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]

2/7*(-a*f+b*e)^2/f^2/(-c*f+d*e)/(f*x+e)^(7/2)-2/5*(-a*f+b*e)*(a*d*f-2*b*c*f+b*d*e)/f^2/(-c*f+d*e)^2/(f*x+e)^(5
/2)+2/3*(-a*d+b*c)^2/(-c*f+d*e)^3/(f*x+e)^(3/2)-2*d^(3/2)*(-a*d+b*c)^2*arctanh(d^(1/2)*(f*x+e)^(1/2)/(-c*f+d*e
)^(1/2))/(-c*f+d*e)^(9/2)+2*d*(-a*d+b*c)^2/(-c*f+d*e)^4/(f*x+e)^(1/2)

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Rubi [A]  time = 0.28, antiderivative size = 207, normalized size of antiderivative = 1.00, 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 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 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 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*}

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Mathematica [C]  time = 0.13, 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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fricas [B]  time = 0.79, size = 1799, normalized size = 8.69 \[ \text {result too large to display} \]

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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giac [B]  time = 1.37, size = 695, normalized size = 3.36 \[ \frac {2 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {c d f - d^{2} e}}\right )}{{\left (c^{4} f^{4} - 4 \, c^{3} d f^{3} e + 6 \, c^{2} d^{2} f^{2} e^{2} - 4 \, c d^{3} f e^{3} + d^{4} e^{4}\right )} \sqrt {c d f - d^{2} e}} + \frac {2 \, {\left (105 \, {\left (f x + e\right )}^{3} b^{2} c^{2} d f^{2} - 210 \, {\left (f x + e\right )}^{3} a b c d^{2} f^{2} + 105 \, {\left (f x + e\right )}^{3} a^{2} d^{3} f^{2} - 35 \, {\left (f x + e\right )}^{2} b^{2} c^{3} f^{3} + 70 \, {\left (f x + e\right )}^{2} a b c^{2} d f^{3} - 35 \, {\left (f x + e\right )}^{2} a^{2} c d^{2} f^{3} - 42 \, {\left (f x + e\right )} a b c^{3} f^{4} + 21 \, {\left (f x + e\right )} a^{2} c^{2} d f^{4} - 15 \, a^{2} c^{3} f^{5} + 35 \, {\left (f x + e\right )}^{2} b^{2} c^{2} d f^{2} e - 70 \, {\left (f x + e\right )}^{2} a b c d^{2} f^{2} e + 35 \, {\left (f x + e\right )}^{2} a^{2} d^{3} f^{2} e + 42 \, {\left (f x + e\right )} b^{2} c^{3} f^{3} e + 84 \, {\left (f x + e\right )} a b c^{2} d f^{3} e - 42 \, {\left (f x + e\right )} a^{2} c d^{2} f^{3} e + 30 \, a b c^{3} f^{4} e + 45 \, a^{2} c^{2} d f^{4} e - 105 \, {\left (f x + e\right )} b^{2} c^{2} d f^{2} e^{2} - 42 \, {\left (f x + e\right )} a b c d^{2} f^{2} e^{2} + 21 \, {\left (f x + e\right )} a^{2} d^{3} f^{2} e^{2} - 15 \, b^{2} c^{3} f^{3} e^{2} - 90 \, a b c^{2} d f^{3} e^{2} - 45 \, a^{2} c d^{2} f^{3} e^{2} + 84 \, {\left (f x + e\right )} b^{2} c d^{2} f e^{3} + 45 \, b^{2} c^{2} d f^{2} e^{3} + 90 \, a b c d^{2} f^{2} e^{3} + 15 \, a^{2} d^{3} f^{2} e^{3} - 21 \, {\left (f x + e\right )} b^{2} d^{3} e^{4} - 45 \, b^{2} c d^{2} f e^{4} - 30 \, a b d^{3} f e^{4} + 15 \, b^{2} d^{3} e^{5}\right )}}{105 \, {\left (c^{4} f^{6} - 4 \, c^{3} d f^{5} e + 6 \, c^{2} d^{2} f^{4} e^{2} - 4 \, c d^{3} f^{3} e^{3} + d^{4} f^{2} e^{4}\right )} {\left (f x + e\right )}^{\frac {7}{2}}} \]

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))

