3.1781 \(\int \frac{(a+b x)^2}{(c+d x) (e+f x)^{9/2}} \, dx\)

Optimal. Leaf size=207 \[ -\frac{2 d^{3/2} (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{(d e-c f)^{9/2}}-\frac{2 (b e-a f) (a d f-2 b c f+b d e)}{5 f^2 (e+f x)^{5/2} (d e-c f)^2}+\frac{2 (b e-a f)^2}{7 f^2 (e+f x)^{7/2} (d e-c f)}+\frac{2 d (b c-a d)^2}{\sqrt{e+f x} (d e-c f)^4}+\frac{2 (b c-a d)^2}{3 (e+f x)^{3/2} (d e-c f)^3} \]

[Out]

(2*(b*e - a*f)^2)/(7*f^2*(d*e - c*f)*(e + f*x)^(7/2)) - (2*(b*e - a*f)*(b*d*e - 2*b*c*f + a*d*f))/(5*f^2*(d*e
- c*f)^2*(e + f*x)^(5/2)) + (2*(b*c - a*d)^2)/(3*(d*e - c*f)^3*(e + f*x)^(3/2)) + (2*d*(b*c - a*d)^2)/((d*e -
c*f)^4*Sqrt[e + f*x]) - (2*d^(3/2)*(b*c - a*d)^2*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d*e - c*f)
^(9/2)

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Rubi [A]  time = 0.277009, antiderivative size = 207, 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} \[ -\frac{2 d^{3/2} (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{(d e-c f)^{9/2}}-\frac{2 (b e-a f) (a d f-2 b c f+b d e)}{5 f^2 (e+f x)^{5/2} (d e-c f)^2}+\frac{2 (b e-a f)^2}{7 f^2 (e+f x)^{7/2} (d e-c f)}+\frac{2 d (b c-a d)^2}{\sqrt{e+f x} (d e-c f)^4}+\frac{2 (b c-a d)^2}{3 (e+f x)^{3/2} (d e-c f)^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^2/((c + d*x)*(e + f*x)^(9/2)),x]

[Out]

(2*(b*e - a*f)^2)/(7*f^2*(d*e - c*f)*(e + f*x)^(7/2)) - (2*(b*e - a*f)*(b*d*e - 2*b*c*f + a*d*f))/(5*f^2*(d*e
- c*f)^2*(e + f*x)^(5/2)) + (2*(b*c - a*d)^2)/(3*(d*e - c*f)^3*(e + f*x)^(3/2)) + (2*d*(b*c - a*d)^2)/((d*e -
c*f)^4*Sqrt[e + f*x]) - (2*d^(3/2)*(b*c - a*d)^2*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/(d*e - c*f)
^(9/2)

