3.17.95 \(\int \frac {(d+e x)^{9/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\)

Optimal. Leaf size=144 \[ -\frac {5 e \left (c d^2-a e^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{7/2} d^{7/2}}+\frac {5 e \sqrt {d+e x} \left (c d^2-a e^2\right )}{c^3 d^3}-\frac {(d+e x)^{5/2}}{c d (a e+c d x)}+\frac {5 e (d+e x)^{3/2}}{3 c^2 d^2} \]

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Rubi [A]  time = 0.10, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {626, 47, 50, 63, 208} \begin {gather*} \frac {5 e \sqrt {d+e x} \left (c d^2-a e^2\right )}{c^3 d^3}-\frac {5 e \left (c d^2-a e^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{7/2} d^{7/2}}-\frac {(d+e x)^{5/2}}{c d (a e+c d x)}+\frac {5 e (d+e x)^{3/2}}{3 c^2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(9/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]

[Out]

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

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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)^{9/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx &=\int \frac {(d+e x)^{5/2}}{(a e+c d x)^2} \, dx\\ &=-\frac {(d+e x)^{5/2}}{c d (a e+c d x)}+\frac {(5 e) \int \frac {(d+e x)^{3/2}}{a e+c d x} \, dx}{2 c d}\\ &=\frac {5 e (d+e x)^{3/2}}{3 c^2 d^2}-\frac {(d+e x)^{5/2}}{c d (a e+c d x)}+\frac {\left (5 e \left (c d^2-a e^2\right )\right ) \int \frac {\sqrt {d+e x}}{a e+c d x} \, dx}{2 c^2 d^2}\\ &=\frac {5 e \left (c d^2-a e^2\right ) \sqrt {d+e x}}{c^3 d^3}+\frac {5 e (d+e x)^{3/2}}{3 c^2 d^2}-\frac {(d+e x)^{5/2}}{c d (a e+c d x)}+\frac {\left (5 e \left (c d^2-a e^2\right )^2\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{2 c^3 d^3}\\ &=\frac {5 e \left (c d^2-a e^2\right ) \sqrt {d+e x}}{c^3 d^3}+\frac {5 e (d+e x)^{3/2}}{3 c^2 d^2}-\frac {(d+e x)^{5/2}}{c d (a e+c d x)}+\frac {\left (5 \left (c d^2-a e^2\right )^2\right ) \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^3 d^3}\\ &=\frac {5 e \left (c d^2-a e^2\right ) \sqrt {d+e x}}{c^3 d^3}+\frac {5 e (d+e x)^{3/2}}{3 c^2 d^2}-\frac {(d+e x)^{5/2}}{c d (a e+c d x)}-\frac {5 e \left (c d^2-a e^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{7/2} d^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 59, normalized size = 0.41 \begin {gather*} \frac {2 e (d+e x)^{7/2} \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};-\frac {c d (d+e x)}{a e^2-c d^2}\right )}{7 \left (a e^2-c d^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(9/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.43, size = 232, normalized size = 1.61 \begin {gather*} \frac {e \sqrt {d+e x} \left (15 a^2 e^4-30 a c d^2 e^2+10 a c d e^2 (d+e x)+15 c^2 d^4-10 c^2 d^3 (d+e x)-2 c^2 d^2 (d+e x)^2\right )}{3 c^3 d^3 \left (-a e^2+c d^2-c d (d+e x)\right )}+\frac {5 \left (-a^3 e^7+3 a^2 c d^2 e^5-3 a c^2 d^4 e^3+c^3 d^6 e\right ) \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^{7/2} d^{7/2} \left (a e^2-c d^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^(9/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]

[Out]

(e*Sqrt[d + e*x]*(15*c^2*d^4 - 30*a*c*d^2*e^2 + 15*a^2*e^4 - 10*c^2*d^3*(d + e*x) + 10*a*c*d*e^2*(d + e*x) - 2
*c^2*d^2*(d + e*x)^2))/(3*c^3*d^3*(c*d^2 - a*e^2 - c*d*(d + e*x))) + (5*(c^3*d^6*e - 3*a*c^2*d^4*e^3 + 3*a^2*c
*d^2*e^5 - a^3*e^7)*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[-(c*d^2) + a*e^2]*Sqrt[d + e*x])/(c*d^2 - a*e^2)])/(c^(7/2)*d
^(7/2)*(-(c*d^2) + a*e^2)^(3/2))

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fricas [A]  time = 0.44, size = 420, normalized size = 2.92 \begin {gather*} \left [\frac {15 \, {\left (a c d^{2} e^{2} - a^{2} e^{4} + {\left (c^{2} d^{3} e - a c d e^{3}\right )} x\right )} \sqrt {\frac {c d^{2} - a e^{2}}{c d}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {e x + d} c d \sqrt {\frac {c d^{2} - a e^{2}}{c d}}}{c d x + a e}\right ) + 2 \, {\left (2 \, c^{2} d^{2} e^{2} x^{2} - 3 \, c^{2} d^{4} + 20 \, a c d^{2} e^{2} - 15 \, a^{2} e^{4} + 2 \, {\left (7 \, c^{2} d^{3} e - 5 \, a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{6 \, {\left (c^{4} d^{4} x + a c^{3} d^{3} e\right )}}, -\frac {15 \, {\left (a c d^{2} e^{2} - a^{2} e^{4} + {\left (c^{2} d^{3} e - a c d e^{3}\right )} x\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac {\sqrt {e x + d} c d \sqrt {-\frac {c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) - {\left (2 \, c^{2} d^{2} e^{2} x^{2} - 3 \, c^{2} d^{4} + 20 \, a c d^{2} e^{2} - 15 \, a^{2} e^{4} + 2 \, {\left (7 \, c^{2} d^{3} e - 5 \, a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (c^{4} d^{4} x + a c^{3} d^{3} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(9/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="fricas")

