3.17.27 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^6} \, dx\)

Optimal. Leaf size=218 \[ \frac {c^{5/2} d^{5/2} \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{e^{7/2}}-\frac {2 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e^3 (d+e x)}-\frac {2 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e^2 (d+e x)^3}-\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 e (d+e x)^5} \]

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Rubi [A]  time = 0.13, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {662, 621, 206} \begin {gather*} -\frac {2 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e^3 (d+e x)}+\frac {c^{5/2} d^{5/2} \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{e^{7/2}}-\frac {2 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e^2 (d+e x)^3}-\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 e (d+e x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^6,x]

[Out]

(-2*c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(e^3*(d + e*x)) - (2*c*d*(a*d*e + (c*d^2 + a*e^2)*x +
c*d*e*x^2)^(3/2))/(3*e^2*(d + e*x)^3) - (2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(5*e*(d + e*x)^5) +
(c^(5/2)*d^(5/2)*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*
x + c*d*e*x^2])])/e^(7/2)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 662

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

Rubi steps

\begin {align*} \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^6} \, dx &=-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e (d+e x)^5}+\frac {(c d) \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^4} \, dx}{e}\\ &=-\frac {2 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 e^2 (d+e x)^3}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e (d+e x)^5}+\frac {\left (c^2 d^2\right ) \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^2} \, dx}{e^2}\\ &=-\frac {2 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^3 (d+e x)}-\frac {2 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 e^2 (d+e x)^3}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e (d+e x)^5}+\frac {\left (c^3 d^3\right ) \int \frac {1}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{e^3}\\ &=-\frac {2 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^3 (d+e x)}-\frac {2 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 e^2 (d+e x)^3}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e (d+e x)^5}+\frac {\left (2 c^3 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{4 c d e-x^2} \, dx,x,\frac {c d^2+a e^2+2 c d e x}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{e^3}\\ &=-\frac {2 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^3 (d+e x)}-\frac {2 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 e^2 (d+e x)^3}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 e (d+e x)^5}+\frac {c^{5/2} d^{5/2} \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{e^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 1.12, size = 217, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {(d+e x) (a e+c d x)} \left (\frac {15 c^{5/2} d^{5/2} \sqrt {c d} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d} \sqrt {c d^2-a e^2}}\right )}{\sqrt {c d^2-a e^2} \sqrt {a e+c d x} \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}}}-\frac {\sqrt {e} \left (3 a^2 e^4+a c d e^2 (5 d+11 e x)+c^2 d^2 \left (15 d^2+35 d e x+23 e^2 x^2\right )\right )}{(d+e x)^3}\right )}{15 e^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^6,x]

[Out]

(2*Sqrt[(a*e + c*d*x)*(d + e*x)]*(-((Sqrt[e]*(3*a^2*e^4 + a*c*d*e^2*(5*d + 11*e*x) + c^2*d^2*(15*d^2 + 35*d*e*
x + 23*e^2*x^2)))/(d + e*x)^3) + (15*c^(5/2)*d^(5/2)*Sqrt[c*d]*ArcSinh[(Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d
*x])/(Sqrt[c*d]*Sqrt[c*d^2 - a*e^2])])/(Sqrt[c*d^2 - a*e^2]*Sqrt[a*e + c*d*x]*Sqrt[(c*d*(d + e*x))/(c*d^2 - a*
e^2)])))/(15*e^(7/2))

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IntegrateAlgebraic [F]  time = 180.01, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(d + e*x)^6,x]

[Out]

$Aborted

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fricas [A]  time = 1.83, size = 578, normalized size = 2.65 \begin {gather*} \left [\frac {15 \, {\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\right )} \sqrt {\frac {c d}{e}} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} + 4 \, {\left (2 \, c d e^{2} x + c d^{2} e + a e^{3}\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {\frac {c d}{e}} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right ) - 4 \, {\left (23 \, c^{2} d^{2} e^{2} x^{2} + 15 \, c^{2} d^{4} + 5 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4} + {\left (35 \, c^{2} d^{3} e + 11 \, a c d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{30 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}}, -\frac {15 \, {\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\right )} \sqrt {-\frac {c d}{e}} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {-\frac {c d}{e}}}{2 \, {\left (c^{2} d^{2} e x^{2} + a c d^{2} e + {\left (c^{2} d^{3} + a c d e^{2}\right )} x\right )}}\right ) + 2 \, {\left (23 \, c^{2} d^{2} e^{2} x^{2} + 15 \, c^{2} d^{4} + 5 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4} + {\left (35 \, c^{2} d^{3} e + 11 \, a c d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{15 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^6,x, algorithm="fricas")

[Out]

