3.17.47 \(\int \frac {1}{(d+e x)^2 (a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=181 \[ -\frac {16 c^2 d^2 \left (a e^2+c d^2+2 c d e x\right )}{5 \left (c d^2-a e^2\right )^4 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac {4 c d}{5 (d+e x) \left (c d^2-a e^2\right )^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac {2}{5 (d+e x)^2 \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {658, 613} \begin {gather*} -\frac {16 c^2 d^2 \left (a e^2+c d^2+2 c d e x\right )}{5 \left (c d^2-a e^2\right )^4 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac {4 c d}{5 (d+e x) \left (c d^2-a e^2\right )^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac {2}{5 (d+e x)^2 \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]

[Out]

2/(5*(c*d^2 - a*e^2)*(d + e*x)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (4*c*d)/(5*(c*d^2 - a*e^2)^2*(
d + e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (16*c^2*d^2*(c*d^2 + a*e^2 + 2*c*d*e*x))/(5*(c*d^2 - a
*e^2)^4*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 658

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))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(2*c*d -
b*e)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], 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] && ILtQ[Simplify[m + 2*p + 2], 0]

Rubi steps

\begin {align*} \int \frac {1}{(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=\frac {2}{5 \left (c d^2-a e^2\right ) (d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {(6 c d) \int \frac {1}{(d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{5 \left (c d^2-a e^2\right )}\\ &=\frac {2}{5 \left (c d^2-a e^2\right ) (d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {4 c d}{5 \left (c d^2-a e^2\right )^2 (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {\left (8 c^2 d^2\right ) \int \frac {1}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{5 \left (c d^2-a e^2\right )^2}\\ &=\frac {2}{5 \left (c d^2-a e^2\right ) (d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {4 c d}{5 \left (c d^2-a e^2\right )^2 (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {16 c^2 d^2 \left (c d^2+a e^2+2 c d e x\right )}{5 \left (c d^2-a e^2\right )^4 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 136, normalized size = 0.75 \begin {gather*} -\frac {2 \left (a^3 e^6-a^2 c d e^4 (5 d+2 e x)+a c^2 d^2 e^2 \left (15 d^2+20 d e x+8 e^2 x^2\right )+c^3 d^3 \left (5 d^3+30 d^2 e x+40 d e^2 x^2+16 e^3 x^3\right )\right )}{5 (d+e x)^2 \left (c d^2-a e^2\right )^4 \sqrt {(d+e x) (a e+c d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]

[Out]

(-2*(a^3*e^6 - a^2*c*d*e^4*(5*d + 2*e*x) + a*c^2*d^2*e^2*(15*d^2 + 20*d*e*x + 8*e^2*x^2) + c^3*d^3*(5*d^3 + 30
*d^2*e*x + 40*d*e^2*x^2 + 16*e^3*x^3)))/(5*(c*d^2 - a*e^2)^4*(d + e*x)^2*Sqrt[(a*e + c*d*x)*(d + e*x)])

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IntegrateAlgebraic [F]  time = 180.02, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/((d + e*x)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]

[Out]