[Out]

[1/6*(15*(a*c*d^2*e^2 - a^2*e^4 + (c^2*d^3*e - a*c*d*e^3)*x)*sqrt((c*d^2 - a*e^2)/(c*d))*log((c*d*e*x + 2*c*d^
2 - a*e^2 - 2*sqrt(e*x + d)*c*d*sqrt((c*d^2 - a*e^2)/(c*d)))/(c*d*x + a*e)) + 2*(2*c^2*d^2*e^2*x^2 - 3*c^2*d^4
+ 20*a*c*d^2*e^2 - 15*a^2*e^4 + 2*(7*c^2*d^3*e - 5*a*c*d*e^3)*x)*sqrt(e*x + d))/(c^4*d^4*x + a*c^3*d^3*e), -1
/3*(15*(a*c*d^2*e^2 - a^2*e^4 + (c^2*d^3*e - a*c*d*e^3)*x)*sqrt(-(c*d^2 - a*e^2)/(c*d))*arctan(-sqrt(e*x + d)*
c*d*sqrt(-(c*d^2 - a*e^2)/(c*d))/(c*d^2 - a*e^2)) - (2*c^2*d^2*e^2*x^2 - 3*c^2*d^4 + 20*a*c*d^2*e^2 - 15*a^2*e
^4 + 2*(7*c^2*d^3*e - 5*a*c*d*e^3)*x)*sqrt(e*x + d))/(c^4*d^4*x + a*c^3*d^3*e)]

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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)^(9/2)/(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 [B]  time = 0.08, size = 314, normalized size = 2.18 \begin {gather*} \frac {5 a^{2} e^{5} \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^{3} d^{3}}-\frac {10 a \,e^{3} \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^{2} d}+\frac {5 d 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}-\frac {\sqrt {e x +d}\, a^{2} e^{5}}{\left (c d e x +a \,e^{2}\right ) c^{3} d^{3}}+\frac {2 \sqrt {e x +d}\, a \,e^{3}}{\left (c d e x +a \,e^{2}\right ) c^{2} d}-\frac {\sqrt {e x +d}\, d e}{\left (c d e x +a \,e^{2}\right ) c}-\frac {4 \sqrt {e x +d}\, a \,e^{3}}{c^{3} d^{3}}+\frac {4 \sqrt {e x +d}\, e}{c^{2} d}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} e}{3 c^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(9/2)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^2,x)

[Out]

2/3*e*(e*x+d)^(3/2)/c^2/d^2-4/c^3/d^3*a*e^3*(e*x+d)^(1/2)+4*e/c^2/d*(e*x+d)^(1/2)-1/c^3/d^3*(e*x+d)^(1/2)/(c*d
*e*x+a*e^2)*a^2*e^5+2/c^2/d*(e*x+d)^(1/2)/(c*d*e*x+a*e^2)*a*e^3-e/c*d*(e*x+d)^(1/2)/(c*d*e*x+a*e^2)+5/c^3/d^3/
((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)*a^2*e^5-10/c^2/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)*a*e^3+5*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)^(9/2)/(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 = 0.70, size = 200, normalized size = 1.39 \begin {gather*} \frac {2\,e\,{\left (d+e\,x\right )}^{3/2}}{3\,c^2\,d^2}-\frac {\sqrt {d+e\,x}\,\left (a^2\,e^5-2\,a\,c\,d^2\,e^3+c^2\,d^4\,e\right )}{c^4\,d^4\,\left (d+e\,x\right )-c^4\,d^5+a\,c^3\,d^3\,e^2}+\frac {2\,e\,\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )\,\sqrt {d+e\,x}}{c^4\,d^4}+\frac {5\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,e\,{\left (a\,e^2-c\,d^2\right )}^{3/2}\,\sqrt {d+e\,x}}{a^2\,e^5-2\,a\,c\,d^2\,e^3+c^2\,d^4\,e}\right )\,{\left (a\,e^2-c\,d^2\right )}^{3/2}}{c^{7/2}\,d^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(9/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^2,x)

[Out]

(2*e*(d + e*x)^(3/2))/(3*c^2*d^2) - ((d + e*x)^(1/2)*(a^2*e^5 + c^2*d^4*e - 2*a*c*d^2*e^3))/(c^4*d^4*(d + e*x)
- c^4*d^5 + a*c^3*d^3*e^2) + (2*e*(2*c^2*d^3 - 2*a*c*d*e^2)*(d + e*x)^(1/2))/(c^4*d^4) + (5*e*atan((c^(1/2)*d
^(1/2)*e*(a*e^2 - c*d^2)^(3/2)*(d + e*x)^(1/2))/(a^2*e^5 + c^2*d^4*e - 2*a*c*d^2*e^3))*(a*e^2 - c*d^2)^(3/2))/
(c^(7/2)*d^(7/2))

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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)**(9/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2,x)

[Out]

Timed out

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