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)^{9/2}} \, dx &=\int \left (\frac{(-b e+a f)^2}{f (-d e+c f) (e+f x)^{9/2}}+\frac{(-b e+a f) (-b d e+2 b c f-a d f)}{f (-d e+c f)^2 (e+f x)^{7/2}}+\frac{(b c-a d)^2 f}{(-d e+c f)^3 (e+f x)^{5/2}}-\frac{d (-b c+a d)^2 f}{(-d e+c f)^4 (e+f x)^{3/2}}+\frac{d^2 (-b c+a d)^2}{(d e-c f)^4 (c+d x) \sqrt{e+f x}}\right ) \, dx\\ &=\frac{2 (b e-a f)^2}{7 f^2 (d e-c f) (e+f x)^{7/2}}-\frac{2 (b e-a f) (b d e-2 b c f+a d f)}{5 f^2 (d e-c f)^2 (e+f x)^{5/2}}+\frac{2 (b c-a d)^2}{3 (d e-c f)^3 (e+f x)^{3/2}}+\frac{2 d (b c-a d)^2}{(d e-c f)^4 \sqrt{e+f x}}+\frac{\left (d^2 (b c-a d)^2\right ) \int \frac{1}{(c+d x) \sqrt{e+f x}} \, dx}{(d e-c f)^4}\\ &=\frac{2 (b e-a f)^2}{7 f^2 (d e-c f) (e+f x)^{7/2}}-\frac{2 (b e-a f) (b d e-2 b c f+a d f)}{5 f^2 (d e-c f)^2 (e+f x)^{5/2}}+\frac{2 (b c-a d)^2}{3 (d e-c f)^3 (e+f x)^{3/2}}+\frac{2 d (b c-a d)^2}{(d e-c f)^4 \sqrt{e+f x}}+\frac{\left (2 d^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)^4}\\ &=\frac{2 (b e-a f)^2}{7 f^2 (d e-c f) (e+f x)^{7/2}}-\frac{2 (b e-a f) (b d e-2 b c f+a d f)}{5 f^2 (d e-c f)^2 (e+f x)^{5/2}}+\frac{2 (b c-a d)^2}{3 (d e-c f)^3 (e+f x)^{3/2}}+\frac{2 d (b c-a d)^2}{(d e-c f)^4 \sqrt{e+f x}}-\frac{2 d^{3/2} (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{(d e-c f)^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0886527, size = 102, normalized size = 0.49 \[ \frac{2 \left (b (d e-c f) (10 a d f+b (-5 c f+2 d e+7 d f x))-5 f^2 (b c-a d)^2 \, _2F_1\left (-\frac{7}{2},1;-\frac{5}{2};\frac{d (e+f x)}{d e-c f}\right )\right )}{35 d^2 f^2 (e+f x)^{7/2} (c f-d e)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^2/((c + d*x)*(e + f*x)^(9/2)),x]

[Out]

(2*(b*(d*e - c*f)*(10*a*d*f + b*(2*d*e - 5*c*f + 7*d*f*x)) - 5*(b*c - a*d)^2*f^2*Hypergeometric2F1[-7/2, 1, -5
/2, (d*(e + f*x))/(d*e - c*f)]))/(35*d^2*f^2*(-(d*e) + c*f)*(e + f*x)^(7/2))

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Maple [B]  time = 0.019, size = 486, normalized size = 2.4 \begin{align*} -{\frac{2\,{a}^{2}}{7\,cf-7\,de} \left ( fx+e \right ) ^{-{\frac{7}{2}}}}+{\frac{4\,aeb}{7\, \left ( cf-de \right ) f} \left ( fx+e \right ) ^{-{\frac{7}{2}}}}-{\frac{2\,{b}^{2}{e}^{2}}{7\,{f}^{2} \left ( cf-de \right ) } \left ( fx+e \right ) ^{-{\frac{7}{2}}}}+{\frac{2\,{a}^{2}d}{5\, \left ( cf-de \right ) ^{2}} \left ( fx+e \right ) ^{-{\frac{5}{2}}}}-{\frac{4\,abc}{5\, \left ( cf-de \right ) ^{2}} \left ( fx+e \right ) ^{-{\frac{5}{2}}}}+{\frac{4\,ce{b}^{2}}{5\,f \left ( cf-de \right ) ^{2}} \left ( fx+e \right ) ^{-{\frac{5}{2}}}}-{\frac{2\,{b}^{2}d{e}^{2}}{5\,{f}^{2} \left ( cf-de \right ) ^{2}} \left ( fx+e \right ) ^{-{\frac{5}{2}}}}-{\frac{2\,{a}^{2}{d}^{2}}{3\, \left ( cf-de \right ) ^{3}} \left ( fx+e \right ) ^{-{\frac{3}{2}}}}+{\frac{4\,abcd}{3\, \left ( cf-de \right ) ^{3}} \left ( fx+e \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,{b}^{2}{c}^{2}}{3\, \left ( cf-de \right ) ^{3}} \left ( fx+e \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{{d}^{3}{a}^{2}}{ \left ( cf-de \right ) ^{4}\sqrt{fx+e}}}-4\,{\frac{a{d}^{2}bc}{ \left ( cf-de \right ) ^{4}\sqrt{fx+e}}}+2\,{\frac{{b}^{2}d{c}^{2}}{ \left ( cf-de \right ) ^{4}\sqrt{fx+e}}}+2\,{\frac{{d}^{4}{a}^{2}}{ \left ( cf-de \right ) ^{4}\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }-4\,{\frac{{d}^{3}abc}{ \left ( cf-de \right ) ^{4}\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }+2\,{\frac{{b}^{2}{d}^{2}{c}^{2}}{ \left ( cf-de \right ) ^{4}\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)^(9/2),x)

