3.842 \(\int \frac{\sqrt{d+e x} (a+b x+c x^2)}{(e+f x)^{3/2}} \, dx\)

Optimal. Leaf size=249 \[ \frac{\sqrt{d+e x} \sqrt{e+f x} \left (4 e f \left (-2 a e f-b d f+3 b e^2\right )-c \left (-d^2 f^2-6 d e^2 f+15 e^4\right )\right )}{4 e f^3 \left (e^2-d f\right )}-\frac{\tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{d+e x}}{\sqrt{e} \sqrt{e+f x}}\right ) \left (4 e f \left (-2 a e f-b d f+3 b e^2\right )-c \left (-d^2 f^2-6 d e^2 f+15 e^4\right )\right )}{4 e^{3/2} f^{7/2}}+\frac{2 (d+e x)^{3/2} \left (a+\frac{e (c e-b f)}{f^2}\right )}{\left (e^2-d f\right ) \sqrt{e+f x}}+\frac{c (d+e x)^{3/2} \sqrt{e+f x}}{2 e f^2} \]

[Out]

(2*(a + (e*(c*e - b*f))/f^2)*(d + e*x)^(3/2))/((e^2 - d*f)*Sqrt[e + f*x]) + ((4*e*f*(3*b*e^2 - b*d*f - 2*a*e*f
) - c*(15*e^4 - 6*d*e^2*f - d^2*f^2))*Sqrt[d + e*x]*Sqrt[e + f*x])/(4*e*f^3*(e^2 - d*f)) + (c*(d + e*x)^(3/2)*
Sqrt[e + f*x])/(2*e*f^2) - ((4*e*f*(3*b*e^2 - b*d*f - 2*a*e*f) - c*(15*e^4 - 6*d*e^2*f - d^2*f^2))*ArcTanh[(Sq
rt[f]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[e + f*x])])/(4*e^(3/2)*f^(7/2))

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Rubi [A]  time = 0.280444, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {949, 80, 50, 63, 217, 206} \[ \frac{\sqrt{d+e x} \sqrt{e+f x} \left (4 e f \left (-2 a e f-b d f+3 b e^2\right )-c \left (-d^2 f^2-6 d e^2 f+15 e^4\right )\right )}{4 e f^3 \left (e^2-d f\right )}-\frac{\tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{d+e x}}{\sqrt{e} \sqrt{e+f x}}\right ) \left (4 e f \left (-2 a e f-b d f+3 b e^2\right )-c \left (-d^2 f^2-6 d e^2 f+15 e^4\right )\right )}{4 e^{3/2} f^{7/2}}+\frac{2 (d+e x)^{3/2} \left (a+\frac{e (c e-b f)}{f^2}\right )}{\left (e^2-d f\right ) \sqrt{e+f x}}+\frac{c (d+e x)^{3/2} \sqrt{e+f x}}{2 e f^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[d + e*x]*(a + b*x + c*x^2))/(e + f*x)^(3/2),x]

[Out]

(2*(a + (e*(c*e - b*f))/f^2)*(d + e*x)^(3/2))/((e^2 - d*f)*Sqrt[e + f*x]) + ((4*e*f*(3*b*e^2 - b*d*f - 2*a*e*f
) - c*(15*e^4 - 6*d*e^2*f - d^2*f^2))*Sqrt[d + e*x]*Sqrt[e + f*x])/(4*e*f^3*(e^2 - d*f)) + (c*(d + e*x)^(3/2)*
Sqrt[e + f*x])/(2*e*f^2) - ((4*e*f*(3*b*e^2 - b*d*f - 2*a*e*f) - c*(15*e^4 - 6*d*e^2*f - d^2*f^2))*ArcTanh[(Sq
rt[f]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[e + f*x])])/(4*e^(3/2)*f^(7/2))

