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

Optimal. Leaf size=196 \[ -\frac {3 c d \sqrt {d+e x}}{\left (c d^2-a e^2\right )^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac {1}{\sqrt {d+e x} \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {3 c d \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{5/2}} \]

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Rubi [A]  time = 0.13, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {672, 666, 660, 205} \begin {gather*} -\frac {3 c d \sqrt {d+e x}}{\left (c d^2-a e^2\right )^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac {1}{\sqrt {d+e x} \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}}-\frac {3 c d \sqrt {e} \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 )}{\left (c d^2-a e^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/((c*d^2 - a*e^2)*Sqrt[d + e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (3*c*d*Sqrt[d + e*x])/((c*d^2
- a*e^2)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (3*c*d*Sqrt[e]*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])])/(c*d^2 - a*e^2)^(5/2)

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 666

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

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}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=\frac {1}{\left (c d^2-a e^2\right ) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {(3 c d) \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{2 \left (c d^2-a e^2\right )}\\ &=\frac {1}{\left (c d^2-a e^2\right ) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {3 c d \sqrt {d+e x}}{\left (c d^2-a e^2\right )^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {(3 c d e) \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 )^2}\\ &=\frac {1}{\left (c d^2-a e^2\right ) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {3 c d \sqrt {d+e x}}{\left (c d^2-a e^2\right )^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {\left (3 c d e^2\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 )}{\left (c d^2-a e^2\right )^2}\\ &=\frac {1}{\left (c d^2-a e^2\right ) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {3 c d \sqrt {d+e x}}{\left (c d^2-a e^2\right )^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {3 c d \sqrt {e} \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 )}{\left (c d^2-a e^2\right )^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 77, normalized size = 0.39 \begin {gather*} -\frac {2 c d \sqrt {d+e x} \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {e (a e+c d x)}{a e^2-c d^2}\right )}{\left (c d^2-a e^2\right )^2 \sqrt {(d+e x) (a e+c d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*d*Sqrt[d + e*x]*Hypergeometric2F1[-1/2, 2, 1/2, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)])/((c*d^2 - a*e^2)^
2*Sqrt[(a*e + c*d*x)*(d + e*x)])

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IntegrateAlgebraic [A]  time = 2.47, size = 179, normalized size = 0.91 \begin {gather*} \frac {(d+e x)^{3/2} (a e+c d x)^{3/2} \left (-\frac {c d \left (3 e (a e+c d x)-2 a e^2+2 c d^2\right )}{\left (c d^2-a e^2\right )^2 \sqrt {a e+c d x} \left (e (a e+c d x)-a e^2+c d^2\right )}-\frac {3 c d \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{5/2}}\right )}{((d+e x) (a e+c d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a*e + c*d*x)^(3/2)*(d + e*x)^(3/2)*(-((c*d*(2*c*d^2 - 2*a*e^2 + 3*e*(a*e + c*d*x)))/((c*d^2 - a*e^2)^2*Sqrt[
a*e + c*d*x]*(c*d^2 - a*e^2 + e*(a*e + c*d*x)))) - (3*c*d*Sqrt[e]*ArcTan[(Sqrt[e]*Sqrt[a*e + c*d*x])/Sqrt[c*d^
2 - a*e^2]])/(c*d^2 - a*e^2)^(5/2)))/((a*e + c*d*x)*(d + e*x))^(3/2)

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fricas [B]  time = 0.44, size = 764, normalized size = 3.90 \begin {gather*} \left [\frac {3 \, {\left (c^{2} d^{2} e^{2} x^{3} + a c d^{3} e + {\left (2 \, c^{2} d^{3} e + a c d e^{3}\right )} x^{2} + {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2}\right )} x\right )} \sqrt {-\frac {e}{c d^{2} - a e^{2}}} \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} {\left (c d^{2} - a e^{2}\right )} \sqrt {e x + d} \sqrt {-\frac {e}{c d^{2} - a e^{2}}}}{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 (3 \, c d e x + 2 \, c d^{2} + a e^{2}\right )} \sqrt {e x + d}}{2 \, {\left (a c^{2} d^{6} e - 2 \, a^{2} c d^{4} e^{3} + a^{3} d^{2} e^{5} + {\left (c^{3} d^{5} e^{2} - 2 \, a c^{2} d^{3} e^{4} + a^{2} c d e^{6}\right )} x^{3} + {\left (2 \, c^{3} d^{6} e - 3 \, a c^{2} d^{4} e^{3} + a^{3} e^{7}\right )} x^{2} + {\left (c^{3} d^{7} - 3 \, a^{2} c d^{3} e^{4} + 2 \, a^{3} d e^{6}\right )} x\right )}}, -\frac {3 \, {\left (c^{2} d^{2} e^{2} x^{3} + a c d^{3} e + {\left (2 \, c^{2} d^{3} e + a c d e^{3}\right )} x^{2} + {\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2}\right )} x\right )} \sqrt {\frac {e}{c d^{2} - a e^{2}}} \arctan \left (-\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (c d^{2} - a e^{2}\right )} \sqrt {e x + d} \sqrt {\frac {e}{c d^{2} - a e^{2}}}}{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 (3 \, c d e x + 2 \, c d^{2} + a e^{2}\right )} \sqrt {e x + d}}{a c^{2} d^{6} e - 2 \, a^{2} c d^{4} e^{3} + a^{3} d^{2} e^{5} + {\left (c^{3} d^{5} e^{2} - 2 \, a c^{2} d^{3} e^{4} + a^{2} c d e^{6}\right )} x^{3} + {\left (2 \, c^{3} d^{6} e - 3 \, a c^{2} d^{4} e^{3} + a^{3} e^{7}\right )} x^{2} + {\left (c^{3} d^{7} - 3 \, a^{2} c d^{3} e^{4} + 2 \, a^{3} d e^{6}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to transpose Error: Bad Argument Value

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maple [A]  time = 0.08, size = 235, normalized size = 1.20 \begin {gather*} \frac {\sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, \left (3 \sqrt {c d x +a e}\, c d \,e^{2} x \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )+3 \sqrt {c d x +a e}\, c \,d^{2} e \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )-3 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, c d e x -\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, a \,e^{2}-2 \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, c \,d^{2}\right )}{\left (e x +d \right )^{\frac {3}{2}} \left (c d x +a e \right ) \left (a \,e^{2}-c \,d^{2}\right )^{2} \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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