[Out]

-2/7/(c*f-d*e)/(f*x+e)^(7/2)*a^2+4/7/f/(c*f-d*e)/(f*x+e)^(7/2)*a*b*e-2/7/f^2/(c*f-d*e)/(f*x+e)^(7/2)*b^2*e^2+2
/5/(c*f-d*e)^2/(f*x+e)^(5/2)*a^2*d-4/5/(c*f-d*e)^2/(f*x+e)^(5/2)*a*b*c+4/5/f/(c*f-d*e)^2/(f*x+e)^(5/2)*b^2*c*e
-2/5/f^2/(c*f-d*e)^2/(f*x+e)^(5/2)*b^2*d*e^2-2/3/(c*f-d*e)^3/(f*x+e)^(3/2)*a^2*d^2+4/3/(c*f-d*e)^3/(f*x+e)^(3/
2)*a*b*c*d-2/3/(c*f-d*e)^3/(f*x+e)^(3/2)*b^2*c^2+2/(c*f-d*e)^4*d^3/(f*x+e)^(1/2)*a^2-4/(c*f-d*e)^4*d^2/(f*x+e)
^(1/2)*a*b*c+2/(c*f-d*e)^4*d/(f*x+e)^(1/2)*b^2*c^2+2*d^4/(c*f-d*e)^4/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*
d/((c*f-d*e)*d)^(1/2))*a^2-4*d^3/(c*f-d*e)^4/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a
*b*c+2*d^2/(c*f-d*e)^4/((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)^(9/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.58588, size = 3681, normalized size = 17.78 \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)^(9/2),x, algorithm="fricas")

[Out]

