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

Optimal. Leaf size=249 \[ -\frac {2 \sqrt {d+e x} (x (2 c d-b e)+b d)}{b^2 \sqrt {b x+c x^2}}-\frac {4 d \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {\frac {e x}{d}+1} (c d-b e) F\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{(-b)^{3/2} \sqrt {c} \sqrt {b x+c x^2} \sqrt {d+e x}}+\frac {2 \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {d+e x} (2 c d-b e) E\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{(-b)^{3/2} \sqrt {c} \sqrt {b x+c x^2} \sqrt {\frac {e x}{d}+1}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {738, 843, 715, 112, 110, 117, 116} \[ -\frac {2 \sqrt {d+e x} (x (2 c d-b e)+b d)}{b^2 \sqrt {b x+c x^2}}-\frac {4 d \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {\frac {e x}{d}+1} (c d-b e) F\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{(-b)^{3/2} \sqrt {c} \sqrt {b x+c x^2} \sqrt {d+e x}}+\frac {2 \sqrt {x} \sqrt {\frac {c x}{b}+1} \sqrt {d+e x} (2 c d-b e) E\left (\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {-b}}\right )|\frac {b e}{c d}\right )}{(-b)^{3/2} \sqrt {c} \sqrt {b x+c x^2} \sqrt {\frac {e x}{d}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[d + e*x]*(b*d + (2*c*d - b*e)*x))/(b^2*Sqrt[b*x + c*x^2]) + (2*(2*c*d - b*e)*Sqrt[x]*Sqrt[1 + (c*x)/b
]*Sqrt[d + e*x]*EllipticE[ArcSin[(Sqrt[c]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/((-b)^(3/2)*Sqrt[c]*Sqrt[1 + (e*x)
/d]*Sqrt[b*x + c*x^2]) - (4*d*(c*d - b*e)*Sqrt[x]*Sqrt[1 + (c*x)/b]*Sqrt[1 + (e*x)/d]*EllipticF[ArcSin[(Sqrt[c
]*Sqrt[x])/Sqrt[-b]], (b*e)/(c*d)])/((-b)^(3/2)*Sqrt[c]*Sqrt[d + e*x]*Sqrt[b*x + c*x^2])

Rule 110

Int[Sqrt[(e_) + (f_.)*(x_)]/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[(2*Sqrt[e]*Rt[-(b/d)
, 2]*EllipticE[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[-(b/d), 2])], (c*f)/(d*e)])/b, x] /; FreeQ[{b, c, d, e, f}, x] &&
NeQ[d*e - c*f, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !LtQ[-(b/d), 0]

Rule 112

Int[Sqrt[(e_) + (f_.)*(x_)]/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Dist[(Sqrt[e + f*x]*Sqrt[
1 + (d*x)/c])/(Sqrt[c + d*x]*Sqrt[1 + (f*x)/e]), Int[Sqrt[1 + (f*x)/e]/(Sqrt[b*x]*Sqrt[1 + (d*x)/c]), x], x] /
; FreeQ[{b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] &&  !(GtQ[c, 0] && GtQ[e, 0])

Rule 116

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*Rt[-(b/d), 2]*E
llipticF[ArcSin[Sqrt[b*x]/(Sqrt[c]*Rt[-(b/d), 2])], (c*f)/(d*e)])/(b*Sqrt[e]), x] /; FreeQ[{b, c, d, e, f}, x]
&& GtQ[c, 0] && GtQ[e, 0] && (PosQ[-(b/d)] || NegQ[-(b/f)])

Rule 117

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Dist[(Sqrt[1 + (d*x)/c]
*Sqrt[1 + (f*x)/e])/(Sqrt[c + d*x]*Sqrt[e + f*x]), Int[1/(Sqrt[b*x]*Sqrt[1 + (d*x)/c]*Sqrt[1 + (f*x)/e]), x],
x] /; FreeQ[{b, c, d, e, f}, x] &&  !(GtQ[c, 0] && GtQ[e, 0])

