3.2014 \(\int \frac {(d+e x)^{3/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\)

Optimal. Leaf size=101 \[ \frac {e \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {c} \sqrt {d} \left (c d^2-a e^2\right )^{3/2}}-\frac {\sqrt {d+e x}}{\left (c d^2-a e^2\right ) (a e+c d x)} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 101, 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 = {626, 51, 63, 208} \[ \frac {e \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {c} \sqrt {d} \left (c d^2-a e^2\right )^{3/2}}-\frac {\sqrt {d+e x}}{\left (c d^2-a e^2\right ) (a e+c d x)} \]

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)^2,x]

[Out]

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

Rule 51

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 63

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 208

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

Rule 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^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}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx &=\int \frac {1}{(a e+c d x)^2 \sqrt {d+e x}} \, dx\\ &=-\frac {\sqrt {d+e x}}{\left (c d^2-a e^2\right ) (a e+c d x)}-\frac {e \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{2 \left (c d^2-a e^2\right )}\\ &=-\frac {\sqrt {d+e x}}{\left (c d^2-a e^2\right ) (a e+c d x)}-\frac {\operatorname {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 )}{c d^2-a e^2}\\ &=-\frac {\sqrt {d+e x}}{\left (c d^2-a e^2\right ) (a e+c d x)}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {c} \sqrt {d} \left (c d^2-a e^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 101, normalized size = 1.00 \[ \frac {e \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {a e^2-c d^2}}\right )}{\sqrt {c} \sqrt {d} \left (a e^2-c d^2\right )^{3/2}}-\frac {\sqrt {d+e x}}{\left (c d^2-a e^2\right ) (a e+c d x)} \]

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)^2,x]

[Out]

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

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fricas [A]  time = 1.00, size = 353, normalized size = 3.50 \[ \left [-\frac {\sqrt {c^{2} d^{3} - a c d e^{2}} {\left (c d e x + a e^{2}\right )} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {c^{2} d^{3} - a c d e^{2}} \sqrt {e x + d}}{c d x + a e}\right ) + 2 \, {\left (c^{2} d^{3} - a c d e^{2}\right )} \sqrt {e x + d}}{2 \, {\left (a c^{3} d^{5} e - 2 \, a^{2} c^{2} d^{3} e^{3} + a^{3} c d e^{5} + {\left (c^{4} d^{6} - 2 \, a c^{3} d^{4} e^{2} + a^{2} c^{2} d^{2} e^{4}\right )} x\right )}}, -\frac {\sqrt {-c^{2} d^{3} + a c d e^{2}} {\left (c d e x + a e^{2}\right )} \arctan \left (\frac {\sqrt {-c^{2} d^{3} + a c d e^{2}} \sqrt {e x + d}}{c d e x + c d^{2}}\right ) + {\left (c^{2} d^{3} - a c d e^{2}\right )} \sqrt {e x + d}}{a c^{3} d^{5} e - 2 \, a^{2} c^{2} d^{3} e^{3} + a^{3} c d e^{5} + {\left (c^{4} d^{6} - 2 \, a c^{3} d^{4} e^{2} + a^{2} c^{2} d^{2} e^{4}\right )} x}\right ] \]

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)^2,x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.06, size = 99, normalized size = 0.98 \[ \frac {e \arctan \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}\right )}{\left (a \,e^{2}-c \,d^{2}\right ) \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}+\frac {\sqrt {e x +d}\, e}{\left (a \,e^{2}-c \,d^{2}\right ) \left (c d e x +a \,e^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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)^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(a*e^2-c*d^2>0)', see `assume?`
for more details)Is a*e^2-c*d^2 positive or negative?

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mupad [B]  time = 0.66, size = 97, normalized size = 0.96 \[ \frac {e\,\mathrm {atan}\left (\frac {c\,d\,\sqrt {d+e\,x}}{\sqrt {c\,d}\,\sqrt {a\,e^2-c\,d^2}}\right )}{\sqrt {c\,d}\,{\left (a\,e^2-c\,d^2\right )}^{3/2}}+\frac {e\,\sqrt {d+e\,x}}{\left (a\,e^2-c\,d^2\right )\,\left (a\,e^2-c\,d^2+c\,d\,\left (d+e\,x\right )\right )} \]

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)^2,x)

[Out]

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

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sympy [B]  time = 146.59, size = 309, normalized size = 3.06 \[ - \frac {e \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} \log {\left (- a^{2} e^{4} \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} + 2 a c d^{2} e^{2} \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} - c^{2} d^{4} \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} + \sqrt {d + e x} \right )}}{2} + \frac {e \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} \log {\left (a^{2} e^{4} \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} - 2 a c d^{2} e^{2} \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} + c^{2} d^{4} \sqrt {- \frac {1}{c d \left (a e^{2} - c d^{2}\right )^{3}}} + \sqrt {d + e x} \right )}}{2} + \frac {2 e \sqrt {d + e x}}{2 a^{2} e^{4} - 2 a c d^{2} e^{2} + 2 a c d e^{3} x - 2 c^{2} d^{3} e x} \]

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)**2,x)

[Out]

-e*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3))*log(-a**2*e**4*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3)) + 2*a*c*d**2*e**2*sq
rt(-1/(c*d*(a*e**2 - c*d**2)**3)) - c**2*d**4*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3)) + sqrt(d + e*x))/2 + e*sqrt(
-1/(c*d*(a*e**2 - c*d**2)**3))*log(a**2*e**4*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3)) - 2*a*c*d**2*e**2*sqrt(-1/(c*
d*(a*e**2 - c*d**2)**3)) + c**2*d**4*sqrt(-1/(c*d*(a*e**2 - c*d**2)**3)) + sqrt(d + e*x))/2 + 2*e*sqrt(d + e*x
)/(2*a**2*e**4 - 2*a*c*d**2*e**2 + 2*a*c*d*e**3*x - 2*c**2*d**3*e*x)

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