[1/105*(105*((b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*f^6*x^4 + 4*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e*f^5*x^3 + 6
*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e^2*f^4*x^2 + 4*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e^3*f^3*x + (b^2*c^2*
d - 2*a*b*c*d^2 + a^2*d^3)*e^4*f^2)*sqrt(d/(d*e - c*f))*log((d*f*x + 2*d*e - c*f - 2*(d*e - c*f)*sqrt(f*x + e)
*sqrt(d/(d*e - c*f)))/(d*x + c)) - 2*(6*b^2*d^3*e^5 + 15*a^2*c^3*f^5 - 105*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)
*f^5*x^3 - 3*(13*b^2*c*d^2 - 10*a*b*d^3)*e^4*f - 8*(10*b^2*c^2*d - 29*a*b*c*d^2 + 22*a^2*d^3)*e^3*f^2 + 2*(4*b
^2*c^3 - 32*a*b*c^2*d + 61*a^2*c*d^2)*e^2*f^3 + 6*(2*a*b*c^3 - 11*a^2*c^2*d)*e*f^4 - 35*(10*(b^2*c^2*d - 2*a*b
*c*d^2 + a^2*d^3)*e*f^4 - (b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*f^5)*x^2 + 7*(3*b^2*d^3*e^4*f - 12*b^2*c*d^2*e^3
*f^2 - 2*(20*b^2*c^2*d - 58*a*b*c*d^2 + 29*a^2*d^3)*e^2*f^3 + 4*(b^2*c^3 - 8*a*b*c^2*d + 4*a^2*c*d^2)*e*f^4 +
3*(2*a*b*c^3 - a^2*c^2*d)*f^5)*x)*sqrt(f*x + e))/(d^4*e^8*f^2 - 4*c*d^3*e^7*f^3 + 6*c^2*d^2*e^6*f^4 - 4*c^3*d*
e^5*f^5 + c^4*e^4*f^6 + (d^4*e^4*f^6 - 4*c*d^3*e^3*f^7 + 6*c^2*d^2*e^2*f^8 - 4*c^3*d*e*f^9 + c^4*f^10)*x^4 + 4
*(d^4*e^5*f^5 - 4*c*d^3*e^4*f^6 + 6*c^2*d^2*e^3*f^7 - 4*c^3*d*e^2*f^8 + c^4*e*f^9)*x^3 + 6*(d^4*e^6*f^4 - 4*c*
d^3*e^5*f^5 + 6*c^2*d^2*e^4*f^6 - 4*c^3*d*e^3*f^7 + c^4*e^2*f^8)*x^2 + 4*(d^4*e^7*f^3 - 4*c*d^3*e^6*f^4 + 6*c^
2*d^2*e^5*f^5 - 4*c^3*d*e^4*f^6 + c^4*e^3*f^7)*x), -2/105*(105*((b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*f^6*x^4 +
4*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e*f^5*x^3 + 6*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e^2*f^4*x^2 + 4*(b^2*c
^2*d - 2*a*b*c*d^2 + a^2*d^3)*e^3*f^3*x + (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e^4*f^2)*sqrt(-d/(d*e - c*f))*ar
ctan(-(d*e - c*f)*sqrt(f*x + e)*sqrt(-d/(d*e - c*f))/(d*f*x + d*e)) + (6*b^2*d^3*e^5 + 15*a^2*c^3*f^5 - 105*(b
^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*f^5*x^3 - 3*(13*b^2*c*d^2 - 10*a*b*d^3)*e^4*f - 8*(10*b^2*c^2*d - 29*a*b*c*d
^2 + 22*a^2*d^3)*e^3*f^2 + 2*(4*b^2*c^3 - 32*a*b*c^2*d + 61*a^2*c*d^2)*e^2*f^3 + 6*(2*a*b*c^3 - 11*a^2*c^2*d)*
e*f^4 - 35*(10*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*e*f^4 - (b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*f^5)*x^2 + 7*(3
*b^2*d^3*e^4*f - 12*b^2*c*d^2*e^3*f^2 - 2*(20*b^2*c^2*d - 58*a*b*c*d^2 + 29*a^2*d^3)*e^2*f^3 + 4*(b^2*c^3 - 8*
a*b*c^2*d + 4*a^2*c*d^2)*e*f^4 + 3*(2*a*b*c^3 - a^2*c^2*d)*f^5)*x)*sqrt(f*x + e))/(d^4*e^8*f^2 - 4*c*d^3*e^7*f
^3 + 6*c^2*d^2*e^6*f^4 - 4*c^3*d*e^5*f^5 + c^4*e^4*f^6 + (d^4*e^4*f^6 - 4*c*d^3*e^3*f^7 + 6*c^2*d^2*e^2*f^8 -
4*c^3*d*e*f^9 + c^4*f^10)*x^4 + 4*(d^4*e^5*f^5 - 4*c*d^3*e^4*f^6 + 6*c^2*d^2*e^3*f^7 - 4*c^3*d*e^2*f^8 + c^4*e
*f^9)*x^3 + 6*(d^4*e^6*f^4 - 4*c*d^3*e^5*f^5 + 6*c^2*d^2*e^4*f^6 - 4*c^3*d*e^3*f^7 + c^4*e^2*f^8)*x^2 + 4*(d^4
*e^7*f^3 - 4*c*d^3*e^6*f^4 + 6*c^2*d^2*e^5*f^5 - 4*c^3*d*e^4*f^6 + c^4*e^3*f^7)*x)]

