3.17.13 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{(d+e x)^5} \, dx\)

Optimal. Leaf size=54 \[ \frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 (d+e x)^5 \left (c d^2-a e^2\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {650} \begin {gather*} \frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 (d+e x)^5 \left (c d^2-a e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(d + e*x)^5,x]

[Out]

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

Rule 650

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] &&
EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

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 )^{3/2}}{(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}}{5 \left (c d^2-a e^2\right ) (d+e x)^5}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 43, normalized size = 0.80 \begin {gather*} \frac {2 ((d+e x) (a e+c d x))^{5/2}}{5 (d+e x)^5 \left (c d^2-a e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(d + e*x)^5,x]

[Out]

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

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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)^(3/2)/(d + e*x)^5,x]

[Out]

$Aborted

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fricas [B]  time = 1.22, size = 130, normalized size = 2.41 \begin {gather*} \frac {2 \, {\left (c^{2} d^{2} x^{2} + 2 \, a c d e x + a^{2} e^{2}\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{5 \, {\left (c d^{5} - a d^{3} e^{2} + {\left (c d^{2} e^{3} - a e^{5}\right )} x^{3} + 3 \, {\left (c d^{3} e^{2} - a d e^{4}\right )} x^{2} + 3 \, {\left (c d^{4} e - a d^{2} e^{3}\right )} x\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)^(3/2)/(e*x+d)^5,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \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)^(3/2)/(e*x+d)^5,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: exp(1)*(2*(-(exp(1)*x+d)^-1/exp(1)*(-(-6
*c^4*d^7*sign((exp(1)*x+d)^-1)*exp(1)^6+18*a*c^3*d^5*sign((exp(1)*x+d)^-1)*exp(1)^8-18*a^2*c^2*d^3*sign((exp(1
)*x+d)^-1)*exp(1)^10+6*a^3*c*d*sign((exp(1)*x+d)^-1)*exp(1)^12)/(-15*a^3*exp(1)^15+15*c^3*d^6*exp(1)^9-45*a*c^
2*d^4*exp(1)^11+45*a^2*c*d^2*exp(1)^13)+(exp(1)*x+d)^-1/exp(1)*(-3*a^4*sign((exp(1)*x+d)^-1)*exp(1)^15-3*c^4*d
^8*sign((exp(1)*x+d)^-1)*exp(1)^7+12*a*c^3*d^6*sign((exp(1)*x+d)^-1)*exp(1)^9-18*a^2*c^2*d^4*sign((exp(1)*x+d)
^-1)*exp(1)^11+12*a^3*c*d^2*sign((exp(1)*x+d)^-1)*exp(1)^13)/(-15*a^3*exp(1)^15+15*c^3*d^6*exp(1)^9-45*a*c^2*d
^4*exp(1)^11+45*a^2*c*d^2*exp(1)^13))-(-3*c^4*d^6*sign((exp(1)*x+d)^-1)*exp(1)^5+6*a*c^3*d^4*sign((exp(1)*x+d)
^-1)*exp(1)^7-3*a^2*c^2*d^2*sign((exp(1)*x+d)^-1)*exp(1)^9)/(-15*a^3*exp(1)^15+15*c^3*d^6*exp(1)^9-45*a*c^2*d^
4*exp(1)^11+45*a^2*c*d^2*exp(1)^13)-C_0*(30*a^2*exp(1)^11+30*c^2*d^4*exp(1)^7-60*a*c*d^2*exp(1)^9)/(-15*a^3*ex
p(1)^15+15*c^3*d^6*exp(1)^9-45*a*c^2*d^4*exp(1)^11+45*a^2*c*d^2*exp(1)^13))*sqrt(c*d*exp(1)+a*d*(-(exp(1)*x+d)
^-1/exp(1))^2*exp(1)^5+a*(exp(1)*x+d)^-1/exp(1)*exp(1)^2*exp(2)-c*d^2*(exp(1)*x+d)^-1/exp(1)*exp(1)^2-a*d*(-(e
xp(1)*x+d)^-1/exp(1))^2*exp(1)^3*exp(2))-4*C_0*sqrt(a*d*exp(1)^3-a*d*exp(1)*exp(2))*ln(abs(a*sqrt(a*d*exp(1)^3
-a*d*exp(1)*exp(2))*exp(2)-c*d^2*sqrt(a*d*exp(1)^3-a*d*exp(1)*exp(2))))/(2*a*d*exp(1)^4-2*a*d*exp(2)^2)+2*c^2*
d^2*sqrt(c*d*exp(1))/(5*a*exp(1)^6-5*c*d^2*exp(1)^4)*sign((exp(1)*x+d)^-1))