[1/30*(15*(c^2*d^2*e^3*x^3 + 3*c^2*d^3*e^2*x^2 + 3*c^2*d^4*e*x + c^2*d^5)*sqrt(c*d/e)*log(8*c^2*d^2*e^2*x^2 +
c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4 + 4*(2*c*d*e^2*x + c*d^2*e + a*e^3)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)
*x)*sqrt(c*d/e) + 8*(c^2*d^3*e + a*c*d*e^3)*x) - 4*(23*c^2*d^2*e^2*x^2 + 15*c^2*d^4 + 5*a*c*d^2*e^2 + 3*a^2*e^
4 + (35*c^2*d^3*e + 11*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(e^6*x^3 + 3*d*e^5*x^2 + 3*d
^2*e^4*x + d^3*e^3), -1/15*(15*(c^2*d^2*e^3*x^3 + 3*c^2*d^3*e^2*x^2 + 3*c^2*d^4*e*x + c^2*d^5)*sqrt(-c*d/e)*ar
ctan(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(-c*d/e)/(c^2*d^2*e*x^2 +
a*c*d^2*e + (c^2*d^3 + a*c*d*e^2)*x)) + 2*(23*c^2*d^2*e^2*x^2 + 15*c^2*d^4 + 5*a*c*d^2*e^2 + 3*a^2*e^4 + (35*
c^2*d^3*e + 11*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*e^4*x
+ d^3*e^3)]

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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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^6,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.06, size = 1764, normalized size = 8.09

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(5/2)/(e*x+d)^6,x)

[Out]

-2/5/e^6/(a*e^2-c*d^2)/(x+d/e)^6*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(7/2)-4/15/e^5*c*d/(a*e^2-c*d^2)^2/(x
+d/e)^5*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(7/2)-16/15/e^4*c^2*d^2/(a*e^2-c*d^2)^3/(x+d/e)^4*((x+d/e)^2*c
*d*e+(a*e^2-c*d^2)*(x+d/e))^(7/2)+32/5/e^3*c^3*d^3/(a*e^2-c*d^2)^4/(x+d/e)^3*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x
+d/e))^(7/2)-256/15/e^2*c^4*d^4/(a*e^2-c*d^2)^5/(x+d/e)^2*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(7/2)+256/15
/e*c^5*d^5/(a*e^2-c*d^2)^5*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(5/2)+2/e^3*c^7*d^11/(a*e^2-c*d^2)^5*((x+d/
e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)-32/3/e*c^6*d^7/(a*e^2-c*d^2)^5*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))
^(3/2)*x-16/3/e^2*c^6*d^8/(a*e^2-c*d^2)^5*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)+4*e^3*c^4*d^5/(a*e^2-c
*d^2)^5*a^3*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)+32/3*e*c^5*d^5/(a*e^2-c*d^2)^5*a*((x+d/e)^2*c*d*e+(a
*e^2-c*d^2)*(x+d/e))^(3/2)*x+5/e*c^7*d^11/(a*e^2-c*d^2)^5*a*ln((1/2*a*e^2-1/2*c*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/
2)+((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2))/(c*d*e)^(1/2)-10*e*c^6*d^9/(a*e^2-c*d^2)^5*a^2*ln((1/2*a*e^2
-1/2*c*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2))/(c*d*e)^(1/2)+e^7*c^3*d
^3/(a*e^2-c*d^2)^5*a^5*ln((1/2*a*e^2-1/2*c*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+
d/e))^(1/2))/(c*d*e)^(1/2)-12*c^6*d^8/(a*e^2-c*d^2)^5*a*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x-4*e^4*
c^4*d^4/(a*e^2-c*d^2)^5*a^3*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x+16/3*e^2*c^4*d^4/(a*e^2-c*d^2)^5*a
^2*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)+12*e^2*c^5*d^6/(a*e^2-c*d^2)^5*a^2*((x+d/e)^2*c*d*e+(a*e^2-c*
d^2)*(x+d/e))^(1/2)*x-2*e^5*c^3*d^3/(a*e^2-c*d^2)^5*a^4*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)+4/e^2*c^
7*d^10/(a*e^2-c*d^2)^5*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x-1/e^3*c^8*d^13/(a*e^2-c*d^2)^5*ln((1/2*
a*e^2-1/2*c*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2))/(c*d*e)^(1/2)-4/e*
c^6*d^9/(a*e^2-c*d^2)^5*a*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)-5*e^5*c^4*d^5/(a*e^2-c*d^2)^5*a^4*ln((
1/2*a*e^2-1/2*c*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2))/(c*d*e)^(1/2)+
10*e^3*c^5*d^7/(a*e^2-c*d^2)^5*a^3*ln((1/2*a*e^2-1/2*c*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((x+d/e)^2*c*d*e+(a*e^
2-c*d^2)*(x+d/e))^(1/2))/(c*d*e)^(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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^6,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 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/2}}{{\left (d+e\,x\right )}^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/(d + e*x)^6,x)

[Out]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/(d + e*x)^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {5}{2}}}{\left (d + e x\right )^{6}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2)/(e*x+d)**6,x)

[Out]

Integral(((d + e*x)*(a*e + c*d*x))**(5/2)/(d + e*x)**6, x)

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