$Aborted

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fricas [B]  time = 6.42, size = 494, normalized size = 2.73 \begin {gather*} -\frac {2 \, {\left (16 \, c^{3} d^{3} e^{3} x^{3} + 5 \, c^{3} d^{6} + 15 \, a c^{2} d^{4} e^{2} - 5 \, a^{2} c d^{2} e^{4} + a^{3} e^{6} + 8 \, {\left (5 \, c^{3} d^{4} e^{2} + a c^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (15 \, c^{3} d^{5} e + 10 \, a c^{2} d^{3} e^{3} - a^{2} c d e^{5}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{5 \, {\left (a c^{4} d^{11} e - 4 \, a^{2} c^{3} d^{9} e^{3} + 6 \, a^{3} c^{2} d^{7} e^{5} - 4 \, a^{4} c d^{5} e^{7} + a^{5} d^{3} e^{9} + {\left (c^{5} d^{9} e^{3} - 4 \, a c^{4} d^{7} e^{5} + 6 \, a^{2} c^{3} d^{5} e^{7} - 4 \, a^{3} c^{2} d^{3} e^{9} + a^{4} c d e^{11}\right )} x^{4} + {\left (3 \, c^{5} d^{10} e^{2} - 11 \, a c^{4} d^{8} e^{4} + 14 \, a^{2} c^{3} d^{6} e^{6} - 6 \, a^{3} c^{2} d^{4} e^{8} - a^{4} c d^{2} e^{10} + a^{5} e^{12}\right )} x^{3} + 3 \, {\left (c^{5} d^{11} e - 3 \, a c^{4} d^{9} e^{3} + 2 \, a^{2} c^{3} d^{7} e^{5} + 2 \, a^{3} c^{2} d^{5} e^{7} - 3 \, a^{4} c d^{3} e^{9} + a^{5} d e^{11}\right )} x^{2} + {\left (c^{5} d^{12} - a c^{4} d^{10} e^{2} - 6 \, a^{2} c^{3} d^{8} e^{4} + 14 \, a^{3} c^{2} d^{6} e^{6} - 11 \, a^{4} c d^{4} e^{8} + 3 \, a^{5} d^{2} e^{10}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

-2/5*(16*c^3*d^3*e^3*x^3 + 5*c^3*d^6 + 15*a*c^2*d^4*e^2 - 5*a^2*c*d^2*e^4 + a^3*e^6 + 8*(5*c^3*d^4*e^2 + a*c^2
*d^2*e^4)*x^2 + 2*(15*c^3*d^5*e + 10*a*c^2*d^3*e^3 - a^2*c*d*e^5)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*
x)/(a*c^4*d^11*e - 4*a^2*c^3*d^9*e^3 + 6*a^3*c^2*d^7*e^5 - 4*a^4*c*d^5*e^7 + a^5*d^3*e^9 + (c^5*d^9*e^3 - 4*a*
c^4*d^7*e^5 + 6*a^2*c^3*d^5*e^7 - 4*a^3*c^2*d^3*e^9 + a^4*c*d*e^11)*x^4 + (3*c^5*d^10*e^2 - 11*a*c^4*d^8*e^4 +
14*a^2*c^3*d^6*e^6 - 6*a^3*c^2*d^4*e^8 - a^4*c*d^2*e^10 + a^5*e^12)*x^3 + 3*(c^5*d^11*e - 3*a*c^4*d^9*e^3 + 2
*a^2*c^3*d^7*e^5 + 2*a^3*c^2*d^5*e^7 - 3*a^4*c*d^3*e^9 + a^5*d*e^11)*x^2 + (c^5*d^12 - a*c^4*d^10*e^2 - 6*a^2*
c^3*d^8*e^4 + 14*a^3*c^2*d^6*e^6 - 11*a^4*c*d^4*e^8 + 3*a^5*d^2*e^10)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.06, size = 216, normalized size = 1.19 \begin {gather*} -\frac {2 \left (c d x +a e \right ) \left (16 c^{3} d^{3} e^{3} x^{3}+8 a \,c^{2} d^{2} e^{4} x^{2}+40 c^{3} d^{4} e^{2} x^{2}-2 a^{2} c d \,e^{5} x +20 a \,c^{2} d^{3} e^{3} x +30 c^{3} d^{5} e x +a^{3} e^{6}-5 a^{2} c \,d^{2} e^{4}+15 a \,c^{2} d^{4} e^{2}+5 c^{3} d^{6}\right )}{5 \left (e x +d \right ) \left (a^{4} e^{8}-4 a^{3} c \,d^{2} e^{6}+6 a^{2} c^{2} d^{4} e^{4}-4 a \,c^{3} d^{6} e^{2}+c^{4} d^{8}\right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(c*d*x+a*e)*(16*c^3*d^3*e^3*x^3+8*a*c^2*d^2*e^4*x^2+40*c^3*d^4*e^2*x^2-2*a^2*c*d*e^5*x+20*a*c^2*d^3*e^3*x
+30*c^3*d^5*e*x+a^3*e^6-5*a^2*c*d^2*e^4+15*a*c^2*d^4*e^2+5*c^3*d^6)/(e*x+d)/(a^4*e^8-4*a^3*c*d^2*e^6+6*a^2*c^2
*d^4*e^4-4*a*c^3*d^6*e^2+c^4*d^8)/(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(1/(e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),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 zero or nonzero?