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maple [A]  time = 0.05, size = 58, normalized size = 1.07 \begin {gather*} -\frac {2 \left (c d x +a e \right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}}{5 \left (e x +d \right )^{4} \left (a \,e^{2}-c \,d^{2}\right )} \end {gather*}

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)^(3/2)/(e*x+d)^5,x)

[Out]

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

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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)^(3/2)/(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 [B]  time = 1.93, size = 901, normalized size = 16.69 \begin {gather*} \frac {\left (\frac {d\,\left (\frac {4\,c^3\,d^4}{5\,\left (a\,e^2-c\,d^2\right )\,\left (3\,a\,e^3-3\,c\,d^2\,e\right )}-\frac {2\,c^2\,d^2\,\left (5\,a\,e^2-c\,d^2\right )}{5\,\left (a\,e^2-c\,d^2\right )\,\left (3\,a\,e^3-3\,c\,d^2\,e\right )}\right )}{e}+\frac {4\,a^2\,c\,d\,e^4+2\,a\,c^2\,d^3\,e^2-2\,c^3\,d^5}{5\,e\,\left (a\,e^2-c\,d^2\right )\,\left (3\,a\,e^3-3\,c\,d^2\,e\right )}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{{\left (d+e\,x\right )}^2}-\frac {\left (\frac {d\,\left (\frac {40\,c^4\,d^5-56\,a\,c^3\,d^3\,e^2}{15\,e\,{\left (a\,e^2-c\,d^2\right )}^3}+\frac {8\,c^4\,d^5}{15\,e\,{\left (a\,e^2-c\,d^2\right )}^3}\right )}{e}+\frac {8\,a\,c^2\,d^2\,\left (6\,a\,e^2-5\,c\,d^2\right )}{15\,{\left (a\,e^2-c\,d^2\right )}^3}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{d+e\,x}-\frac {\left (\frac {2\,a^2\,e^3}{5\,a\,e^3-5\,c\,d^2\,e}+\frac {d\,\left (\frac {2\,c^2\,d^3}{5\,a\,e^3-5\,c\,d^2\,e}-\frac {4\,a\,c\,d\,e^2}{5\,a\,e^3-5\,c\,d^2\,e}\right )}{e}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{{\left (d+e\,x\right )}^3}-\frac {\left (\frac {10\,c^3\,d^4-22\,a\,c^2\,d^2\,e^2}{15\,e^2\,{\left (a\,e^2-c\,d^2\right )}^2}+\frac {4\,c^3\,d^4}{5\,e^2\,{\left (a\,e^2-c\,d^2\right )}^2}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{d+e\,x}-\frac {\left (\frac {d\,\left (\frac {12\,c^3\,d^4-20\,a\,c^2\,d^2\,e^2}{5\,\left (a\,e^2-c\,d^2\right )\,\left (3\,a\,e^3-3\,c\,d^2\,e\right )}+\frac {4\,c^3\,d^4}{5\,\left (a\,e^2-c\,d^2\right )\,\left (3\,a\,e^3-3\,c\,d^2\,e\right )}\right )}{e}+\frac {4\,a\,c\,d\,e\,\left (4\,a\,e^2-3\,c\,d^2\right )}{5\,\left (a\,e^2-c\,d^2\right )\,\left (3\,a\,e^3-3\,c\,d^2\,e\right )}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{{\left (d+e\,x\right )}^2}+\frac {\left (\frac {d\,\left (\frac {8\,c^4\,d^5}{15\,e\,{\left (a\,e^2-c\,d^2\right )}^3}-\frac {4\,c^3\,d^3\,\left (11\,a\,e^2-7\,c\,d^2\right )}{15\,e\,{\left (a\,e^2-c\,d^2\right )}^3}\right )}{e}+\frac {20\,a^2\,c^2\,d^2\,e^4+4\,a\,c^3\,d^4\,e^2-16\,c^4\,d^6}{15\,e^2\,{\left (a\,e^2-c\,d^2\right )}^3}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{d+e\,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)^(3/2)/(d + e*x)^5,x)