Rule 949

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{Qx = PolynomialQuotient[(a + b*x + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p,
d + e*x, x]}, Simp[(R*(d + e*x)^(m + 1)*(f + g*x)^(n + 1))/((m + 1)*(e*f - d*g)), x] + Dist[1/((m + 1)*(e*f -
d*g)), Int[(d + e*x)^(m + 1)*(f + g*x)^n*ExpandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /;
FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&& IGtQ[p, 0] && LtQ[m, -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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sqrt{d+e x} \left (a+b x+c x^2\right )}{(e+f x)^{3/2}} \, dx &=\frac{2 \left (a+\frac{e (c e-b f)}{f^2}\right ) (d+e x)^{3/2}}{\left (e^2-d f\right ) \sqrt{e+f x}}+\frac{2 \int \frac{\sqrt{d+e x} \left (\frac{f \left (3 b e^2-b d f-2 a e f\right )-c \left (3 e^3-d e f\right )}{2 f^2}-\frac{1}{2} c \left (d-\frac{e^2}{f}\right ) x\right )}{\sqrt{e+f x}} \, dx}{e^2-d f}\\ &=\frac{2 \left (a+\frac{e (c e-b f)}{f^2}\right ) (d+e x)^{3/2}}{\left (e^2-d f\right ) \sqrt{e+f x}}+\frac{c (d+e x)^{3/2} \sqrt{e+f x}}{2 e f^2}+\frac{\left (4 e f \left (3 b e^2-b d f-2 a e f\right )-c \left (15 e^4-6 d e^2 f-d^2 f^2\right )\right ) \int \frac{\sqrt{d+e x}}{\sqrt{e+f x}} \, dx}{4 e f^2 \left (e^2-d f\right )}\\ &=\frac{2 \left (a+\frac{e (c e-b f)}{f^2}\right ) (d+e x)^{3/2}}{\left (e^2-d f\right ) \sqrt{e+f x}}+\frac{\left (4 e f \left (3 b e^2-b d f-2 a e f\right )-c \left (15 e^4-6 d e^2 f-d^2 f^2\right )\right ) \sqrt{d+e x} \sqrt{e+f x}}{4 e f^3 \left (e^2-d f\right )}+\frac{c (d+e x)^{3/2} \sqrt{e+f x}}{2 e f^2}-\frac{\left (4 e f \left (3 b e^2-b d f-2 a e f\right )-c \left (15 e^4-6 d e^2 f-d^2 f^2\right )\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{e+f x}} \, dx}{8 e f^3}\\ &=\frac{2 \left (a+\frac{e (c e-b f)}{f^2}\right ) (d+e x)^{3/2}}{\left (e^2-d f\right ) \sqrt{e+f x}}+\frac{\left (4 e f \left (3 b e^2-b d f-2 a e f\right )-c \left (15 e^4-6 d e^2 f-d^2 f^2\right )\right ) \sqrt{d+e x} \sqrt{e+f x}}{4 e f^3 \left (e^2-d f\right )}+\frac{c (d+e x)^{3/2} \sqrt{e+f x}}{2 e f^2}-\frac{\left (4 e f \left (3 b e^2-b d f-2 a e f\right )-c \left (15 e^4-6 d e^2 f-d^2 f^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{e-\frac{d f}{e}+\frac{f x^2}{e}}} \, dx,x,\sqrt{d+e x}\right )}{4 e^2 f^3}\\ &=\frac{2 \left (a+\frac{e (c e-b f)}{f^2}\right ) (d+e x)^{3/2}}{\left (e^2-d f\right ) \sqrt{e+f x}}+\frac{\left (4 e f \left (3 b e^2-b d f-2 a e f\right )-c \left (15 e^4-6 d e^2 f-d^2 f^2\right )\right ) \sqrt{d+e x} \sqrt{e+f x}}{4 e f^3 \left (e^2-d f\right )}+\frac{c (d+e x)^{3/2} \sqrt{e+f x}}{2 e f^2}-\frac{\left (4 e f \left (3 b e^2-b d f-2 a e f\right )-c \left (15 e^4-6 d e^2 f-d^2 f^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{f x^2}{e}} \, dx,x,\frac{\sqrt{d+e x}}{\sqrt{e+f x}}\right )}{4 e^2 f^3}\\ &=\frac{2 \left (a+\frac{e (c e-b f)}{f^2}\right ) (d+e x)^{3/2}}{\left (e^2-d f\right ) \sqrt{e+f x}}+\frac{\left (4 e f \left (3 b e^2-b d f-2 a e f\right )-c \left (15 e^4-6 d e^2 f-d^2 f^2\right )\right ) \sqrt{d+e x} \sqrt{e+f x}}{4 e f^3 \left (e^2-d f\right )}+\frac{c (d+e x)^{3/2} \sqrt{e+f x}}{2 e f^2}-\frac{\left (4 e f \left (3 b e^2-b d f-2 a e f\right )-c \left (15 e^4-6 d e^2 f-d^2 f^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{f} \sqrt{d+e x}}{\sqrt{e} \sqrt{e+f x}}\right )}{4 e^{3/2} f^{7/2}}\\ \end{align*}

