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

Optimal. Leaf size=180 \[ -\frac {2 \left (c d^2-a e^2\right )^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{9/2} d^{9/2}}+\frac {2 \sqrt {d+e x} \left (c d^2-a e^2\right )^3}{c^4 d^4}+\frac {2 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2}{3 c^3 d^3}+\frac {2 (d+e x)^{5/2} \left (c d^2-a e^2\right )}{5 c^2 d^2}+\frac {2 (d+e x)^{7/2}}{7 c d} \]

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Rubi [A]  time = 0.24, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {626, 50, 63, 208} \begin {gather*} \frac {2 \sqrt {d+e x} \left (c d^2-a e^2\right )^3}{c^4 d^4}+\frac {2 (d+e x)^{3/2} \left (c d^2-a e^2\right )^2}{3 c^3 d^3}+\frac {2 (d+e x)^{5/2} \left (c d^2-a e^2\right )}{5 c^2 d^2}-\frac {2 \left (c d^2-a e^2\right )^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{9/2} d^{9/2}}+\frac {2 (d+e x)^{7/2}}{7 c d} \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),x]

[Out]

(2*(c*d^2 - a*e^2)^3*Sqrt[d + e*x])/(c^4*d^4) + (2*(c*d^2 - a*e^2)^2*(d + e*x)^(3/2))/(3*c^3*d^3) + (2*(c*d^2
- a*e^2)*(d + e*x)^(5/2))/(5*c^2*d^2) + (2*(d + e*x)^(7/2))/(7*c*d) - (2*(c*d^2 - a*e^2)^(7/2)*ArcTanh[(Sqrt[c
]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(c^(9/2)*d^(9/2))

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}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx &=\int \frac {(d+e x)^{7/2}}{a e+c d x} \, dx\\ &=\frac {2 (d+e x)^{7/2}}{7 c d}+\frac {\left (c d^2-a e^2\right ) \int \frac {(d+e x)^{5/2}}{a e+c d x} \, dx}{c d}\\ &=\frac {2 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}{5 c^2 d^2}+\frac {2 (d+e x)^{7/2}}{7 c d}+\frac {\left (c d^2-a e^2\right )^2 \int \frac {(d+e x)^{3/2}}{a e+c d x} \, dx}{c^2 d^2}\\ &=\frac {2 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}{3 c^3 d^3}+\frac {2 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}{5 c^2 d^2}+\frac {2 (d+e x)^{7/2}}{7 c d}+\frac {\left (c d^2-a e^2\right )^3 \int \frac {\sqrt {d+e x}}{a e+c d x} \, dx}{c^3 d^3}\\ &=\frac {2 \left (c d^2-a e^2\right )^3 \sqrt {d+e x}}{c^4 d^4}+\frac {2 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}{3 c^3 d^3}+\frac {2 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}{5 c^2 d^2}+\frac {2 (d+e x)^{7/2}}{7 c d}+\frac {\left (c d^2-a e^2\right )^4 \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{c^4 d^4}\\ &=\frac {2 \left (c d^2-a e^2\right )^3 \sqrt {d+e x}}{c^4 d^4}+\frac {2 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}{3 c^3 d^3}+\frac {2 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}{5 c^2 d^2}+\frac {2 (d+e x)^{7/2}}{7 c d}+\frac {\left (2 \left (c d^2-a e^2\right )^4\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^4 d^4 e}\\ &=\frac {2 \left (c d^2-a e^2\right )^3 \sqrt {d+e x}}{c^4 d^4}+\frac {2 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}{3 c^3 d^3}+\frac {2 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}{5 c^2 d^2}+\frac {2 (d+e x)^{7/2}}{7 c d}-\frac {2 \left (c d^2-a e^2\right )^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{9/2} d^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 175, normalized size = 0.97 \begin {gather*} \frac {2 \left (c d^2-a e^2\right ) \left (5 \left (c d^2-a e^2\right ) \left (\sqrt {c} \sqrt {d} \sqrt {d+e x} \left (c d (4 d+e x)-3 a e^2\right )-3 \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 )\right )+3 c^{5/2} d^{5/2} (d+e x)^{5/2}\right )}{15 c^{9/2} d^{9/2}}+\frac {2 (d+e x)^{7/2}}{7 c d} \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),x]

[Out]

(2*(d + e*x)^(7/2))/(7*c*d) + (2*(c*d^2 - a*e^2)*(3*c^(5/2)*d^(5/2)*(d + e*x)^(5/2) + 5*(c*d^2 - a*e^2)*(Sqrt[
c]*Sqrt[d]*Sqrt[d + e*x]*(-3*a*e^2 + c*d*(4*d + e*x)) - 3*(c*d^2 - a*e^2)^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[
d + e*x])/Sqrt[c*d^2 - a*e^2]])))/(15*c^(9/2)*d^(9/2))

