3.1779 \(\int \frac{(a+b x)^2}{(c+d x) (e+f x)^{5/2}} \, dx\)

Optimal. Leaf size=140 \[ -\frac{2 (b e-a f) (a d f-2 b c f+b d e)}{f^2 \sqrt{e+f x} (d e-c f)^2}+\frac{2 (b e-a f)^2}{3 f^2 (e+f x)^{3/2} (d e-c f)}-\frac{2 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{\sqrt{d} (d e-c f)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.163914, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {87, 63, 208} \[ -\frac{2 (b e-a f) (a d f-2 b c f+b d e)}{f^2 \sqrt{e+f x} (d e-c f)^2}+\frac{2 (b e-a f)^2}{3 f^2 (e+f x)^{3/2} (d e-c f)}-\frac{2 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{\sqrt{d} (d e-c f)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/((c + d*x)*(e + f*x)^(5/2)),x]

[Out]

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

Rule 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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

Mathematica [C]  time = 0.0854331, size = 103, normalized size = 0.74 \[ \frac{2 b (d e-c f) (2 a d f+b (-c f+2 d e+3 d f x))-2 f^2 (b c-a d)^2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{d (e+f x)}{d e-c f}\right )}{3 d^2 f^2 (e+f x)^{3/2} (c f-d e)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/((c + d*x)*(e + f*x)^(5/2)),x]

[Out]

(2*b*(d*e - c*f)*(2*a*d*f + b*(2*d*e - c*f + 3*d*f*x)) - 2*(b*c - a*d)^2*f^2*Hypergeometric2F1[-3/2, 1, -1/2,
(d*(e + f*x))/(d*e - c*f)])/(3*d^2*f^2*(-(d*e) + c*f)*(e + f*x)^(3/2))

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Maple [B]  time = 0.013, size = 332, normalized size = 2.4 \begin{align*} -{\frac{2\,{a}^{2}}{3\,cf-3\,de} \left ( fx+e \right ) ^{-{\frac{3}{2}}}}+{\frac{4\,aeb}{3\, \left ( cf-de \right ) f} \left ( fx+e \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,{b}^{2}{e}^{2}}{3\,{f}^{2} \left ( cf-de \right ) } \left ( fx+e \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{{a}^{2}d}{ \left ( cf-de \right ) ^{2}\sqrt{fx+e}}}-4\,{\frac{abc}{ \left ( cf-de \right ) ^{2}\sqrt{fx+e}}}+4\,{\frac{ce{b}^{2}}{f \left ( cf-de \right ) ^{2}\sqrt{fx+e}}}-2\,{\frac{{b}^{2}d{e}^{2}}{{f}^{2} \left ( cf-de \right ) ^{2}\sqrt{fx+e}}}+2\,{\frac{{a}^{2}{d}^{2}}{ \left ( cf-de \right ) ^{2}\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }-4\,{\frac{abcd}{ \left ( cf-de \right ) ^{2}\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }+2\,{\frac{{b}^{2}{c}^{2}}{ \left ( cf-de \right ) ^{2}\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)/(f*x+e)^(5/2),x)

[Out]

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

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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)/(f*x+e)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.49613, size = 1956, normalized size = 13.97 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(d*x+c)/(f*x+e)^(5/2),x, algorithm="fricas")

[Out]

