3.21.32 \(\int \frac {\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}}{(d+e x)^{3/2}} \, dx\) [2032]

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

[Out]

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

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Rubi [A]
time = 0.06, 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 = {678, 674, 211} \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} \text {ArcTan}\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 211

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

Rule 674

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 678

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 ) \text {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 = 0.10, size = 118, normalized size = 0.92 \begin {gather*} \frac {2 \sqrt {a e+c d x} \sqrt {d+e x} \left (\sqrt {e} \sqrt {a e+c d x}-\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 )\right )}{e^{3/2} \sqrt {(a e+c d x) (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]*Sqrt[d + e*x]*(Sqrt[e]*Sqrt[a*e + c*d*x] - Sqrt[c*d^2 - a*e^2]*ArcTan[(Sqrt[e]*Sqrt[a*e +
c*d*x])/Sqrt[c*d^2 - a*e^2]]))/(e^(3/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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Maple [A]
time = 0.73, size = 153, normalized size = 1.20

method	result	size
default	\(-\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (\arctanh \left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\right ) a \,e^{2}-\arctanh \left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\right ) c \,d^{2}-\sqrt {c d x +a e}\, \sqrt {\left (e^{2} a -c \,d^{2}\right ) e}\right )}{\sqrt {e x +d}\, \sqrt {c d x +a e}\, e \sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\)	\(153\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((c*d*x+a*e)*(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))/(e*x+d)^(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*} \text {Failed to integrate} \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*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)/(x*e + d)^(3/2), x)

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Fricas [A]
time = 3.72, size = 302, normalized size = 2.36 \begin {gather*} \left [\frac {{\left (x e + d\right )} \sqrt {-{\left (c d^{2} - a e^{2}\right )} e^{\left (-1\right )}} \log \left (\frac {c d^{3} - 2 \, a x e^{3} + 2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d} \sqrt {-{\left (c d^{2} - a e^{2}\right )} e^{\left (-1\right )}} e - {\left (c d x^{2} + 2 \, a d\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) + 2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d}}{x e^{2} + d e}, \frac {2 \, {\left (\sqrt {c d^{2} - a e^{2}} {\left (x e + d\right )} \arctan \left (\frac {\sqrt {c d^{2} - a e^{2}} \sqrt {x e + d} e^{\left (-\frac {1}{2}\right )}}{\sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}}\right ) e^{\left (-\frac {1}{2}\right )} + \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {x e + d}\right )}}{x e^{2} + 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]

[((x*e + d)*sqrt(-(c*d^2 - a*e^2)*e^(-1))*log((c*d^3 - 2*a*x*e^3 + 2*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*
e)*sqrt(x*e + d)*sqrt(-(c*d^2 - a*e^2)*e^(-1))*e - (c*d*x^2 + 2*a*d)*e^2)/(x^2*e^2 + 2*d*x*e + d^2)) + 2*sqrt(
c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(x*e + d))/(x*e^2 + d*e), 2*(sqrt(c*d^2 - a*e^2)*(x*e + d)*arctan(s
qrt(c*d^2 - a*e^2)*sqrt(x*e + d)*e^(-1/2)/sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e))*e^(-1/2) + sqrt(c*d^2*x
+ a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(x*e + d))/(x*e^2 + d*e)]

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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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (111) = 222\).
time = 1.47, size = 233, normalized size = 1.82 \begin {gather*} -2 \, {\left (\frac {{\left (c d^{2} - a e^{2}\right )} \arctan \left (\frac {\sqrt {{\left (x e + d\right )} c d e - c d^{2} e + a e^{3}}}{\sqrt {c d^{2} e - a e^{3}}}\right )}{\sqrt {c d^{2} e - a e^{3}}} - \sqrt {{\left (x e + d\right )} c d e - c d^{2} e + a e^{3}} e^{\left (-1\right )} - \frac {{\left (c d^{2} \arctan \left (\frac {\sqrt {-c d^{2} e + a e^{3}}}{\sqrt {c d^{2} e - a e^{3}}}\right ) e - a \arctan \left (\frac {\sqrt {-c d^{2} e + a e^{3}}}{\sqrt {c d^{2} e - a e^{3}}}\right ) e^{3} - \sqrt {c d^{2} e - a e^{3}} \sqrt {-c d^{2} e + a e^{3}}\right )} e^{\left (-1\right )}}{\sqrt {c d^{2} e - a e^{3}}}\right )} e^{\left (-1\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="giac")

[Out]

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

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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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