Mathematica [A]  time = 1.17084, size = 196, normalized size = 0.79 \[ \frac{\frac{\sqrt{e^2-d f} \sqrt{\frac{e (e+f x)}{e^2-d f}} \sinh ^{-1}\left (\frac{\sqrt{f} \sqrt{d+e x}}{\sqrt{e^2-d f}}\right ) \left (4 e f \left (2 a e f+b d f-3 b e^2\right )+c \left (-d^2 f^2-6 d e^2 f+15 e^4\right )\right )}{e}+\sqrt{f} \sqrt{d+e x} \left (4 e f (-2 a f+3 b e+b f x)+c \left (e f \left (d+2 f x^2\right )+d f^2 x-5 e^2 f x-15 e^3\right )\right )}{4 e f^{7/2} \sqrt{e+f x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[d + e*x]*(a + b*x + c*x^2))/(e + f*x)^(3/2),x]

[Out]

(Sqrt[f]*Sqrt[d + e*x]*(4*e*f*(3*b*e - 2*a*f + b*f*x) + c*(-15*e^3 - 5*e^2*f*x + d*f^2*x + e*f*(d + 2*f*x^2)))
+ (Sqrt[e^2 - d*f]*(4*e*f*(-3*b*e^2 + b*d*f + 2*a*e*f) + c*(15*e^4 - 6*d*e^2*f - d^2*f^2))*Sqrt[(e*(e + f*x))
/(e^2 - d*f)]*ArcSinh[(Sqrt[f]*Sqrt[d + e*x])/Sqrt[e^2 - d*f]])/e)/(4*e*f^(7/2)*Sqrt[e + f*x])

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Maple [B]  time = 0.333, size = 834, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(1/2)*(c*x^2+b*x+a)/(f*x+e)^(3/2),x)

[Out]

1/8*(e*x+d)^(1/2)*(8*ln(1/2*(2*e*f*x+2*((e*x+d)*(f*x+e))^(1/2)*(e*f)^(1/2)+d*f+e^2)/(e*f)^(1/2))*x*a*e^2*f^3+4
*ln(1/2*(2*e*f*x+2*((e*x+d)*(f*x+e))^(1/2)*(e*f)^(1/2)+d*f+e^2)/(e*f)^(1/2))*x*b*d*e*f^3-12*ln(1/2*(2*e*f*x+2*
((e*x+d)*(f*x+e))^(1/2)*(e*f)^(1/2)+d*f+e^2)/(e*f)^(1/2))*x*b*e^3*f^2-ln(1/2*(2*e*f*x+2*((e*x+d)*(f*x+e))^(1/2
)*(e*f)^(1/2)+d*f+e^2)/(e*f)^(1/2))*x*c*d^2*f^3-6*ln(1/2*(2*e*f*x+2*((e*x+d)*(f*x+e))^(1/2)*(e*f)^(1/2)+d*f+e^
2)/(e*f)^(1/2))*x*c*d*e^2*f^2+15*ln(1/2*(2*e*f*x+2*((e*x+d)*(f*x+e))^(1/2)*(e*f)^(1/2)+d*f+e^2)/(e*f)^(1/2))*x
*c*e^4*f+4*x^2*c*e*f^2*(e*f)^(1/2)*((e*x+d)*(f*x+e))^(1/2)+8*ln(1/2*(2*e*f*x+2*((e*x+d)*(f*x+e))^(1/2)*(e*f)^(
1/2)+d*f+e^2)/(e*f)^(1/2))*a*e^3*f^2+4*ln(1/2*(2*e*f*x+2*((e*x+d)*(f*x+e))^(1/2)*(e*f)^(1/2)+d*f+e^2)/(e*f)^(1
/2))*b*d*e^2*f^2-12*ln(1/2*(2*e*f*x+2*((e*x+d)*(f*x+e))^(1/2)*(e*f)^(1/2)+d*f+e^2)/(e*f)^(1/2))*b*e^4*f-ln(1/2
*(2*e*f*x+2*((e*x+d)*(f*x+e))^(1/2)*(e*f)^(1/2)+d*f+e^2)/(e*f)^(1/2))*c*d^2*e*f^2-6*ln(1/2*(2*e*f*x+2*((e*x+d)
*(f*x+e))^(1/2)*(e*f)^(1/2)+d*f+e^2)/(e*f)^(1/2))*c*d*e^3*f+15*ln(1/2*(2*e*f*x+2*((e*x+d)*(f*x+e))^(1/2)*(e*f)
^(1/2)+d*f+e^2)/(e*f)^(1/2))*c*e^5+8*x*b*e*f^2*(e*f)^(1/2)*((e*x+d)*(f*x+e))^(1/2)+2*x*c*d*f^2*(e*f)^(1/2)*((e
*x+d)*(f*x+e))^(1/2)-10*x*c*e^2*f*(e*f)^(1/2)*((e*x+d)*(f*x+e))^(1/2)-16*a*e*f^2*(e*f)^(1/2)*((e*x+d)*(f*x+e))
^(1/2)+24*b*e^2*f*(e*f)^(1/2)*((e*x+d)*(f*x+e))^(1/2)+2*c*d*e*f*(e*f)^(1/2)*((e*x+d)*(f*x+e))^(1/2)-30*c*e^3*(
e*f)^(1/2)*((e*x+d)*(f*x+e))^(1/2))/(e*f)^(1/2)/e/((e*x+d)*(f*x+e))^(1/2)/f^3/(f*x+e)^(1/2)

