e + (g^2*x^2 + 4*f*g*x)*e^2 + 2*(d^2*g^2 + 2*d*f*g*e + f^2*e^2)*log(x*e - d))*e^(-3)

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Sympy [A]
time = 0.11, size = 46, normalized size = 0.92 \begin {gather*} - x \left (\frac {d g^{2}}{e^{2}} + \frac {2 f g}{e}\right ) - \frac {g^{2} x^{2}}{2 e} - \frac {\left (d g + e f\right )^{2} \log {\left (- d + e x \right )}}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(g*x+f)**2/(-e**2*x**2+d**2),x)

[Out]

-x*(d*g**2/e**2 + 2*f*g/e) - g**2*x**2/(2*e) - (d*g + e*f)**2*log(-d + e*x)/e**3

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Giac [A]
time = 1.35, size = 64, normalized size = 1.28 \begin {gather*} -{\left (d^{2} g^{2} + 2 \, d f g e + f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | x e - d \right |}\right ) - \frac {1}{2} \, {\left (g^{2} x^{2} e + 2 \, d g^{2} x + 4 \, f g x e\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(g*x+f)^2/(-e^2*x^2+d^2),x, algorithm="giac")

[Out]

-(d^2*g^2 + 2*d*f*g*e + f^2*e^2)*e^(-3)*log(abs(x*e - d)) - 1/2*(g^2*x^2*e + 2*d*g^2*x + 4*f*g*x*e)*e^(-2)

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Mupad [B]
time = 2.61, size = 65, normalized size = 1.30 \begin {gather*} -x\,\left (\frac {d\,g^2}{e^2}+\frac {2\,f\,g}{e}\right )-\frac {\ln \left (e\,x-d\right )\,\left (d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2\right )}{e^3}-\frac {g^2\,x^2}{2\,e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)^2*(d + e*x))/(d^2 - e^2*x^2),x)

[Out]

- x*((d*g^2)/e^2 + (2*f*g)/e) - (log(e*x - d)*(d^2*g^2 + e^2*f^2 + 2*d*e*f*g))/e^3 - (g^2*x^2)/(2*e)

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