3.21.3 \(\int \frac {(d+e x)^{3/2}}{a d e+(c d^2+a e^2) x+c d e x^2} \, dx\) [2003]

Optimal. Leaf size=83 \[ \frac {2 \sqrt {d+e x}}{c d}-\frac {2 \sqrt {c d^2-a e^2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{3/2} d^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {640, 52, 65, 214} \begin {gather*} \frac {2 \sqrt {d+e x}}{c d}-\frac {2 \sqrt {c d^2-a e^2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{3/2} d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 {(d+e x)^{3/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx &=\int \frac {\sqrt {d+e x}}{a e+c d x} \, dx\\ &=\frac {2 \sqrt {d+e x}}{c d}+\frac {\left (c d^2-a e^2\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{c d}\\ &=\frac {2 \sqrt {d+e x}}{c d}+\left (2 \left (\frac {d}{e}-\frac {a e}{c d}\right )\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 )\\ &=\frac {2 \sqrt {d+e x}}{c d}-\frac {2 \sqrt {c d^2-a e^2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{3/2} d^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 83, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {d+e x}}{c d}-\frac {2 \sqrt {-c d^2+a e^2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{c^{3/2} d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.75, size = 82, normalized size = 0.99

method	result	size
derivativedivides	\(\frac {2 \sqrt {e x +d}}{c d}+\frac {2 \left (-e^{2} a +c \,d^{2}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{c d \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\)	\(82\)
default	\(\frac {2 \sqrt {e x +d}}{c d}+\frac {2 \left (-e^{2} a +c \,d^{2}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{c d \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\)	\(82\)
risch	\(\frac {2 \textit {\_O1} \sqrt {e x +d}}{d}-\frac {2 \arctan \left (\frac {\sqrt {e x +d}\, d}{\sqrt {\left (\textit {\_O1} a \,e^{2}-d^{2}\right ) d}}\right ) e^{2} a}{c^{2} d \sqrt {\left (\textit {\_O1} a \,e^{2}-d^{2}\right ) d}}+\frac {2 d \arctan \left (\frac {\sqrt {e x +d}\, d}{\sqrt {\left (\textit {\_O1} a \,e^{2}-d^{2}\right ) d}}\right )}{c \sqrt {\left (\textit {\_O1} a \,e^{2}-d^{2}\right ) d}}\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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*(e*x+d)^(1/2)/c/d+2*(-a*e^2+c*d^2)/c/d/((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((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 [A]
time = 2.87, size = 191, normalized size = 2.30 \begin {gather*} \left [\frac {\sqrt {\frac {c d^{2} - a e^{2}}{c d}} \log \left (\frac {c d x e + 2 \, c d^{2} - 2 \, \sqrt {x e + d} c d \sqrt {\frac {c d^{2} - a e^{2}}{c d}} - a e^{2}}{c d x + a e}\right ) + 2 \, \sqrt {x e + d}}{c d}, -\frac {2 \, {\left (\sqrt {-\frac {c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac {\sqrt {x e + d} c d \sqrt {-\frac {c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) - \sqrt {x e + d}\right )}}{c d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 4.35, size = 80, normalized size = 0.96 \begin {gather*} \frac {2 \left (\frac {e \sqrt {d + e x}}{c d} - \frac {e \left (a e^{2} - c d^{2}\right ) \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e^{2} - c d^{2}}{c d}}} \right )}}{c^{2} d^{2} \sqrt {\frac {a e^{2} - c d^{2}}{c d}}}\right )}{e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.53, size = 82, normalized size = 0.99 \begin {gather*} \frac {2 \, {\left (c d^{2} - a e^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{\sqrt {-c^{2} d^{3} + a c d e^{2}} c d} + \frac {2 \, \sqrt {x e + d}}{c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [B]
time = 0.07, size = 67, normalized size = 0.81 \begin {gather*} \frac {2\,\sqrt {d+e\,x}}{c\,d}-\frac {2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}}{\sqrt {a\,e^2-c\,d^2}}\right )\,\sqrt {a\,e^2-c\,d^2}}{c^{3/2}\,d^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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