[1/3*(3*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^4*x^2 + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e*f^3*x + (b^2*c^2 - 2*a*
b*c*d + a^2*d^2)*e^2*f^2)*sqrt(d^2*e - c*d*f)*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*d^3*e^4 - a^2*c^2*d*f^4 - (7*b^2*c*d^2 - 2*a*b*d^3)*e^3*f + (5*b^2*c^2*d + 2*a*b*c*d^2 -
4*a^2*d^3)*e^2*f^2 - (4*a*b*c^2*d - 5*a^2*c*d^2)*e*f^3 + 3*(b^2*d^3*e^3*f - 3*b^2*c*d^2*e^2*f^2 + (2*b^2*c^2*
d + 2*a*b*c*d^2 - a^2*d^3)*e*f^3 - (2*a*b*c^2*d - a^2*c*d^2)*f^4)*x)*sqrt(f*x + e))/(d^4*e^5*f^2 - 3*c*d^3*e^4
*f^3 + 3*c^2*d^2*e^3*f^4 - c^3*d*e^2*f^5 + (d^4*e^3*f^4 - 3*c*d^3*e^2*f^5 + 3*c^2*d^2*e*f^6 - c^3*d*f^7)*x^2 +
2*(d^4*e^4*f^3 - 3*c*d^3*e^3*f^4 + 3*c^2*d^2*e^2*f^5 - c^3*d*e*f^6)*x), 2/3*(3*((b^2*c^2 - 2*a*b*c*d + a^2*d^
2)*f^4*x^2 + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e*f^3*x + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*e^2*f^2)*sqrt(-d^2*e
+ c*d*f)*arctan(sqrt(-d^2*e + c*d*f)*sqrt(f*x + e)/(d*f*x + d*e)) - (2*b^2*d^3*e^4 - a^2*c^2*d*f^4 - (7*b^2*c*
d^2 - 2*a*b*d^3)*e^3*f + (5*b^2*c^2*d + 2*a*b*c*d^2 - 4*a^2*d^3)*e^2*f^2 - (4*a*b*c^2*d - 5*a^2*c*d^2)*e*f^3 +
3*(b^2*d^3*e^3*f - 3*b^2*c*d^2*e^2*f^2 + (2*b^2*c^2*d + 2*a*b*c*d^2 - a^2*d^3)*e*f^3 - (2*a*b*c^2*d - a^2*c*d
^2)*f^4)*x)*sqrt(f*x + e))/(d^4*e^5*f^2 - 3*c*d^3*e^4*f^3 + 3*c^2*d^2*e^3*f^4 - c^3*d*e^2*f^5 + (d^4*e^3*f^4 -
3*c*d^3*e^2*f^5 + 3*c^2*d^2*e*f^6 - c^3*d*f^7)*x^2 + 2*(d^4*e^4*f^3 - 3*c*d^3*e^3*f^4 + 3*c^2*d^2*e^2*f^5 - c
^3*d*e*f^6)*x)]

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Sympy [A]  time = 56.7451, size = 129, normalized size = 0.92 \begin{align*} \frac{2 \left (a f - b e\right ) \left (a d f - 2 b c f + b d e\right )}{f^{2} \sqrt{e + f x} \left (c f - d e\right )^{2}} - \frac{2 \left (a f - b e\right )^{2}}{3 f^{2} \left (e + f x\right )^{\frac{3}{2}} \left (c f - d e\right )} + \frac{2 \left (a d - b c\right )^{2} \operatorname{atan}{\left (\frac{\sqrt{e + f x}}{\sqrt{\frac{c f - d e}{d}}} \right )}}{d \sqrt{\frac{c f - d e}{d}} \left (c f - d e\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2/(d*x+c)/(f*x+e)**(5/2),x)

[Out]

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

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Giac [A]  time = 2.44413, size = 319, normalized size = 2.28 \begin{align*} \frac{2 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac{\sqrt{f x + e} d}{\sqrt{c d f - d^{2} e}}\right )}{{\left (c^{2} f^{2} - 2 \, c d f e + d^{2} e^{2}\right )} \sqrt{c d f - d^{2} e}} - \frac{2 \,{\left (6 \,{\left (f x + e\right )} a b c f^{2} - 3 \,{\left (f x + e\right )} a^{2} d f^{2} + a^{2} c f^{3} - 6 \,{\left (f x + e\right )} b^{2} c f e - 2 \, a b c f^{2} e - a^{2} d f^{2} e + 3 \,{\left (f x + e\right )} b^{2} d e^{2} + b^{2} c f e^{2} + 2 \, a b d f e^{2} - b^{2} d e^{3}\right )}}{3 \,{\left (c^{2} f^{4} - 2 \, c d f^{3} e + d^{2} f^{2} e^{2}\right )}{\left (f x + e\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2/(d*x+c)/(f*x+e)^(5/2),x, algorithm="giac")

[Out]

2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*arctan(sqrt(f*x + e)*d/sqrt(c*d*f - d^2*e))/((c^2*f^2 - 2*c*d*f*e + d^2*e^2)
*sqrt(c*d*f - d^2*e)) - 2/3*(6*(f*x + e)*a*b*c*f^2 - 3*(f*x + e)*a^2*d*f^2 + a^2*c*f^3 - 6*(f*x + e)*b^2*c*f*e
- 2*a*b*c*f^2*e - a^2*d*f^2*e + 3*(f*x + e)*b^2*d*e^2 + b^2*c*f*e^2 + 2*a*b*d*f*e^2 - b^2*d*e^3)/((c^2*f^4 -
2*c*d*f^3*e + d^2*f^2*e^2)*(f*x + e)^(3/2))