3.21.33 \(\int \frac {\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}}{(d+e x)^{5/2}} \, dx\) [2033]

Optimal. Leaf size=129 \[ -\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e (d+e x)^{3/2}}+\frac {c d \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c d^2-a e^2} \sqrt {d+e x}}\right )}{e^{3/2} \sqrt {c d^2-a e^2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 129, 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 = {676, 674, 211} \begin {gather*} \frac {c d \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} \sqrt {c d^2-a e^2}}-\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e (d+e x)^{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)^(5/2),x]

[Out]

-(Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]/(e*(d + e*x)^(3/2))) + (c*d*ArcTan[(Sqrt[e]*Sqrt[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)*Sqrt[c*d^2 - a*e^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 676

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 + p + 1))), x] - Dist[c*(p/(e^2*(m + p + 1))), Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + 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)^{5/2}} \, dx &=-\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e (d+e x)^{3/2}}+\frac {(c d) \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}{2 e}\\ &=-\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e (d+e x)^{3/2}}+(c d) \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 {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e (d+e x)^{3/2}}+\frac {c d \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} \sqrt {c d^2-a e^2}}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 112, normalized size = 0.87 \begin {gather*} \frac {\sqrt {(a e+c d x) (d+e x)} \left (-\sqrt {e}+\frac {c d (d+e x) \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {c d^2-a e^2} \sqrt {a e+c d x}}\right )}{e^{3/2} (d+e x)^{3/2}} \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)^(5/2),x]

[Out]

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

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Maple [A]
time = 0.79, size = 153, normalized size = 1.19

method	result	size
default	\(\frac {\left (-\arctanh \left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\right ) c d e x -\arctanh \left (\frac {e \sqrt {c d x +a e}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\right ) c \,d^{2}-\sqrt {c d x +a e}\, \sqrt {\left (e^{2} a -c \,d^{2}\right ) e}\right ) \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}}{\left (e x +d \right )^{\frac {3}{2}} \sqrt {c d x +a e}\, e \sqrt {\left (e^{2} a -c \,d^{2}\right ) e}}\)	\(153\)

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)^(5/2),x,method=_RETURNVERBOSE)

[Out]

(-arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*c*d*e*x-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)*(e*x+d))^(1/2)/(e*x+d)^(3/2)/(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)^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)/(x*e + d)^(5/2), x)

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Fricas [A]
time = 3.58, size = 468, normalized size = 3.63 \begin {gather*} \left [-\frac {{\left (c d x^{2} e^{2} + 2 \, c d^{2} x e + c d^{3}\right )} \sqrt {-c d^{2} e + a e^{3}} \log \left (\frac {c d^{3} - 2 \, a x e^{3} - {\left (c d x^{2} + 2 \, a d\right )} e^{2} + 2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {-c d^{2} e + a e^{3}} \sqrt {x e + d}}{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} {\left (c d^{2} e - a e^{3}\right )} \sqrt {x e + d}}{2 \, {\left (2 \, c d^{3} x e^{3} + c d^{4} e^{2} - a x^{2} e^{6} - 2 \, a d x e^{5} + {\left (c d^{2} x^{2} - a d^{2}\right )} e^{4}\right )}}, -\frac {{\left (c d x^{2} e^{2} + 2 \, c d^{2} x e + c d^{3}\right )} \sqrt {c d^{2} e - a e^{3}} \arctan \left (\frac {\sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {c d^{2} e - a e^{3}} \sqrt {x e + d}}{c d^{2} x e + a x e^{3} + {\left (c d x^{2} + a d\right )} e^{2}}\right ) + \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (c d^{2} e - a e^{3}\right )} \sqrt {x e + d}}{2 \, c d^{3} x e^{3} + c d^{4} e^{2} - a x^{2} e^{6} - 2 \, a d x e^{5} + {\left (c d^{2} x^{2} - a d^{2}\right )} e^{4}}\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)^(5/2),x, algorithm="fricas")

[Out]

[-1/2*((c*d*x^2*e^2 + 2*c*d^2*x*e + c*d^3)*sqrt(-c*d^2*e + a*e^3)*log((c*d^3 - 2*a*x*e^3 - (c*d*x^2 + 2*a*d)*e
^2 + 2*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(-c*d^2*e + a*e^3)*sqrt(x*e + d))/(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)*(c*d^2*e - a*e^3)*sqrt(x*e + d))/(2*c*d^3*x*e^3 + c*d^4*
e^2 - a*x^2*e^6 - 2*a*d*x*e^5 + (c*d^2*x^2 - a*d^2)*e^4), -((c*d*x^2*e^2 + 2*c*d^2*x*e + c*d^3)*sqrt(c*d^2*e -
a*e^3)*arctan(sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(c*d^2*e - a*e^3)*sqrt(x*e + d)/(c*d^2*x*e + a*
x*e^3 + (c*d*x^2 + a*d)*e^2)) + sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*(c*d^2*e - a*e^3)*sqrt(x*e + d))/(
2*c*d^3*x*e^3 + c*d^4*e^2 - a*x^2*e^6 - 2*a*d*x*e^5 + (c*d^2*x^2 - a*d^2)*e^4)]

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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 {5}{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)**(5/2),x)

[Out]

Integral(sqrt((d + e*x)*(a*e + c*d*x))/(d + e*x)**(5/2), x)

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Giac [A]
time = 1.90, size = 115, normalized size = 0.89 \begin {gather*} \frac {{\left (\frac {c^{2} d^{2} \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 ) e}{\sqrt {c d^{2} e - a e^{3}}} - \frac {\sqrt {{\left (x e + d\right )} c d e - c d^{2} e + a e^{3}} c d}{x e + d}\right )} e^{\left (-2\right )}}{c d} \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)^(5/2),x, algorithm="giac")

[Out]

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

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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 )}^{5/2}} \,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)^(5/2),x)

[Out]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)/(d + e*x)^(5/2), x)

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