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maple [B]  time = 0.02, size = 486, normalized size = 2.35 \[ \frac {2 a^{2} d^{4} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{4} \sqrt {\left (c f -d e \right ) d}}-\frac {4 a b c \,d^{3} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{4} \sqrt {\left (c f -d e \right ) d}}+\frac {2 b^{2} c^{2} d^{2} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{4} \sqrt {\left (c f -d e \right ) d}}+\frac {2 a^{2} d^{3}}{\left (c f -d e \right )^{4} \sqrt {f x +e}}-\frac {4 a b c \,d^{2}}{\left (c f -d e \right )^{4} \sqrt {f x +e}}+\frac {2 b^{2} c^{2} d}{\left (c f -d e \right )^{4} \sqrt {f x +e}}-\frac {2 a^{2} d^{2}}{3 \left (c f -d e \right )^{3} \left (f x +e \right )^{\frac {3}{2}}}+\frac {4 a b c d}{3 \left (c f -d e \right )^{3} \left (f x +e \right )^{\frac {3}{2}}}-\frac {2 b^{2} c^{2}}{3 \left (c f -d e \right )^{3} \left (f x +e \right )^{\frac {3}{2}}}+\frac {2 a^{2} d}{5 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {5}{2}}}-\frac {4 a b c}{5 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {5}{2}}}+\frac {4 b^{2} c e}{5 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {5}{2}} f}-\frac {2 b^{2} d \,e^{2}}{5 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {5}{2}} f^{2}}-\frac {2 a^{2}}{7 \left (c f -d e \right ) \left (f x +e \right )^{\frac {7}{2}}}+\frac {4 a b e}{7 \left (c f -d e \right ) \left (f x +e \right )^{\frac {7}{2}} f}-\frac {2 b^{2} e^{2}}{7 \left (c f -d e \right ) \left (f x +e \right )^{\frac {7}{2}} f^{2}} \]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c*f-d*e>0)', see `assume?` for
more details)Is c*f-d*e positive or negative?

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mupad [B]  time = 1.47, size = 327, normalized size = 1.58 \[ \frac {2\,d^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,{\left (a\,d-b\,c\right )}^2\,\left (c^4\,f^4-4\,c^3\,d\,e\,f^3+6\,c^2\,d^2\,e^2\,f^2-4\,c\,d^3\,e^3\,f+d^4\,e^4\right )}{{\left (c\,f-d\,e\right )}^{9/2}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}\right )\,{\left (a\,d-b\,c\right )}^2}{{\left (c\,f-d\,e\right )}^{9/2}}-\frac {\frac {2\,\left (a^2\,f^2-2\,a\,b\,e\,f+b^2\,e^2\right )}{7\,\left (c\,f-d\,e\right )}+\frac {2\,{\left (e+f\,x\right )}^2\,\left (a^2\,d^2\,f^2-2\,a\,b\,c\,d\,f^2+b^2\,c^2\,f^2\right )}{3\,{\left (c\,f-d\,e\right )}^3}-\frac {2\,\left (e+f\,x\right )\,\left (d\,a^2\,f^2-2\,c\,a\,b\,f^2-d\,b^2\,e^2+2\,c\,b^2\,e\,f\right )}{5\,{\left (c\,f-d\,e\right )}^2}-\frac {2\,d\,{\left (e+f\,x\right )}^3\,\left (a^2\,d^2\,f^2-2\,a\,b\,c\,d\,f^2+b^2\,c^2\,f^2\right )}{{\left (c\,f-d\,e\right )}^4}}{f^2\,{\left (e+f\,x\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*d^(3/2)*atan((d^(1/2)*(e + f*x)^(1/2)*(a*d - b*c)^2*(c^4*f^4 + d^4*e^4 + 6*c^2*d^2*e^2*f^2 - 4*c*d^3*e^3*f
- 4*c^3*d*e*f^3))/((c*f - d*e)^(9/2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)))*(a*d - b*c)^2)/(c*f - d*e)^(9/2) - ((2*
(a^2*f^2 + b^2*e^2 - 2*a*b*e*f))/(7*(c*f - d*e)) + (2*(e + f*x)^2*(a^2*d^2*f^2 + b^2*c^2*f^2 - 2*a*b*c*d*f^2))
/(3*(c*f - d*e)^3) - (2*(e + f*x)*(a^2*d*f^2 - b^2*d*e^2 - 2*a*b*c*f^2 + 2*b^2*c*e*f))/(5*(c*f - d*e)^2) - (2*
d*(e + f*x)^3*(a^2*d^2*f^2 + b^2*c^2*f^2 - 2*a*b*c*d*f^2))/(c*f - d*e)^4)/(f^2*(e + f*x)^(7/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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