3.2062 \(\int \frac {1}{(d+e x)^{3/2} \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\)

Optimal. Leaf size=139 \[ \frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{(d+e x)^{3/2} \left (c d^2-a e^2\right )}+\frac {c d \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d^2-a e^2}}\right )}{\sqrt {e} \left (c d^2-a e^2\right )^{3/2}} \]

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

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Rubi [A]  time = 0.08, antiderivative size = 139, 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 = {672, 660, 205} \[ \frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{(d+e x)^{3/2} \left (c d^2-a e^2\right )}+\frac {c d \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d^2-a e^2}}\right )}{\sqrt {e} \left (c d^2-a e^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^(3/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]),x]

[Out]

Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]/((c*d^2 - a*e^2)*(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])])/(Sqrt[e]*(c*d^2 - a*e^2)^(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 672

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 135, normalized size = 0.97 \[ \frac {\sqrt {e} \sqrt {c d^2-a e^2} (a e+c d x)+c d (d+e x) \sqrt {a e+c d x} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {e} \sqrt {d+e x} \left (c d^2-a e^2\right )^{3/2} \sqrt {(d+e x) (a e+c d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^(3/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]),x]

[Out]

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

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fricas [B]  time = 0.89, size = 558, normalized size = 4.01 \[ \left [\frac {{\left (c d e^{2} x^{2} + 2 \, c d^{2} e x + c d^{3}\right )} \sqrt {-c d^{2} e + a e^{3}} \log \left (-\frac {c d e^{2} x^{2} + 2 \, a e^{3} x - c d^{3} + 2 \, a d e^{2} + 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {-c d^{2} e + a e^{3}} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (c d^{2} e - a e^{3}\right )} \sqrt {e x + d}}{2 \, {\left (c^{2} d^{6} e - 2 \, a c d^{4} e^{3} + a^{2} d^{2} e^{5} + {\left (c^{2} d^{4} e^{3} - 2 \, a c d^{2} e^{5} + a^{2} e^{7}\right )} x^{2} + 2 \, {\left (c^{2} d^{5} e^{2} - 2 \, a c d^{3} e^{4} + a^{2} d e^{6}\right )} x\right )}}, -\frac {{\left (c d e^{2} x^{2} + 2 \, c d^{2} e x + c d^{3}\right )} \sqrt {c d^{2} e - a e^{3}} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {c d^{2} e - a e^{3}} \sqrt {e x + d}}{c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x}\right ) - \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (c d^{2} e - a e^{3}\right )} \sqrt {e x + d}}{c^{2} d^{6} e - 2 \, a c d^{4} e^{3} + a^{2} d^{2} e^{5} + {\left (c^{2} d^{4} e^{3} - 2 \, a c d^{2} e^{5} + a^{2} e^{7}\right )} x^{2} + 2 \, {\left (c^{2} d^{5} e^{2} - 2 \, a c d^{3} e^{4} + a^{2} d e^{6}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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]

[1/2*((c*d*e^2*x^2 + 2*c*d^2*e*x + c*d^3)*sqrt(-c*d^2*e + a*e^3)*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(-c*d^2*e + a*e^3)*sqrt(e*x + d))/(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)*(c*d^2*e - a*e^3)*sqrt(e*x + d))/(c^2*d^6*e - 2*a*c*d^
4*e^3 + a^2*d^2*e^5 + (c^2*d^4*e^3 - 2*a*c*d^2*e^5 + a^2*e^7)*x^2 + 2*(c^2*d^5*e^2 - 2*a*c*d^3*e^4 + a^2*d*e^6
)*x), -((c*d*e^2*x^2 + 2*c*d^2*e*x + c*d^3)*sqrt(c*d^2*e - a*e^3)*arctan(sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e
^2)*x)*sqrt(c*d^2*e - a*e^3)*sqrt(e*x + d)/(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x)) - sqrt(c*d*e*x^2 + a
*d*e + (c*d^2 + a*e^2)*x)*(c*d^2*e - a*e^3)*sqrt(e*x + d))/(c^2*d^6*e - 2*a*c*d^4*e^3 + a^2*d^2*e^5 + (c^2*d^4
*e^3 - 2*a*c*d^2*e^5 + a^2*e^7)*x^2 + 2*(c^2*d^5*e^2 - 2*a*c*d^3*e^4 + a^2*d*e^6)*x)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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(1/(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 = 0.07, size = 172, normalized size = 1.24 \[ \frac {\sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, \left (c d e x \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )+c \,d^{2} \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )-\sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\right )}{\left (e x +d \right )^{\frac {3}{2}} \sqrt {c d x +a e}\, \left (a \,e^{2}-c \,d^{2}\right ) \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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]

integrate(1/(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 \[ \int \frac {1}{{\left (d+e\,x\right )}^{3/2}\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\left (d + e x\right ) \left (a e + c d x\right )} \left (d + e x\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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