3.2035 \(\int \frac{(a+b x)^2}{(c+d x)^2 \sqrt{e+f x}} \, dx\)

Optimal. Leaf size=132 \[ -\frac{\sqrt{e+f x} (b c-a d)^2}{d^2 (c+d x) (d e-c f)}+\frac{(b c-a d) (-a d f-3 b c f+4 b d e) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{5/2} (d e-c f)^{3/2}}+\frac{2 b^2 \sqrt{e+f x}}{d^2 f} \]

[Out]

(2*b^2*Sqrt[e + f*x])/(d^2*f) - ((b*c - a*d)^2*Sqrt[e + f*x])/(d^2*(d*e - c*f)*(c + d*x)) + ((b*c - a*d)*(4*b*
d*e - 3*b*c*f - a*d*f)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d^(5/2)*(d*e - c*f)^(3/2))

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Rubi [A]  time = 0.143111, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {89, 80, 63, 208} \[ -\frac{\sqrt{e+f x} (b c-a d)^2}{d^2 (c+d x) (d e-c f)}+\frac{(b c-a d) (-a d f-3 b c f+4 b d e) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{5/2} (d e-c f)^{3/2}}+\frac{2 b^2 \sqrt{e+f x}}{d^2 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/((c + d*x)^2*Sqrt[e + f*x]),x]

[Out]

(2*b^2*Sqrt[e + f*x])/(d^2*f) - ((b*c - a*d)^2*Sqrt[e + f*x])/(d^2*(d*e - c*f)*(c + d*x)) + ((b*c - a*d)*(4*b*
d*e - 3*b*c*f - a*d*f)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d^(5/2)*(d*e - c*f)^(3/2))

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{(a+b x)^2}{(c+d x)^2 \sqrt{e+f x}} \, dx &=-\frac{(b c-a d)^2 \sqrt{e+f x}}{d^2 (d e-c f) (c+d x)}+\frac{\int \frac{\frac{1}{2} \left (-a^2 d^2 f-b^2 c (2 d e-c f)+2 a b d (2 d e-c f)\right )+b^2 d (d e-c f) x}{(c+d x) \sqrt{e+f x}} \, dx}{d^2 (d e-c f)}\\ &=\frac{2 b^2 \sqrt{e+f x}}{d^2 f}-\frac{(b c-a d)^2 \sqrt{e+f x}}{d^2 (d e-c f) (c+d x)}-\frac{((b c-a d) (4 b d e-3 b c f-a d f)) \int \frac{1}{(c+d x) \sqrt{e+f x}} \, dx}{2 d^2 (d e-c f)}\\ &=\frac{2 b^2 \sqrt{e+f x}}{d^2 f}-\frac{(b c-a d)^2 \sqrt{e+f x}}{d^2 (d e-c f) (c+d x)}-\frac{((b c-a d) (4 b d e-3 b c f-a d f)) \operatorname{Subst}\left (\int \frac{1}{c-\frac{d e}{f}+\frac{d x^2}{f}} \, dx,x,\sqrt{e+f x}\right )}{d^2 f (d e-c f)}\\ &=\frac{2 b^2 \sqrt{e+f x}}{d^2 f}-\frac{(b c-a d)^2 \sqrt{e+f x}}{d^2 (d e-c f) (c+d x)}+\frac{(b c-a d) (4 b d e-3 b c f-a d f) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{5/2} (d e-c f)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.181852, size = 132, normalized size = 1. \[ -\frac{\sqrt{e+f x} (b c-a d)^2}{d^2 (c+d x) (d e-c f)}-\frac{(b c-a d) (a d f+3 b c f-4 b d e) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{5/2} (d e-c f)^{3/2}}+\frac{2 b^2 \sqrt{e+f x}}{d^2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/((c + d*x)^2*Sqrt[e + f*x]),x]

[Out]

(2*b^2*Sqrt[e + f*x])/(d^2*f) - ((b*c - a*d)^2*Sqrt[e + f*x])/(d^2*(d*e - c*f)*(c + d*x)) - ((b*c - a*d)*(-4*b
*d*e + 3*b*c*f + a*d*f)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d^(5/2)*(d*e - c*f)^(3/2))

