3.2059 \(\int \frac {(d+e x)^{3/2}}{\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\)

Optimal. Leaf size=109 \[ \frac {4 \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c^2 d^2 \sqrt {d+e x}}+\frac {2 \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {656, 648} \[ \frac {4 \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c^2 d^2 \sqrt {d+e x}}+\frac {2 \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(c*d^2 - a*e^2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(3*c^2*d^2*Sqrt[d + e*x]) + (2*Sqrt[d + e*x]*S
qrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(3*c*d)

Rule 648

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

Rule 656

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 54, normalized size = 0.50 \[ \frac {2 \sqrt {(d+e x) (a e+c d x)} \left (c d (3 d+e x)-2 a e^2\right )}{3 c^2 d^2 \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a*e + c*d*x)*(d + e*x)]*(-2*a*e^2 + c*d*(3*d + e*x)))/(3*c^2*d^2*Sqrt[d + e*x])

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fricas [A]  time = 0.86, size = 73, normalized size = 0.67 \[ \frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (c d e x + 3 \, c d^{2} - 2 \, a e^{2}\right )} \sqrt {e x + d}}{3 \, {\left (c^{2} d^{2} e x + c^{2} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(c*d*e*x + 3*c*d^2 - 2*a*e^2)*sqrt(e*x + d)/(c^2*d^2*e*x + c^2
*d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 69, normalized size = 0.63 \[ -\frac {2 \left (c d x +a e \right ) \left (-c d e x +2 a \,e^{2}-3 c \,d^{2}\right ) \sqrt {e x +d}}{3 \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, c^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(c*d*x+a*e)*(-c*d*e*x+2*a*e^2-3*c*d^2)*(e*x+d)^(1/2)/c^2/d^2/(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)

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maxima [A]  time = 1.25, size = 65, normalized size = 0.60 \[ \frac {2 \, {\left (c^{2} d^{2} e x^{2} + 3 \, a c d^{2} e - 2 \, a^{2} e^{3} + {\left (3 \, c^{2} d^{3} - a c d e^{2}\right )} x\right )}}{3 \, \sqrt {c d x + a e} c^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(c^2*d^2*e*x^2 + 3*a*c*d^2*e - 2*a^2*e^3 + (3*c^2*d^3 - a*c*d*e^2)*x)/(sqrt(c*d*x + a*e)*c^2*d^2)

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mupad [B]  time = 0.85, size = 85, normalized size = 0.78 \[ \frac {\left (\frac {2\,x\,\sqrt {d+e\,x}}{3\,c\,d}-\frac {\left (4\,a\,e^2-6\,c\,d^2\right )\,\sqrt {d+e\,x}}{3\,c^2\,d^2\,e}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{x+\frac {d}{e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((2*x*(d + e*x)^(1/2))/(3*c*d) - ((4*a*e^2 - 6*c*d^2)*(d + e*x)^(1/2))/(3*c^2*d^2*e))*(x*(a*e^2 + c*d^2) + a*
d*e + c*d*e*x^2)^(1/2))/(x + d/e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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