3.21.6 \(\int \frac {1}{(d+e x)^{3/2} (a d e+(c d^2+a e^2) x+c d e x^2)} \, dx\) [2006]

Optimal. Leaf size=120 \[ \frac {2}{3 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}+\frac {2 c d}{\left (c d^2-a e^2\right )^2 \sqrt {d+e x}}-\frac {2 c^{3/2} d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {640, 53, 65, 214} \begin {gather*} -\frac {2 c^{3/2} d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{5/2}}+\frac {2 c d}{\sqrt {d+e x} \left (c d^2-a e^2\right )^2}+\frac {2}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rule 640

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

Rubi steps

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

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Mathematica [A]
time = 0.21, size = 111, normalized size = 0.92 \begin {gather*} \frac {2 \left (-a e^2+c d (4 d+3 e x)\right )}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}+\frac {2 c^{3/2} d^{3/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \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]

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

[Out]

(2*(-(a*e^2) + c*d*(4*d + 3*e*x)))/(3*(c*d^2 - a*e^2)^2*(d + e*x)^(3/2)) + (2*c^(3/2)*d^(3/2)*ArcTan[(Sqrt[c]*
Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]])/(-(c*d^2) + a*e^2)^(5/2)

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Maple [A]
time = 0.72, size = 117, normalized size = 0.98

method	result	size
derivativedivides	\(-\frac {2}{3 \left (e^{2} a -c \,d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}+\frac {2 c d}{\left (e^{2} a -c \,d^{2}\right )^{2} \sqrt {e x +d}}+\frac {2 c^{2} d^{2} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right )^{2} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\)	\(117\)
default	\(-\frac {2}{3 \left (e^{2} a -c \,d^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}+\frac {2 c d}{\left (e^{2} a -c \,d^{2}\right )^{2} \sqrt {e x +d}}+\frac {2 c^{2} d^{2} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right )^{2} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c*d^2-%e^2*a>0)', see `assume?
` for more d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (99) = 198\).
time = 3.18, size = 463, normalized size = 3.86 \begin {gather*} \left [\frac {3 \, {\left (c d x^{2} e^{2} + 2 \, c d^{2} x e + c d^{3}\right )} \sqrt {\frac {c d}{c d^{2} - a e^{2}}} \log \left (\frac {c d x e + 2 \, c d^{2} - 2 \, {\left (c d^{2} - a e^{2}\right )} \sqrt {x e + d} \sqrt {\frac {c d}{c d^{2} - a e^{2}}} - a e^{2}}{c d x + a e}\right ) + 2 \, {\left (3 \, c d x e + 4 \, c d^{2} - a e^{2}\right )} \sqrt {x e + d}}{3 \, {\left (2 \, c^{2} d^{5} x e + c^{2} d^{6} - 4 \, a c d^{3} x e^{3} + a^{2} x^{2} e^{6} + 2 \, a^{2} d x e^{5} - {\left (2 \, a c d^{2} x^{2} - a^{2} d^{2}\right )} e^{4} + {\left (c^{2} d^{4} x^{2} - 2 \, a c d^{4}\right )} e^{2}\right )}}, -\frac {2 \, {\left (3 \, {\left (c d x^{2} e^{2} + 2 \, c d^{2} x e + c d^{3}\right )} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}} \arctan \left (-\frac {{\left (c d^{2} - a e^{2}\right )} \sqrt {x e + d} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}}}{c d x e + c d^{2}}\right ) - {\left (3 \, c d x e + 4 \, c d^{2} - a e^{2}\right )} \sqrt {x e + d}\right )}}{3 \, {\left (2 \, c^{2} d^{5} x e + c^{2} d^{6} - 4 \, a c d^{3} x e^{3} + a^{2} x^{2} e^{6} + 2 \, a^{2} d x e^{5} - {\left (2 \, a c d^{2} x^{2} - a^{2} d^{2}\right )} e^{4} + {\left (c^{2} d^{4} x^{2} - 2 \, a c d^{4}\right )} e^{2}\right )}}\right ] \end {gather*}

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),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 5.34, size = 107, normalized size = 0.89 \begin {gather*} \frac {2 c d}{\sqrt {d + e x} \left (a e^{2} - c d^{2}\right )^{2}} + \frac {2 c d \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e^{2} - c d^{2}}{c d}}} \right )}}{\sqrt {\frac {a e^{2} - c d^{2}}{c d}} \left (a e^{2} - c d^{2}\right )^{2}} - \frac {2}{3 \left (d + e x\right )^{\frac {3}{2}} \left (a e^{2} - c d^{2}\right )} \end {gather*}

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),x)

[Out]

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

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Giac [A]
time = 1.30, size = 136, normalized size = 1.13 \begin {gather*} \frac {2 \, c^{2} d^{2} \arctan \left (\frac {\sqrt {x e + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} + \frac {2 \, {\left (3 \, {\left (x e + d\right )} c d + c d^{2} - a e^{2}\right )}}{3 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} {\left (x e + d\right )}^{\frac {3}{2}}} \end {gather*}

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),x, algorithm="giac")

[Out]

2*c^2*d^2*arctan(sqrt(x*e + d)*c*d/sqrt(-c^2*d^3 + a*c*d*e^2))/((c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c^2*
d^3 + a*c*d*e^2)) + 2/3*(3*(x*e + d)*c*d + c*d^2 - a*e^2)/((c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*(x*e + d)^(3/2)
)

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Mupad [B]
time = 0.69, size = 127, normalized size = 1.06 \begin {gather*} \frac {2\,c^{3/2}\,d^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}\,\left (a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4\right )}{{\left (a\,e^2-c\,d^2\right )}^{5/2}}\right )}{{\left (a\,e^2-c\,d^2\right )}^{5/2}}-\frac {\frac {2}{3\,\left (a\,e^2-c\,d^2\right )}-\frac {2\,c\,d\,\left (d+e\,x\right )}{{\left (a\,e^2-c\,d^2\right )}^2}}{{\left (d+e\,x\right )}^{3/2}} \end {gather*}

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)),x)

[Out]

(2*c^(3/2)*d^(3/2)*atan((c^(1/2)*d^(1/2)*(d + e*x)^(1/2)*(a^2*e^4 + c^2*d^4 - 2*a*c*d^2*e^2))/(a*e^2 - c*d^2)^
(5/2)))/(a*e^2 - c*d^2)^(5/2) - (2/(3*(a*e^2 - c*d^2)) - (2*c*d*(d + e*x))/(a*e^2 - c*d^2)^2)/(d + e*x)^(3/2)

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