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

Optimal. Leaf size=226 \[ -\frac {5 c^{3/2} d^{3/2} \left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{2 e^{7/2}}+\frac {5 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e^3}-\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 e (d+e x)^4}-\frac {10 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e^2 (d+e x)^2} \]

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Rubi [A]  time = 0.15, antiderivative size = 226, 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 = {662, 664, 621, 206} \begin {gather*} \frac {5 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e^3}-\frac {5 c^{3/2} d^{3/2} \left (c d^2-a e^2\right ) \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 )}{2 e^{7/2}}-\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 e (d+e x)^4}-\frac {10 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)^2} \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)^5,x]

[Out]

(5*c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/e^3 - (10*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)^2) - (2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(3*e*(d + e*x)^4) - (5*c^(3/2)*
d^(3/2)*(c*d^2 - a*e^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])])/(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]

Rule 664

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 + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(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] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*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)^5} \, dx &=-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^4}+\frac {(5 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)^3} \, dx}{3 e}\\ &=-\frac {10 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)^2}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^4}+\frac {\left (5 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} \, dx}{e^2}\\ &=\frac {5 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^3}-\frac {10 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)^2}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^4}-\frac {\left (5 c^2 d^2 \left (c d^2-a e^2\right )\right ) \int \frac {1}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{2 e^3}\\ &=\frac {5 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^3}-\frac {10 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)^2}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^4}-\frac {\left (5 c^2 d^2 \left (c d^2-a e^2\right )\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 {5 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^3}-\frac {10 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)^2}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^4}-\frac {5 c^{3/2} d^{3/2} \left (c d^2-a e^2\right ) \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 )}{2 e^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 112, normalized size = 0.50 \begin {gather*} \frac {2 c^2 d^2 (a e+c d x)^3 \sqrt {(d+e x) (a e+c d x)} \, _2F_1\left (\frac {5}{2},\frac {7}{2};\frac {9}{2};\frac {e (a e+c d x)}{a e^2-c d^2}\right )}{7 \left (c d^2-a e^2\right )^3 \sqrt {\frac {c d (d+e x)}{c d^2-a e^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)^5,x]

[Out]

(2*c^2*d^2*(a*e + c*d*x)^3*Sqrt[(a*e + c*d*x)*(d + e*x)]*Hypergeometric2F1[5/2, 7/2, 9/2, (e*(a*e + c*d*x))/(-
(c*d^2) + a*e^2)])/(7*(c*d^2 - a*e^2)^3*Sqrt[(c*d*(d + e*x))/(c*d^2 - a*e^2)])

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IntegrateAlgebraic [B]  time = 13.01, size = 13728, normalized size = 60.74 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

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fricas [A]  time = 0.85, size = 594, normalized size = 2.63 \begin {gather*} \left [\frac {15 \, {\left (c^{2} d^{5} - a c d^{3} e^{2} + {\left (c^{2} d^{3} e^{2} - a c d e^{4}\right )} x^{2} + 2 \, {\left (c^{2} d^{4} e - a c d^{2} e^{3}\right )} x\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 (3 \, c^{2} d^{2} e^{2} x^{2} + 15 \, c^{2} d^{4} - 10 \, a c d^{2} e^{2} - 2 \, a^{2} e^{4} + 2 \, {\left (10 \, c^{2} d^{3} e - 7 \, 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}}{12 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}}, \frac {15 \, {\left (c^{2} d^{5} - a c d^{3} e^{2} + {\left (c^{2} d^{3} e^{2} - a c d e^{4}\right )} x^{2} + 2 \, {\left (c^{2} d^{4} e - a c d^{2} e^{3}\right )} x\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 (3 \, c^{2} d^{2} e^{2} x^{2} + 15 \, c^{2} d^{4} - 10 \, a c d^{2} e^{2} - 2 \, a^{2} e^{4} + 2 \, {\left (10 \, c^{2} d^{3} e - 7 \, 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}}{6 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} 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)^5,x, algorithm="fricas")

[Out]

[1/12*(15*(c^2*d^5 - a*c*d^3*e^2 + (c^2*d^3*e^2 - a*c*d*e^4)*x^2 + 2*(c^2*d^4*e - a*c*d^2*e^3)*x)*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*(3*c^2*d^2*e^2*x^2 + 15*c^2*d^4 - 1
0*a*c*d^2*e^2 - 2*a^2*e^4 + 2*(10*c^2*d^3*e - 7*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(e^
5*x^2 + 2*d*e^4*x + d^2*e^3), 1/6*(15*(c^2*d^5 - a*c*d^3*e^2 + (c^2*d^3*e^2 - a*c*d*e^4)*x^2 + 2*(c^2*d^4*e -
a*c*d^2*e^3)*x)*sqrt(-c*d/e)*arctan(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*(3*c^2*d^2*e^2*x^2 + 15*c^2*d^4 - 10
*a*c*d^2*e^2 - 2*a^2*e^4 + 2*(10*c^2*d^3*e - 7*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(e^5
*x^2 + 2*d*e^4*x + d^2*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)^5,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.06, size = 1695, normalized size = 7.50

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)^5,x)

[Out]

-2/3/e^5/(a*e^2-c*d^2)/(x+d/e)^5*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(7/2)-8/3/e^4*c*d/(a*e^2-c*d^2)^2/(x+
d/e)^4*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(7/2)+16/e^3*c^2*d^2/(a*e^2-c*d^2)^3/(x+d/e)^3*((x+d/e)^2*c*d*e
+(a*e^2-c*d^2)*(x+d/e))^(7/2)-128/3/e^2*c^3*d^3/(a*e^2-c*d^2)^4/(x+d/e)^2*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/
e))^(7/2)+128/3/e*c^4*d^4/(a*e^2-c*d^2)^4*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(5/2)+5/e^3*c^6*d^10/(a*e^2-
c*d^2)^4*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)-80/3/e*c^5*d^6/(a*e^2-c*d^2)^4*((x+d/e)^2*c*d*e+(a*e^2-
c*d^2)*(x+d/e))^(3/2)*x-40/3/e^2*c^5*d^7/(a*e^2-c*d^2)^4*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)+80/3*e*
c^4*d^4/(a*e^2-c*d^2)^4*a*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)*x+10*e^3*c^3*d^4/(a*e^2-c*d^2)^4*a^3*(
(x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)-25*e*c^5*d^8/(a*e^2-c*d^2)^4*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)+25/2/e*c^6*d^10/(a*e^2-c*d^2
)^4*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^4*c^3*d^3/(a*e^2-c*d^2)^4*a^3*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x-30*c^5*d^7/(a*e^2
-c*d^2)^4*a*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x+5/2*e^7*c^2*d^2/(a*e^2-c*d^2)^4*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)-5/2/e^3*c^
7*d^12/(a*e^2-c*d^2)^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)-25/2*e^5*c^3*d^4/(a*e^2-c*d^2)^4*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)+25*e^3*c^4*d^6/(a*e^2-c*d^2)^4*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)+10/
e^2*c^6*d^9/(a*e^2-c*d^2)^4*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)*x+40/3*e^2*c^3*d^3/(a*e^2-c*d^2)^4*a
^2*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)+30*e^2*c^4*d^5/(a*e^2-c*d^2)^4*a^2*((x+d/e)^2*c*d*e+(a*e^2-c*
d^2)*(x+d/e))^(1/2)*x-5*e^5*c^2*d^2/(a*e^2-c*d^2)^4*a^4*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)-10/e*c^5
*d^8/(a*e^2-c*d^2)^4*a*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+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)^5,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 )}^5} \,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)^5,x)

[Out]

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

[Out]

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

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