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mupad [B]  time = 1.52, size = 1005, normalized size = 5.55 \begin {gather*} \frac {\left (\frac {16\,c^3\,d^4\,e}{15\,{\left (a\,e^2-c\,d^2\right )}^5}-\frac {8\,c^2\,d^2\,e\,\left (c\,d^2+a\,e^2\right )}{15\,{\left (a\,e^2-c\,d^2\right )}^5}\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 {e^2\,\left (10\,c^2\,d^3-18\,a\,c\,d\,e^2\right )}{5\,{\left (a\,e^2-c\,d^2\right )}^2\,\left (3\,a^2\,e^5-6\,a\,c\,d^2\,e^3+3\,c^2\,d^4\,e\right )}+\frac {8\,c^2\,d^3\,e^2}{5\,{\left (a\,e^2-c\,d^2\right )}^2\,\left (3\,a^2\,e^5-6\,a\,c\,d^2\,e^3+3\,c^2\,d^4\,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 {2\,e^2\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{{\left (d+e\,x\right )}^3\,\left (5\,a^2\,e^5-10\,a\,c\,d^2\,e^3+5\,c^2\,d^4\,e\right )}-\frac {\left (x\,\left (\frac {32\,a\,c^5\,d^6\,e^4}{15\,{\left (a\,e^2-c\,d^2\right )}^4\,\left (a^2\,c\,d\,e^5-2\,a\,c^2\,d^3\,e^3+c^3\,d^5\,e\right )}-\frac {\left (c\,d^2+a\,e^2\right )\,\left (\frac {16\,c^5\,d^5\,e^3\,\left (c\,d^2+a\,e^2\right )}{15\,{\left (a\,e^2-c\,d^2\right )}^4\,\left (a^2\,c\,d\,e^5-2\,a\,c^2\,d^3\,e^3+c^3\,d^5\,e\right )}+\frac {8\,c^5\,d^5\,e^3\,\left (3\,a\,e^2-11\,c\,d^2\right )}{15\,{\left (a\,e^2-c\,d^2\right )}^4\,\left (a^2\,c\,d\,e^5-2\,a\,c^2\,d^3\,e^3+c^3\,d^5\,e\right )}\right )}{c\,d\,e}+\frac {2\,c^2\,d^2\,e^2\,\left (58\,a^2\,c^2\,d^2\,e^4-104\,a\,c^3\,d^4\,e^2+30\,c^4\,d^6\right )}{15\,{\left (a\,e^2-c\,d^2\right )}^4\,\left (a^2\,c\,d\,e^5-2\,a\,c^2\,d^3\,e^3+c^3\,d^5\,e\right )}+\frac {4\,c^4\,d^4\,e^2\,\left (c\,d^2+a\,e^2\right )\,\left (3\,a\,e^2-11\,c\,d^2\right )}{15\,{\left (a\,e^2-c\,d^2\right )}^4\,\left (a^2\,c\,d\,e^5-2\,a\,c^2\,d^3\,e^3+c^3\,d^5\,e\right )}\right )-\frac {a\,\left (\frac {16\,c^5\,d^5\,e^3\,\left (c\,d^2+a\,e^2\right )}{15\,{\left (a\,e^2-c\,d^2\right )}^4\,\left (a^2\,c\,d\,e^5-2\,a\,c^2\,d^3\,e^3+c^3\,d^5\,e\right )}+\frac {8\,c^5\,d^5\,e^3\,\left (3\,a\,e^2-11\,c\,d^2\right )}{15\,{\left (a\,e^2-c\,d^2\right )}^4\,\left (a^2\,c\,d\,e^5-2\,a\,c^2\,d^3\,e^3+c^3\,d^5\,e\right )}\right )}{c}+\frac {c\,d\,e\,\left (c\,d^2+a\,e^2\right )\,\left (58\,a^2\,c^2\,d^2\,e^4-104\,a\,c^3\,d^4\,e^2+30\,c^4\,d^6\right )}{15\,{\left (a\,e^2-c\,d^2\right )}^4\,\left (a^2\,c\,d\,e^5-2\,a\,c^2\,d^3\,e^3+c^3\,d^5\,e\right )}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{\left (a\,e+c\,d\,x\right )\,\left (d+e\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((16*c^3*d^4*e)/(15*(a*e^2 - c*d^2)^5) - (8*c^2*d^2*e*(a*e^2 + c*d^2))/(15*(a*e^2 - c*d^2)^5))*(x*(a*e^2 + c*
d^2) + a*d*e + c*d*e*x^2)^(1/2))/(d + e*x) - (((e^2*(10*c^2*d^3 - 18*a*c*d*e^2))/(5*(a*e^2 - c*d^2)^2*(3*a^2*e
^5 + 3*c^2*d^4*e - 6*a*c*d^2*e^3)) + (8*c^2*d^3*e^2)/(5*(a*e^2 - c*d^2)^2*(3*a^2*e^5 + 3*c^2*d^4*e - 6*a*c*d^2
*e^3)))*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/(d + e*x)^2 - (2*e^2*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e
*x^2)^(1/2))/((d + e*x)^3*(5*a^2*e^5 + 5*c^2*d^4*e - 10*a*c*d^2*e^3)) - ((x*((32*a*c^5*d^6*e^4)/(15*(a*e^2 - c
*d^2)^4*(c^3*d^5*e - 2*a*c^2*d^3*e^3 + a^2*c*d*e^5)) - ((a*e^2 + c*d^2)*((16*c^5*d^5*e^3*(a*e^2 + c*d^2))/(15*
(a*e^2 - c*d^2)^4*(c^3*d^5*e - 2*a*c^2*d^3*e^3 + a^2*c*d*e^5)) + (8*c^5*d^5*e^3*(3*a*e^2 - 11*c*d^2))/(15*(a*e
^2 - c*d^2)^4*(c^3*d^5*e - 2*a*c^2*d^3*e^3 + a^2*c*d*e^5))))/(c*d*e) + (2*c^2*d^2*e^2*(30*c^4*d^6 - 104*a*c^3*
d^4*e^2 + 58*a^2*c^2*d^2*e^4))/(15*(a*e^2 - c*d^2)^4*(c^3*d^5*e - 2*a*c^2*d^3*e^3 + a^2*c*d*e^5)) + (4*c^4*d^4
*e^2*(a*e^2 + c*d^2)*(3*a*e^2 - 11*c*d^2))/(15*(a*e^2 - c*d^2)^4*(c^3*d^5*e - 2*a*c^2*d^3*e^3 + a^2*c*d*e^5)))
- (a*((16*c^5*d^5*e^3*(a*e^2 + c*d^2))/(15*(a*e^2 - c*d^2)^4*(c^3*d^5*e - 2*a*c^2*d^3*e^3 + a^2*c*d*e^5)) + (
8*c^5*d^5*e^3*(3*a*e^2 - 11*c*d^2))/(15*(a*e^2 - c*d^2)^4*(c^3*d^5*e - 2*a*c^2*d^3*e^3 + a^2*c*d*e^5))))/c + (
c*d*e*(a*e^2 + c*d^2)*(30*c^4*d^6 - 104*a*c^3*d^4*e^2 + 58*a^2*c^2*d^2*e^4))/(15*(a*e^2 - c*d^2)^4*(c^3*d^5*e
- 2*a*c^2*d^3*e^3 + a^2*c*d*e^5)))*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/((a*e + c*d*x)*(d + e*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {3}{2}} \left (d + e x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**2/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2),x)

[Out]

Integral(1/(((d + e*x)*(a*e + c*d*x))**(3/2)*(d + e*x)**2), x)

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