3.1645 \(\int \frac {(b+2 c x) (d+e x)^{3/2}}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=216 \[ \frac {3 \sqrt {2} e \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c \sqrt {a+b x+c x^2} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {2 (d+e x)^{3/2}}{\sqrt {a+b x+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {768, 718, 424} \[ \frac {3 \sqrt {2} e \sqrt {b^2-4 a c} \sqrt {d+e x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} e}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c \sqrt {a+b x+c x^2} \sqrt {\frac {c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {2 (d+e x)^{3/2}}{\sqrt {a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)^(3/2))/Sqrt[a + b*x + c*x^2] + (3*Sqrt[2]*Sqrt[b^2 - 4*a*c]*e*Sqrt[d + e*x]*Sqrt[-((c*(a + b*x +
c*x^2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c]]/Sqrt[2]], (-
2*Sqrt[b^2 - 4*a*c]*e)/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(c*Sqrt[(c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*
a*c])*e)]*Sqrt[a + b*x + c*x^2])

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 718

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(2*Rt[b^2 - 4*a*c, 2]
*(d + e*x)^m*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))])/(c*Sqrt[a + b*x + c*x^2]*((2*c*(d + e*x))/(2*c*d -
b*e - e*Rt[b^2 - 4*a*c, 2]))^m), Subst[Int[(1 + (2*e*Rt[b^2 - 4*a*c, 2]*x^2)/(2*c*d - b*e - e*Rt[b^2 - 4*a*c,
2]))^m/Sqrt[1 - x^2], x], x, Sqrt[(b + Rt[b^2 - 4*a*c, 2] + 2*c*x)/(2*Rt[b^2 - 4*a*c, 2])]], x] /; FreeQ[{a, b
, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m^2, 1/4]

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

Rubi steps

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

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Mathematica [C]  time = 0.78, size = 378, normalized size = 1.75 \[ \frac {-4 (d+e x)^{3/2}+\frac {3 i \sqrt {2} \left (e \left (\sqrt {b^2-4 a c}-b\right )+2 c d\right ) \sqrt {\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{e \left (\sqrt {b^2-4 a c}+b\right )-2 c d}} \sqrt {1-\frac {2 c (d+e x)}{e \left (\sqrt {b^2-4 a c}-b\right )+2 c d}} \left (E\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{\left (b+\sqrt {b^2-4 a c}\right ) e-2 c d}} \sqrt {d+e x}\right )|\frac {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c d+\left (\sqrt {b^2-4 a c}-b\right ) e}\right )-F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{\left (b+\sqrt {b^2-4 a c}\right ) e-2 c d}} \sqrt {d+e x}\right )|\frac {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c d+\left (\sqrt {b^2-4 a c}-b\right ) e}\right )\right )}{c \sqrt {\frac {c}{e \left (\sqrt {b^2-4 a c}+b\right )-2 c d}}}}{2 \sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(d + e*x)^(3/2) + ((3*I)*Sqrt[2]*(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*Sqrt[(e*(b + Sqrt[b^2 - 4*a*c] + 2*c
*x))/(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[1 - (2*c*(d + e*x))/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)]*(Ell
ipticE[I*ArcSinh[Sqrt[2]*Sqrt[c/(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)]*Sqrt[d + e*x]], (2*c*d - (b + Sqrt[b^2 -
4*a*c])*e)/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)] - EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[c/(-2*c*d + (b + Sqrt[b^2
- 4*a*c])*e)]*Sqrt[d + e*x]], (2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)]))/(c*
Sqrt[c/(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)]))/(2*Sqrt[a + x*(b + c*x)])

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (2 \, c e x^{2} + b d + {\left (2 \, c d + b e\right )} x\right )} \sqrt {c x^{2} + b x + a} \sqrt {e x + d}}{c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((2*c*e*x^2 + b*d + (2*c*d + b*e)*x)*sqrt(c*x^2 + b*x + a)*sqrt(e*x + d)/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*
x + (b^2 + 2*a*c)*x^2 + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (2 \, c x + b\right )} {\left (e x + d\right )}^{\frac {3}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.12, size = 1349, normalized size = 6.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(e*x+d)^(1/2)*(c*x^2+b*x+a)^(1/2)*(3*2^(1/2)*EllipticF(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)
^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*a*e^2*(-(e*x+d)/(b*e-2*c*d
+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*
c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)-3*2^(1/2)*EllipticF(2^(1/2)*(-(e*x+d)/(b*e
-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1
/2))*b*d*e*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*
a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)+3*2^(1/2)*E
llipticF(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+
2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*c*d^2*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a
*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2
)^(1/2)*e)*e)^(1/2)-3*2^(1/2)*EllipticE(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c
*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*a*e^2*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/
2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2
)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)+3*2^(1/2)*EllipticE(2^(1/2)*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^
2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*b*d*e*(-(e*x
+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)
*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)-3*2^(1/2)*EllipticE(2^(1/2)*
(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2),(-(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(-b*e+2*c*d+(-4*a*c+b^2
)^(1/2)*e))^(1/2))*c*d^2*(-(e*x+d)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-
b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/2)*((2*c*x+b+(-4*a*c+b^2)^(1/2))/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*e)^(1/
2)-c*e^2*x^2-2*c*d*e*x-c*d^2)/c/(c*e*x^3+b*e*x^2+c*d*x^2+a*e*x+b*d*x+a*d)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (2 \, c x + b\right )} {\left (e x + d\right )}^{\frac {3}{2}}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (b+2\,c\,x\right )\,{\left (d+e\,x\right )}^{3/2}}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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