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

Optimal. Leaf size=128 \[ \frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e \sqrt {d+e x}}-\frac {2 \sqrt {c d^2-a e^2} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d^2-a e^2}}\right )}{e^{3/2}} \]

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Rubi [A]  time = 0.10, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {664, 660, 205} \begin {gather*} \frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e \sqrt {d+e x}}-\frac {2 \sqrt {c d^2-a e^2} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d^2-a e^2}}\right )}{e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 660

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

Rule 664

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 105, normalized size = 0.82 \begin {gather*} \frac {2 \sqrt {(d+e x) (a e+c d x)} \left (\sqrt {e}-\frac {\sqrt {c d^2-a e^2} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {a e+c d x}}\right )}{e^{3/2} \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.46, size = 170, normalized size = 1.33 \begin {gather*} \frac {4 \sqrt {c d^2-a e^2} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {d+e x} \sqrt {c d^2-a e^2}}{e \sqrt {a e (d+e x)-\frac {c d^2 (d+e x)}{e}+\frac {c d (d+e x)^2}{e}}-\sqrt {c d e} (d+e x)}\right )}{e^{3/2}}+\frac {2 \sqrt {a e (d+e x)-\frac {c d^2 (d+e x)}{e}+\frac {c d (d+e x)^2}{e}}}{e \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 309, normalized size = 2.41 \begin {gather*} \left [\frac {{\left (e x + d\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{e}} \log \left (-\frac {c d e^{2} x^{2} + 2 \, a e^{3} x - c d^{3} + 2 \, a d e^{2} - 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d} e \sqrt {-\frac {c d^{2} - a e^{2}}{e}}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{e^{2} x + d e}, \frac {2 \, {\left ({\left (e x + d\right )} \sqrt {\frac {c d^{2} - a e^{2}}{e}} \arctan \left (\frac {\sqrt {e x + d} \sqrt {\frac {c d^{2} - a e^{2}}{e}}}{\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}\right ) + \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}\right )}}{e^{2} x + d e}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[((e*x + d)*sqrt(-(c*d^2 - a*e^2)/e)*log(-(c*d*e^2*x^2 + 2*a*e^3*x - c*d^3 + 2*a*d*e^2 - 2*sqrt(c*d*e*x^2 + a*
d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)*e*sqrt(-(c*d^2 - a*e^2)/e))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*sqrt(c*d*e*x
^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d))/(e^2*x + d*e), 2*((e*x + d)*sqrt((c*d^2 - a*e^2)/e)*arctan(sqrt
(e*x + d)*sqrt((c*d^2 - a*e^2)/e)/sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)) + sqrt(c*d*e*x^2 + a*d*e + (c*d
^2 + a*e^2)*x)*sqrt(e*x + d))/(e^2*x + d*e)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 163, normalized size = 1.27 \begin {gather*} -\frac {2 \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, \left (a \,e^{2} \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )-c \,d^{2} \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )-\sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\right )}{\sqrt {e x +d}\, \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{{\left (d+e\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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