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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((e*x+d)^(1/2)*(c*x^2+b*x+a)/(f*x+e)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 6.99904, size = 1268, normalized size = 5.09 \begin{align*} \left [\frac{{\left (15 \, c e^{5} -{\left (c d^{2} e - 4 \, b d e^{2} - 8 \, a e^{3}\right )} f^{2} - 6 \,{\left (c d e^{3} + 2 \, b e^{4}\right )} f +{\left (15 \, c e^{4} f -{\left (c d^{2} - 4 \, b d e - 8 \, a e^{2}\right )} f^{3} - 6 \,{\left (c d e^{2} + 2 \, b e^{3}\right )} f^{2}\right )} x\right )} \sqrt{e f} \log \left (8 \, e^{2} f^{2} x^{2} + e^{4} + 6 \, d e^{2} f + d^{2} f^{2} + 4 \,{\left (2 \, e f x + e^{2} + d f\right )} \sqrt{e f} \sqrt{e x + d} \sqrt{f x + e} + 8 \,{\left (e^{3} f + d e f^{2}\right )} x\right ) + 4 \,{\left (2 \, c e^{2} f^{3} x^{2} - 15 \, c e^{4} f - 8 \, a e^{2} f^{3} +{\left (c d e^{2} + 12 \, b e^{3}\right )} f^{2} -{\left (5 \, c e^{3} f^{2} -{\left (c d e + 4 \, b e^{2}\right )} f^{3}\right )} x\right )} \sqrt{e x + d} \sqrt{f x + e}}{16 \,{\left (e^{2} f^{5} x + e^{3} f^{4}\right )}}, -\frac{{\left (15 \, c e^{5} -{\left (c d^{2} e - 4 \, b d e^{2} - 8 \, a e^{3}\right )} f^{2} - 6 \,{\left (c d e^{3} + 2 \, b e^{4}\right )} f +{\left (15 \, c e^{4} f -{\left (c d^{2} - 4 \, b d e - 8 \, a e^{2}\right )} f^{3} - 6 \,{\left (c d e^{2} + 2 \, b e^{3}\right )} f^{2}\right )} x\right )} \sqrt{-e f} \arctan \left (\frac{{\left (2 \, e f x + e^{2} + d f\right )} \sqrt{-e f} \sqrt{e x + d} \sqrt{f x + e}}{2 \,{\left (e^{2} f^{2} x^{2} + d e^{2} f +{\left (e^{3} f + d e f^{2}\right )} x\right )}}\right ) - 2 \,{\left (2 \, c e^{2} f^{3} x^{2} - 15 \, c e^{4} f - 8 \, a e^{2} f^{3} +{\left (c d e^{2} + 12 \, b e^{3}\right )} f^{2} -{\left (5 \, c e^{3} f^{2} -{\left (c d e + 4 \, b e^{2}\right )} f^{3}\right )} x\right )} \sqrt{e x + d} \sqrt{f x + e}}{8 \,{\left (e^{2} f^{5} x + e^{3} f^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)*(c*x^2+b*x+a)/(f*x+e)^(3/2),x, algorithm="fricas")

[Out]

