3.17.97 \(\int \frac {(d+e x)^{5/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\)

Optimal. Leaf size=94 \[ -\frac {e \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{3/2} d^{3/2} \sqrt {c d^2-a e^2}}-\frac {\sqrt {d+e x}}{c d (a e+c d x)} \]

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Rubi [A]  time = 0.06, antiderivative size = 94, 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, 47, 63, 208} \begin {gather*} -\frac {e \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} \sqrt {c d^2-a e^2}}-\frac {\sqrt {d+e x}}{c d (a e+c d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(5/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx &=\int \frac {\sqrt {d+e x}}{(a e+c d x)^2} \, dx\\ &=-\frac {\sqrt {d+e x}}{c d (a e+c d x)}+\frac {e \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{2 c d}\\ &=-\frac {\sqrt {d+e x}}{c d (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}\\ &=-\frac {\sqrt {d+e x}}{c d (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 )}{c^{3/2} d^{3/2} \sqrt {c d^2-a e^2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 93, normalized size = 0.99 \begin {gather*} \frac {e \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {a e^2-c d^2}}\right )}{c^{3/2} d^{3/2} \sqrt {a e^2-c d^2}}-\frac {\sqrt {d+e x}}{a c d e+c^2 d^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(5/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.26, size = 121, normalized size = 1.29 \begin {gather*} \frac {e \sqrt {d+e x}}{c d \left (-a e^2+c d^2-c d (d+e x)\right )}-\frac {e \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x} \sqrt {a e^2-c d^2}}{c d^2-a e^2}\right )}{c^{3/2} d^{3/2} \sqrt {a e^2-c d^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^(5/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]

[Out]

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

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fricas [A]  time = 0.41, size = 309, normalized size = 3.29 \begin {gather*} \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^{4} e - a^{2} c^{2} d^{2} e^{3} + {\left (c^{4} d^{5} - a c^{3} d^{3} e^{2}\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^{4} e - a^{2} c^{2} d^{2} e^{3} + {\left (c^{4} d^{5} - a c^{3} d^{3} e^{2}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(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 = 84, normalized size = 0.89 \begin {gather*} \frac {e \arctan \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}\right )}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}\, c d}-\frac {\sqrt {e x +d}\, e}{\left (c d e x +a \,e^{2}\right ) c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(5/2)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^2,x)

[Out]

-e/c/d*(e*x+d)^(1/2)/(c*d*e*x+a*e^2)+e/c/d/((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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(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 = 0.65, size = 81, normalized size = 0.86 \begin {gather*} \frac {e\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}}{\sqrt {a\,e^2-c\,d^2}}\right )}{c^{3/2}\,d^{3/2}\,\sqrt {a\,e^2-c\,d^2}}-\frac {e\,\sqrt {d+e\,x}}{x\,c^2\,d^2\,e+a\,c\,d\,e^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(5/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^2,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(5/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2,x)

[Out]

Timed out

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