Rule 715

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

Rule 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], 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] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.47, size = 210, normalized size = 0.84 \[ \frac {2 (c d-b e) \left (b (d+e x)-i c e x^{3/2} \sqrt {\frac {b}{c}} \sqrt {\frac {b}{c x}+1} \sqrt {\frac {d}{e x}+1} F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {b}{c}}}{\sqrt {x}}\right )|\frac {c d}{b e}\right )\right )-2 i c e x^{3/2} \sqrt {\frac {b}{c}} \sqrt {\frac {b}{c x}+1} \sqrt {\frac {d}{e x}+1} (b e-2 c d) E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {b}{c}}}{\sqrt {x}}\right )|\frac {c d}{b e}\right )}{b^2 c \sqrt {x (b+c x)} \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 1.35, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x} {\left (e x + d\right )}^{\frac {3}{2}}}{c^{2} x^{4} + 2 \, b c x^{3} + b^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.10, size = 451, normalized size = 1.81 \[ \frac {2 \left (b \,c^{2} e^{2} x^{2}-2 c^{3} d e \,x^{2}+\sqrt {\frac {c x +b}{b}}\, \sqrt {-\frac {\left (e x +d \right ) c}{b e -c d}}\, \sqrt {-\frac {c x}{b}}\, b^{3} e^{2} \EllipticE \left (\sqrt {\frac {c x +b}{b}}, \sqrt {\frac {b e}{b e -c d}}\right )-3 \sqrt {\frac {c x +b}{b}}\, \sqrt {-\frac {\left (e x +d \right ) c}{b e -c d}}\, \sqrt {-\frac {c x}{b}}\, b^{2} c d e \EllipticE \left (\sqrt {\frac {c x +b}{b}}, \sqrt {\frac {b e}{b e -c d}}\right )+2 \sqrt {\frac {c x +b}{b}}\, \sqrt {-\frac {\left (e x +d \right ) c}{b e -c d}}\, \sqrt {-\frac {c x}{b}}\, b^{2} c d e \EllipticF \left (\sqrt {\frac {c x +b}{b}}, \sqrt {\frac {b e}{b e -c d}}\right )+2 \sqrt {\frac {c x +b}{b}}\, \sqrt {-\frac {\left (e x +d \right ) c}{b e -c d}}\, \sqrt {-\frac {c x}{b}}\, b \,c^{2} d^{2} \EllipticE \left (\sqrt {\frac {c x +b}{b}}, \sqrt {\frac {b e}{b e -c d}}\right )-2 \sqrt {\frac {c x +b}{b}}\, \sqrt {-\frac {\left (e x +d \right ) c}{b e -c d}}\, \sqrt {-\frac {c x}{b}}\, b \,c^{2} d^{2} \EllipticF \left (\sqrt {\frac {c x +b}{b}}, \sqrt {\frac {b e}{b e -c d}}\right )-2 c^{3} d^{2} x -b \,c^{2} d^{2}\right ) \sqrt {\left (c x +b \right ) x}}{\left (c x +b \right ) \sqrt {e x +d}\, b^{2} c^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(2*b^2*d*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticF(((c*x+b)/b)^(1/2),(1/(b*e
-c*d)*b*e)^(1/2))*e*c-2*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticF(((c*x+b)/b)^
(1/2),(1/(b*e-c*d)*b*e)^(1/2))*b*c^2*d^2+((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*Ellip
ticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*b^3*e^2-3*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-c*d)*c)^(1/2)*(-1/b
*c*x)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*b^2*c*d*e+2*((c*x+b)/b)^(1/2)*(-(e*x+d)/(b*e-
c*d)*c)^(1/2)*(-1/b*c*x)^(1/2)*EllipticE(((c*x+b)/b)^(1/2),(1/(b*e-c*d)*b*e)^(1/2))*b*c^2*d^2+b*c^2*e^2*x^2-2*
c^3*d*e*x^2-2*x*c^3*d^2-c^2*d^2*b)/x*((c*x+b)*x)^(1/2)/(c*x+b)/b^2/c^2/(e*x+d)^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________