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IntegrateAlgebraic [A]  time = 0.21, size = 234, normalized size = 1.30 \begin {gather*} \frac {2 \sqrt {d+e x} \left (-105 a^3 e^6+315 a^2 c d^2 e^4+35 a^2 c d e^4 (d+e x)-315 a c^2 d^4 e^2-70 a c^2 d^3 e^2 (d+e x)-21 a c^2 d^2 e^2 (d+e x)^2+105 c^3 d^6+35 c^3 d^5 (d+e x)+21 c^3 d^4 (d+e x)^2+15 c^3 d^3 (d+e x)^3\right )}{105 c^4 d^4}-\frac {2 \left (a e^2-c d^2\right )^{7/2} \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^{9/2} d^{9/2}} \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),x]

[Out]

(2*Sqrt[d + e*x]*(105*c^3*d^6 - 315*a*c^2*d^4*e^2 + 315*a^2*c*d^2*e^4 - 105*a^3*e^6 + 35*c^3*d^5*(d + e*x) - 7
0*a*c^2*d^3*e^2*(d + e*x) + 35*a^2*c*d*e^4*(d + e*x) + 21*c^3*d^4*(d + e*x)^2 - 21*a*c^2*d^2*e^2*(d + e*x)^2 +
15*c^3*d^3*(d + e*x)^3))/(105*c^4*d^4) - (2*(-(c*d^2) + a*e^2)^(7/2)*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[-(c*d^2) +
a*e^2]*Sqrt[d + e*x])/(c*d^2 - a*e^2)])/(c^(9/2)*d^(9/2))

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fricas [A]  time = 0.43, size = 510, normalized size = 2.83 \begin {gather*} \left [\frac {105 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\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 (15 \, c^{3} d^{3} e^{3} x^{3} + 176 \, c^{3} d^{6} - 406 \, a c^{2} d^{4} e^{2} + 350 \, a^{2} c d^{2} e^{4} - 105 \, a^{3} e^{6} + 3 \, {\left (22 \, c^{3} d^{4} e^{2} - 7 \, a c^{2} d^{2} e^{4}\right )} x^{2} + {\left (122 \, c^{3} d^{5} e - 112 \, a c^{2} d^{3} e^{3} + 35 \, a^{2} c d e^{5}\right )} x\right )} \sqrt {e x + d}}{105 \, c^{4} d^{4}}, -\frac {2 \, {\left (105 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\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 (15 \, c^{3} d^{3} e^{3} x^{3} + 176 \, c^{3} d^{6} - 406 \, a c^{2} d^{4} e^{2} + 350 \, a^{2} c d^{2} e^{4} - 105 \, a^{3} e^{6} + 3 \, {\left (22 \, c^{3} d^{4} e^{2} - 7 \, a c^{2} d^{2} e^{4}\right )} x^{2} + {\left (122 \, c^{3} d^{5} e - 112 \, a c^{2} d^{3} e^{3} + 35 \, a^{2} c d e^{5}\right )} x\right )} \sqrt {e x + d}\right )}}{105 \, c^{4} d^{4}}\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),x, algorithm="fricas")

[Out]

[1/105*(105*(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*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*(15*c^3*d^3*e^3*x^3 + 1
76*c^3*d^6 - 406*a*c^2*d^4*e^2 + 350*a^2*c*d^2*e^4 - 105*a^3*e^6 + 3*(22*c^3*d^4*e^2 - 7*a*c^2*d^2*e^4)*x^2 +
(122*c^3*d^5*e - 112*a*c^2*d^3*e^3 + 35*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(c^4*d^4), -2/105*(105*(c^3*d^6 - 3*a*c
^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*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)) - (15*c^3*d^3*e^3*x^3 + 176*c^3*d^6 - 406*a*c^2*d^4*e^2 + 350*a^2*c*d^2*e^4 - 1
05*a^3*e^6 + 3*(22*c^3*d^4*e^2 - 7*a*c^2*d^2*e^4)*x^2 + (122*c^3*d^5*e - 112*a*c^2*d^3*e^3 + 35*a^2*c*d*e^5)*x
)*sqrt(e*x + d))/(c^4*d^4)]