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Sympy [A]  time = 175.868, size = 189, normalized size = 0.91 \begin{align*} \frac{2 d \left (a d - b c\right )^{2}}{\sqrt{e + f x} \left (c f - d e\right )^{4}} + \frac{2 d \left (a d - b c\right )^{2} \operatorname{atan}{\left (\frac{\sqrt{e + f x}}{\sqrt{\frac{c f - d e}{d}}} \right )}}{\sqrt{\frac{c f - d e}{d}} \left (c f - d e\right )^{4}} - \frac{2 \left (a d - b c\right )^{2}}{3 \left (e + f x\right )^{\frac{3}{2}} \left (c f - d e\right )^{3}} + \frac{2 \left (a f - b e\right ) \left (a d f - 2 b c f + b d e\right )}{5 f^{2} \left (e + f x\right )^{\frac{5}{2}} \left (c f - d e\right )^{2}} - \frac{2 \left (a f - b e\right )^{2}}{7 f^{2} \left (e + f x\right )^{\frac{7}{2}} \left (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)/(f*x+e)**(9/2),x)

[Out]

2*d*(a*d - b*c)**2/(sqrt(e + f*x)*(c*f - d*e)**4) + 2*d*(a*d - b*c)**2*atan(sqrt(e + f*x)/sqrt((c*f - d*e)/d))
/(sqrt((c*f - d*e)/d)*(c*f - d*e)**4) - 2*(a*d - b*c)**2/(3*(e + f*x)**(3/2)*(c*f - d*e)**3) + 2*(a*f - b*e)*(
a*d*f - 2*b*c*f + b*d*e)/(5*f**2*(e + f*x)**(5/2)*(c*f - d*e)**2) - 2*(a*f - b*e)**2/(7*f**2*(e + f*x)**(7/2)*
(c*f - d*e))

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Giac [B]  time = 1.81408, size = 938, normalized size = 4.53 \begin{align*} \frac{2 \,{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} \arctan \left (\frac{\sqrt{f x + e} d}{\sqrt{c d f - d^{2} e}}\right )}{{\left (c^{4} f^{4} - 4 \, c^{3} d f^{3} e + 6 \, c^{2} d^{2} f^{2} e^{2} - 4 \, c d^{3} f e^{3} + d^{4} e^{4}\right )} \sqrt{c d f - d^{2} e}} + \frac{2 \,{\left (105 \,{\left (f x + e\right )}^{3} b^{2} c^{2} d f^{2} - 210 \,{\left (f x + e\right )}^{3} a b c d^{2} f^{2} + 105 \,{\left (f x + e\right )}^{3} a^{2} d^{3} f^{2} - 35 \,{\left (f x + e\right )}^{2} b^{2} c^{3} f^{3} + 70 \,{\left (f x + e\right )}^{2} a b c^{2} d f^{3} - 35 \,{\left (f x + e\right )}^{2} a^{2} c d^{2} f^{3} - 42 \,{\left (f x + e\right )} a b c^{3} f^{4} + 21 \,{\left (f x + e\right )} a^{2} c^{2} d f^{4} - 15 \, a^{2} c^{3} f^{5} + 35 \,{\left (f x + e\right )}^{2} b^{2} c^{2} d f^{2} e - 70 \,{\left (f x + e\right )}^{2} a b c d^{2} f^{2} e + 35 \,{\left (f x + e\right )}^{2} a^{2} d^{3} f^{2} e + 42 \,{\left (f x + e\right )} b^{2} c^{3} f^{3} e + 84 \,{\left (f x + e\right )} a b c^{2} d f^{3} e - 42 \,{\left (f x + e\right )} a^{2} c d^{2} f^{3} e + 30 \, a b c^{3} f^{4} e + 45 \, a^{2} c^{2} d f^{4} e - 105 \,{\left (f x + e\right )} b^{2} c^{2} d f^{2} e^{2} - 42 \,{\left (f x + e\right )} a b c d^{2} f^{2} e^{2} + 21 \,{\left (f x + e\right )} a^{2} d^{3} f^{2} e^{2} - 15 \, b^{2} c^{3} f^{3} e^{2} - 90 \, a b c^{2} d f^{3} e^{2} - 45 \, a^{2} c d^{2} f^{3} e^{2} + 84 \,{\left (f x + e\right )} b^{2} c d^{2} f e^{3} + 45 \, b^{2} c^{2} d f^{2} e^{3} + 90 \, a b c d^{2} f^{2} e^{3} + 15 \, a^{2} d^{3} f^{2} e^{3} - 21 \,{\left (f x + e\right )} b^{2} d^{3} e^{4} - 45 \, b^{2} c d^{2} f e^{4} - 30 \, a b d^{3} f e^{4} + 15 \, b^{2} d^{3} e^{5}\right )}}{105 \,{\left (c^{4} f^{6} - 4 \, c^{3} d f^{5} e + 6 \, c^{2} d^{2} f^{4} e^{2} - 4 \, c d^{3} f^{3} e^{3} + d^{4} f^{2} e^{4}\right )}{\left (f x + e\right )}^{\frac{7}{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)^(9/2),x, algorithm="giac")