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Maple [B]  time = 0.015, size = 387, normalized size = 2.9 \begin{align*} 2\,{\frac{{b}^{2}\sqrt{fx+e}}{{d}^{2}f}}+{\frac{{a}^{2}f}{ \left ( cf-de \right ) \left ( dfx+cf \right ) }\sqrt{fx+e}}-2\,{\frac{f\sqrt{fx+e}abc}{ \left ( cf-de \right ) d \left ( dfx+cf \right ) }}+{\frac{{b}^{2}{c}^{2}f}{{d}^{2} \left ( cf-de \right ) \left ( dfx+cf \right ) }\sqrt{fx+e}}+{\frac{{a}^{2}f}{cf-de}\arctan \left ({d\sqrt{fx+e}{\frac{1}{\sqrt{ \left ( cf-de \right ) d}}}} \right ){\frac{1}{\sqrt{ \left ( cf-de \right ) d}}}}+2\,{\frac{abfc}{ \left ( cf-de \right ) d\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }-4\,{\frac{aeb}{ \left ( cf-de \right ) \sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }-3\,{\frac{{b}^{2}{c}^{2}f}{{d}^{2} \left ( cf-de \right ) \sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }+4\,{\frac{ce{b}^{2}}{ \left ( cf-de \right ) d\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2/(d*x+c)^2/(f*x+e)^(1/2),x)

[Out]

2*b^2*(f*x+e)^(1/2)/d^2/f+f/(c*f-d*e)*(f*x+e)^(1/2)/(d*f*x+c*f)*a^2-2*f/d/(c*f-d*e)*(f*x+e)^(1/2)/(d*f*x+c*f)*
a*b*c+f/d^2/(c*f-d*e)*(f*x+e)^(1/2)/(d*f*x+c*f)*b^2*c^2+f/(c*f-d*e)/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d
/((c*f-d*e)*d)^(1/2))*a^2+2*f/d/(c*f-d*e)/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a*b*
c-4/(c*f-d*e)/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a*b*e-3*f/d^2/(c*f-d*e)/((c*f-d*
e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*b^2*c^2+4/d/(c*f-d*e)/((c*f-d*e)*d)^(1/2)*arctan((f*x+
e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*b^2*c*e

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(d*x+c)^2/(f*x+e)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.74144, size = 1419, normalized size = 10.75 \begin{align*} \left [-\frac{\sqrt{d^{2} e - c d f}{\left (4 \,{\left (b^{2} c^{2} d - a b c d^{2}\right )} e f -{\left (3 \, b^{2} c^{3} - 2 \, a b c^{2} d - a^{2} c d^{2}\right )} f^{2} +{\left (4 \,{\left (b^{2} c d^{2} - a b d^{3}\right )} e f -{\left (3 \, b^{2} c^{2} d - 2 \, a b c d^{2} - a^{2} d^{3}\right )} f^{2}\right )} x\right )} \log \left (\frac{d f x + 2 \, d e - c f - 2 \, \sqrt{d^{2} e - c d f} \sqrt{f x + e}}{d x + c}\right ) - 2 \,{\left (2 \, b^{2} c d^{3} e^{2} -{\left (5 \, b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} e f +{\left (3 \, b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} f^{2} + 2 \,{\left (b^{2} d^{4} e^{2} - 2 \, b^{2} c d^{3} e f + b^{2} c^{2} d^{2} f^{2}\right )} x\right )} \sqrt{f x + e}}{2 \,{\left (c d^{5} e^{2} f - 2 \, c^{2} d^{4} e f^{2} + c^{3} d^{3} f^{3} +{\left (d^{6} e^{2} f - 2 \, c d^{5} e f^{2} + c^{2} d^{4} f^{3}\right )} x\right )}}, -\frac{\sqrt{-d^{2} e + c d f}{\left (4 \,{\left (b^{2} c^{2} d - a b c d^{2}\right )} e f -{\left (3 \, b^{2} c^{3} - 2 \, a b c^{2} d - a^{2} c d^{2}\right )} f^{2} +{\left (4 \,{\left (b^{2} c d^{2} - a b d^{3}\right )} e f -{\left (3 \, b^{2} c^{2} d - 2 \, a b c d^{2} - a^{2} d^{3}\right )} f^{2}\right )} x\right )} \arctan \left (\frac{\sqrt{-d^{2} e + c d f} \sqrt{f x + e}}{d f x + d e}\right ) -{\left (2 \, b^{2} c d^{3} e^{2} -{\left (5 \, b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} e f +{\left (3 \, b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} f^{2} + 2 \,{\left (b^{2} d^{4} e^{2} - 2 \, b^{2} c d^{3} e f + b^{2} c^{2} d^{2} f^{2}\right )} x\right )} \sqrt{f x + e}}{c d^{5} e^{2} f - 2 \, c^{2} d^{4} e f^{2} + c^{3} d^{3} f^{3} +{\left (d^{6} e^{2} f - 2 \, c d^{5} e f^{2} + c^{2} d^{4} f^{3}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(d*x+c)^2/(f*x+e)^(1/2),x, algorithm="fricas")