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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),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.08, size = 455, normalized size = 2.53 \begin {gather*} \frac {2 a^{4} e^{8} \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^{4} d^{4}}-\frac {8 a^{3} e^{6} \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^{2}}+\frac {12 a^{2} e^{4} \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}}-\frac {8 a \,d^{2} e^{2} \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 {2 d^{4} \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}}-\frac {2 \sqrt {e x +d}\, a^{3} e^{6}}{c^{4} d^{4}}+\frac {6 \sqrt {e x +d}\, a^{2} e^{4}}{c^{3} d^{2}}-\frac {6 \sqrt {e x +d}\, a \,e^{2}}{c^{2}}+\frac {2 \sqrt {e x +d}\, d^{2}}{c}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} a^{2} e^{4}}{3 c^{3} d^{3}}-\frac {4 \left (e x +d \right )^{\frac {3}{2}} a \,e^{2}}{3 c^{2} d}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} d}{3 c}-\frac {2 \left (e x +d \right )^{\frac {5}{2}} a \,e^{2}}{5 c^{2} d^{2}}+\frac {2 \left (e x +d \right )^{\frac {5}{2}}}{5 c}+\frac {2 \left (e x +d \right )^{\frac {7}{2}}}{7 c d} \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),x)

[Out]

2/7*(e*x+d)^(7/2)/c/d-2/5/c^2/d^2*(e*x+d)^(5/2)*a*e^2+2/5/c*(e*x+d)^(5/2)+2/3/c^3/d^3*(e*x+d)^(3/2)*a^2*e^4-4/
3/c^2/d*(e*x+d)^(3/2)*a*e^2+2/3/c*d*(e*x+d)^(3/2)-2/c^4/d^4*a^3*e^6*(e*x+d)^(1/2)+6/c^3/d^2*a^2*e^4*(e*x+d)^(1
/2)-6/c^2*a*e^2*(e*x+d)^(1/2)+2/c*d^2*(e*x+d)^(1/2)+2/c^4/d^4/((a*e^2-c*d^2)*c*d)^(1/2)*arctan((e*x+d)^(1/2)*c
*d/((a*e^2-c*d^2)*c*d)^(1/2))*a^4*e^8-8/c^3/d^2/((a*e^2-c*d^2)*c*d)^(1/2)*arctan((e*x+d)^(1/2)*c*d/((a*e^2-c*d
^2)*c*d)^(1/2))*a^3*e^6+12/c^2/((a*e^2-c*d^2)*c*d)^(1/2)*arctan((e*x+d)^(1/2)*c*d/((a*e^2-c*d^2)*c*d)^(1/2))*a
^2*e^4-8/c*d^2/((a*e^2-c*d^2)*c*d)^(1/2)*arctan((e*x+d)^(1/2)*c*d/((a*e^2-c*d^2)*c*d)^(1/2))*a*e^2+2*d^4/((a*e
^2-c*d^2)*c*d)^(1/2)*arctan((e*x+d)^(1/2)*c*d/((a*e^2-c*d^2)*c*d)^(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((e*x+d)^(9/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^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 = 0.63, size = 207, normalized size = 1.15 \begin {gather*} \frac {2\,{\left (d+e\,x\right )}^{7/2}}{7\,c\,d}+\frac {2\,{\left (a\,e^2-c\,d^2\right )}^2\,{\left (d+e\,x\right )}^{3/2}}{3\,c^3\,d^3}-\frac {2\,{\left (a\,e^2-c\,d^2\right )}^3\,\sqrt {d+e\,x}}{c^4\,d^4}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,{\left (a\,e^2-c\,d^2\right )}^{7/2}\,\sqrt {d+e\,x}}{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 (a\,e^2-c\,d^2\right )}^{7/2}}{c^{9/2}\,d^{9/2}}-\frac {2\,\left (a\,e^2-c\,d^2\right )\,{\left (d+e\,x\right )}^{5/2}}{5\,c^2\,d^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),x)

[Out]

(2*(d + e*x)^(7/2))/(7*c*d) + (2*(a*e^2 - c*d^2)^2*(d + e*x)^(3/2))/(3*c^3*d^3) - (2*(a*e^2 - c*d^2)^3*(d + e*
x)^(1/2))/(c^4*d^4) + (2*atan((c^(1/2)*d^(1/2)*(a*e^2 - c*d^2)^(7/2)*(d + e*x)^(1/2))/(a^4*e^8 + c^4*d^8 - 4*a
*c^3*d^6*e^2 - 4*a^3*c*d^2*e^6 + 6*a^2*c^2*d^4*e^4))*(a*e^2 - c*d^2)^(7/2))/(c^(9/2)*d^(9/2)) - (2*(a*e^2 - c*
d^2)*(d + e*x)^(5/2))/(5*c^2*d^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),x)

[Out]

Timed out

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