[Out]

2*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*arctan(sqrt(f*x + e)*d/sqrt(c*d*f - d^2*e))/((c^4*f^4 - 4*c^3*d*f^3*e
+ 6*c^2*d^2*f^2*e^2 - 4*c*d^3*f*e^3 + d^4*e^4)*sqrt(c*d*f - d^2*e)) + 2/105*(105*(f*x + e)^3*b^2*c^2*d*f^2 - 2
10*(f*x + e)^3*a*b*c*d^2*f^2 + 105*(f*x + e)^3*a^2*d^3*f^2 - 35*(f*x + e)^2*b^2*c^3*f^3 + 70*(f*x + e)^2*a*b*c
^2*d*f^3 - 35*(f*x + e)^2*a^2*c*d^2*f^3 - 42*(f*x + e)*a*b*c^3*f^4 + 21*(f*x + e)*a^2*c^2*d*f^4 - 15*a^2*c^3*f
^5 + 35*(f*x + e)^2*b^2*c^2*d*f^2*e - 70*(f*x + e)^2*a*b*c*d^2*f^2*e + 35*(f*x + e)^2*a^2*d^3*f^2*e + 42*(f*x
+ e)*b^2*c^3*f^3*e + 84*(f*x + e)*a*b*c^2*d*f^3*e - 42*(f*x + e)*a^2*c*d^2*f^3*e + 30*a*b*c^3*f^4*e + 45*a^2*c
^2*d*f^4*e - 105*(f*x + e)*b^2*c^2*d*f^2*e^2 - 42*(f*x + e)*a*b*c*d^2*f^2*e^2 + 21*(f*x + e)*a^2*d^3*f^2*e^2 -
15*b^2*c^3*f^3*e^2 - 90*a*b*c^2*d*f^3*e^2 - 45*a^2*c*d^2*f^3*e^2 + 84*(f*x + e)*b^2*c*d^2*f*e^3 + 45*b^2*c^2*
d*f^2*e^3 + 90*a*b*c*d^2*f^2*e^3 + 15*a^2*d^3*f^2*e^3 - 21*(f*x + e)*b^2*d^3*e^4 - 45*b^2*c*d^2*f*e^4 - 30*a*b
*d^3*f*e^4 + 15*b^2*d^3*e^5)/((c^4*f^6 - 4*c^3*d*f^5*e + 6*c^2*d^2*f^4*e^2 - 4*c*d^3*f^3*e^3 + d^4*f^2*e^4)*(f
*x + e)^(7/2))