[Out]

(((d*((4*c^3*d^4)/(5*(a*e^2 - c*d^2)*(3*a*e^3 - 3*c*d^2*e)) - (2*c^2*d^2*(5*a*e^2 - c*d^2))/(5*(a*e^2 - c*d^2)
*(3*a*e^3 - 3*c*d^2*e))))/e + (2*a*c^2*d^3*e^2 - 2*c^3*d^5 + 4*a^2*c*d*e^4)/(5*e*(a*e^2 - c*d^2)*(3*a*e^3 - 3*
c*d^2*e)))*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/(d + e*x)^2 - (((d*((40*c^4*d^5 - 56*a*c^3*d^3*e^2)/
(15*e*(a*e^2 - c*d^2)^3) + (8*c^4*d^5)/(15*e*(a*e^2 - c*d^2)^3)))/e + (8*a*c^2*d^2*(6*a*e^2 - 5*c*d^2))/(15*(a
*e^2 - c*d^2)^3))*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/(d + e*x) - (((2*a^2*e^3)/(5*a*e^3 - 5*c*d^2*
e) + (d*((2*c^2*d^3)/(5*a*e^3 - 5*c*d^2*e) - (4*a*c*d*e^2)/(5*a*e^3 - 5*c*d^2*e)))/e)*(x*(a*e^2 + c*d^2) + a*d
*e + c*d*e*x^2)^(1/2))/(d + e*x)^3 - (((10*c^3*d^4 - 22*a*c^2*d^2*e^2)/(15*e^2*(a*e^2 - c*d^2)^2) + (4*c^3*d^4
)/(5*e^2*(a*e^2 - c*d^2)^2))*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/(d + e*x) - (((d*((12*c^3*d^4 - 20
*a*c^2*d^2*e^2)/(5*(a*e^2 - c*d^2)*(3*a*e^3 - 3*c*d^2*e)) + (4*c^3*d^4)/(5*(a*e^2 - c*d^2)*(3*a*e^3 - 3*c*d^2*
e))))/e + (4*a*c*d*e*(4*a*e^2 - 3*c*d^2))/(5*(a*e^2 - c*d^2)*(3*a*e^3 - 3*c*d^2*e)))*(x*(a*e^2 + c*d^2) + a*d*
e + c*d*e*x^2)^(1/2))/(d + e*x)^2 + (((d*((8*c^4*d^5)/(15*e*(a*e^2 - c*d^2)^3) - (4*c^3*d^3*(11*a*e^2 - 7*c*d^
2))/(15*e*(a*e^2 - c*d^2)^3)))/e + (4*a*c^3*d^4*e^2 - 16*c^4*d^6 + 20*a^2*c^2*d^2*e^4)/(15*e^2*(a*e^2 - c*d^2)
^3))*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/(d + e*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 {3}{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)**(3/2)/(e*x+d)**5,x)

[Out]

Integral(((d + e*x)*(a*e + c*d*x))**(3/2)/(d + e*x)**5, x)

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