[Out]

[-1/2*(sqrt(d^2*e - c*d*f)*(4*(b^2*c^2*d - a*b*c*d^2)*e*f - (3*b^2*c^3 - 2*a*b*c^2*d - a^2*c*d^2)*f^2 + (4*(b^
2*c*d^2 - a*b*d^3)*e*f - (3*b^2*c^2*d - 2*a*b*c*d^2 - a^2*d^3)*f^2)*x)*log((d*f*x + 2*d*e - c*f - 2*sqrt(d^2*e
- c*d*f)*sqrt(f*x + e))/(d*x + c)) - 2*(2*b^2*c*d^3*e^2 - (5*b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*e*f + (3*b^
2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*f^2 + 2*(b^2*d^4*e^2 - 2*b^2*c*d^3*e*f + b^2*c^2*d^2*f^2)*x)*sqrt(f*x + e
))/(c*d^5*e^2*f - 2*c^2*d^4*e*f^2 + c^3*d^3*f^3 + (d^6*e^2*f - 2*c*d^5*e*f^2 + c^2*d^4*f^3)*x), -(sqrt(-d^2*e
+ c*d*f)*(4*(b^2*c^2*d - a*b*c*d^2)*e*f - (3*b^2*c^3 - 2*a*b*c^2*d - a^2*c*d^2)*f^2 + (4*(b^2*c*d^2 - a*b*d^3)
*e*f - (3*b^2*c^2*d - 2*a*b*c*d^2 - a^2*d^3)*f^2)*x)*arctan(sqrt(-d^2*e + c*d*f)*sqrt(f*x + e)/(d*f*x + d*e))
- (2*b^2*c*d^3*e^2 - (5*b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*e*f + (3*b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*c*d^3)*f
^2 + 2*(b^2*d^4*e^2 - 2*b^2*c*d^3*e*f + b^2*c^2*d^2*f^2)*x)*sqrt(f*x + e))/(c*d^5*e^2*f - 2*c^2*d^4*e*f^2 + c^
3*d^3*f^3 + (d^6*e^2*f - 2*c*d^5*e*f^2 + c^2*d^4*f^3)*x)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2/(d*x+c)**2/(f*x+e)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.7552, size = 277, normalized size = 2.1 \begin{align*} -\frac{{\left (3 \, b^{2} c^{2} f - 2 \, a b c d f - a^{2} d^{2} f - 4 \, b^{2} c d e + 4 \, a b d^{2} e\right )} \arctan \left (\frac{\sqrt{f x + e} d}{\sqrt{c d f - d^{2} e}}\right )}{{\left (c d^{2} f - d^{3} e\right )} \sqrt{c d f - d^{2} e}} + \frac{2 \, \sqrt{f x + e} b^{2}}{d^{2} f} + \frac{\sqrt{f x + e} b^{2} c^{2} f - 2 \, \sqrt{f x + e} a b c d f + \sqrt{f x + e} a^{2} d^{2} f}{{\left (c d^{2} f - d^{3} e\right )}{\left ({\left (f x + e\right )} d + c f - d e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(d*x+c)^2/(f*x+e)^(1/2),x, algorithm="giac")

[Out]

-(3*b^2*c^2*f - 2*a*b*c*d*f - a^2*d^2*f - 4*b^2*c*d*e + 4*a*b*d^2*e)*arctan(sqrt(f*x + e)*d/sqrt(c*d*f - d^2*e
))/((c*d^2*f - d^3*e)*sqrt(c*d*f - d^2*e)) + 2*sqrt(f*x + e)*b^2/(d^2*f) + (sqrt(f*x + e)*b^2*c^2*f - 2*sqrt(f
*x + e)*a*b*c*d*f + sqrt(f*x + e)*a^2*d^2*f)/((c*d^2*f - d^3*e)*((f*x + e)*d + c*f - d*e))