[1/16*((15*c*e^5 - (c*d^2*e - 4*b*d*e^2 - 8*a*e^3)*f^2 - 6*(c*d*e^3 + 2*b*e^4)*f + (15*c*e^4*f - (c*d^2 - 4*b*
d*e - 8*a*e^2)*f^3 - 6*(c*d*e^2 + 2*b*e^3)*f^2)*x)*sqrt(e*f)*log(8*e^2*f^2*x^2 + e^4 + 6*d*e^2*f + d^2*f^2 + 4
*(2*e*f*x + e^2 + d*f)*sqrt(e*f)*sqrt(e*x + d)*sqrt(f*x + e) + 8*(e^3*f + d*e*f^2)*x) + 4*(2*c*e^2*f^3*x^2 - 1
5*c*e^4*f - 8*a*e^2*f^3 + (c*d*e^2 + 12*b*e^3)*f^2 - (5*c*e^3*f^2 - (c*d*e + 4*b*e^2)*f^3)*x)*sqrt(e*x + d)*sq
rt(f*x + e))/(e^2*f^5*x + e^3*f^4), -1/8*((15*c*e^5 - (c*d^2*e - 4*b*d*e^2 - 8*a*e^3)*f^2 - 6*(c*d*e^3 + 2*b*e
^4)*f + (15*c*e^4*f - (c*d^2 - 4*b*d*e - 8*a*e^2)*f^3 - 6*(c*d*e^2 + 2*b*e^3)*f^2)*x)*sqrt(-e*f)*arctan(1/2*(2
*e*f*x + e^2 + d*f)*sqrt(-e*f)*sqrt(e*x + d)*sqrt(f*x + e)/(e^2*f^2*x^2 + d*e^2*f + (e^3*f + d*e*f^2)*x)) - 2*
(2*c*e^2*f^3*x^2 - 15*c*e^4*f - 8*a*e^2*f^3 + (c*d*e^2 + 12*b*e^3)*f^2 - (5*c*e^3*f^2 - (c*d*e + 4*b*e^2)*f^3)
*x)*sqrt(e*x + d)*sqrt(f*x + e))/(e^2*f^5*x + e^3*f^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d + e x} \left (a + b x + c x^{2}\right )}{\left (e + f x\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(1/2)*(c*x**2+b*x+a)/(f*x+e)**(3/2),x)

[Out]

Integral(sqrt(d + e*x)*(a + b*x + c*x**2)/(e + f*x)**(3/2), x)

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Giac [A]  time = 1.24471, size = 320, normalized size = 1.29 \begin{align*} \frac{{\left ({\left (x e + d\right )}{\left (\frac{2 \,{\left (x e + d\right )} c e^{\left (-1\right )}}{f} - \frac{{\left (3 \, c d f^{4} e^{2} - 4 \, b f^{4} e^{3} + 5 \, c f^{3} e^{4}\right )} e^{\left (-3\right )}}{f^{5}}\right )} + \frac{{\left (c d^{2} f^{4} e^{2} - 4 \, b d f^{4} e^{3} + 6 \, c d f^{3} e^{4} - 8 \, a f^{4} e^{4} + 12 \, b f^{3} e^{5} - 15 \, c f^{2} e^{6}\right )} e^{\left (-3\right )}}{f^{5}}\right )} \sqrt{x e + d}}{4 \, \sqrt{{\left (x e + d\right )} f e - d f e + e^{3}}} + \frac{{\left (c d^{2} f^{2} - 4 \, b d f^{2} e + 6 \, c d f e^{2} - 8 \, a f^{2} e^{2} + 12 \, b f e^{3} - 15 \, c e^{4}\right )} e^{\left (-\frac{3}{2}\right )} \log \left ({\left | -\sqrt{x e + d} \sqrt{f} e^{\frac{1}{2}} + \sqrt{{\left (x e + d\right )} f e - d f e + e^{3}} \right |}\right )}{4 \, f^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)*(c*x^2+b*x+a)/(f*x+e)^(3/2),x, algorithm="giac")

[Out]

1/4*((x*e + d)*(2*(x*e + d)*c*e^(-1)/f - (3*c*d*f^4*e^2 - 4*b*f^4*e^3 + 5*c*f^3*e^4)*e^(-3)/f^5) + (c*d^2*f^4*
e^2 - 4*b*d*f^4*e^3 + 6*c*d*f^3*e^4 - 8*a*f^4*e^4 + 12*b*f^3*e^5 - 15*c*f^2*e^6)*e^(-3)/f^5)*sqrt(x*e + d)/sqr
t((x*e + d)*f*e - d*f*e + e^3) + 1/4*(c*d^2*f^2 - 4*b*d*f^2*e + 6*c*d*f*e^2 - 8*a*f^2*e^2 + 12*b*f*e^3 - 15*c*
e^4)*e^(-3/2)*log(abs(-sqrt(x*e + d)*sqrt(f)*e^(1/2) + sqrt((x*e + d)*f*e - d*f*e + e^3)))/f^(7/2)