3.21.26 \(\int \frac {1}{\sqrt {d+e x} (a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\) [2026]

Optimal. Leaf size=244 \[ \frac {63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac {9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac {21 c d e^2}{4 \left (c d^2-a e^2\right )^4 (d+e x)^{3/2}}+\frac {63 c^2 d^2 e^2}{4 \left (c d^2-a e^2\right )^5 \sqrt {d+e x}}-\frac {63 c^{5/2} d^{5/2} e^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{11/2}} \]

[Out]

63/20*e^2/(-a*e^2+c*d^2)^3/(e*x+d)^(5/2)-1/2/(-a*e^2+c*d^2)/(c*d*x+a*e)^2/(e*x+d)^(5/2)+9/4*e/(-a*e^2+c*d^2)^2
/(c*d*x+a*e)/(e*x+d)^(5/2)+21/4*c*d*e^2/(-a*e^2+c*d^2)^4/(e*x+d)^(3/2)-63/4*c^(5/2)*d^(5/2)*e^2*arctanh(c^(1/2
)*d^(1/2)*(e*x+d)^(1/2)/(-a*e^2+c*d^2)^(1/2))/(-a*e^2+c*d^2)^(11/2)+63/4*c^2*d^2*e^2/(-a*e^2+c*d^2)^5/(e*x+d)^
(1/2)

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Rubi [A]
time = 0.15, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {640, 44, 53, 65, 214} \begin {gather*} -\frac {63 c^{5/2} d^{5/2} e^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{11/2}}+\frac {63 c^2 d^2 e^2}{4 \sqrt {d+e x} \left (c d^2-a e^2\right )^5}+\frac {21 c d e^2}{4 (d+e x)^{3/2} \left (c d^2-a e^2\right )^4}+\frac {9 e}{4 (d+e x)^{5/2} \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac {1}{2 (d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)^2}+\frac {63 e^2}{20 (d+e x)^{5/2} \left (c d^2-a e^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(63*e^2)/(20*(c*d^2 - a*e^2)^3*(d + e*x)^(5/2)) - 1/(2*(c*d^2 - a*e^2)*(a*e + c*d*x)^2*(d + e*x)^(5/2)) + (9*e
)/(4*(c*d^2 - a*e^2)^2*(a*e + c*d*x)*(d + e*x)^(5/2)) + (21*c*d*e^2)/(4*(c*d^2 - a*e^2)^4*(d + e*x)^(3/2)) + (
63*c^2*d^2*e^2)/(4*(c*d^2 - a*e^2)^5*Sqrt[d + e*x]) - (63*c^(5/2)*d^(5/2)*e^2*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d
+ e*x])/Sqrt[c*d^2 - a*e^2]])/(4*(c*d^2 - a*e^2)^(11/2))

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^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 {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx &=\int \frac {1}{(a e+c d x)^3 (d+e x)^{7/2}} \, dx\\ &=-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}-\frac {(9 e) \int \frac {1}{(a e+c d x)^2 (d+e x)^{7/2}} \, dx}{4 \left (c d^2-a e^2\right )}\\ &=-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac {9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac {\left (63 e^2\right ) \int \frac {1}{(a e+c d x) (d+e x)^{7/2}} \, dx}{8 \left (c d^2-a e^2\right )^2}\\ &=\frac {63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac {9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac {\left (63 c d e^2\right ) \int \frac {1}{(a e+c d x) (d+e x)^{5/2}} \, dx}{8 \left (c d^2-a e^2\right )^3}\\ &=\frac {63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac {9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac {21 c d e^2}{4 \left (c d^2-a e^2\right )^4 (d+e x)^{3/2}}+\frac {\left (63 c^2 d^2 e^2\right ) \int \frac {1}{(a e+c d x) (d+e x)^{3/2}} \, dx}{8 \left (c d^2-a e^2\right )^4}\\ &=\frac {63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac {9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac {21 c d e^2}{4 \left (c d^2-a e^2\right )^4 (d+e x)^{3/2}}+\frac {63 c^2 d^2 e^2}{4 \left (c d^2-a e^2\right )^5 \sqrt {d+e x}}+\frac {\left (63 c^3 d^3 e^2\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{8 \left (c d^2-a e^2\right )^5}\\ &=\frac {63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac {9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac {21 c d e^2}{4 \left (c d^2-a e^2\right )^4 (d+e x)^{3/2}}+\frac {63 c^2 d^2 e^2}{4 \left (c d^2-a e^2\right )^5 \sqrt {d+e x}}+\frac {\left (63 c^3 d^3 e\right ) \text {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 )}{4 \left (c d^2-a e^2\right )^5}\\ &=\frac {63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac {9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac {21 c d e^2}{4 \left (c d^2-a e^2\right )^4 (d+e x)^{3/2}}+\frac {63 c^2 d^2 e^2}{4 \left (c d^2-a e^2\right )^5 \sqrt {d+e x}}-\frac {63 c^{5/2} d^{5/2} e^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{11/2}}\\ \end {align*}

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Mathematica [A]
time = 0.90, size = 257, normalized size = 1.05 \begin {gather*} \frac {8 a^4 e^8-8 a^3 c d e^6 (7 d+3 e x)+24 a^2 c^2 d^2 e^4 \left (12 d^2+17 d e x+7 e^2 x^2\right )+a c^3 d^3 e^2 \left (85 d^3+831 d^2 e x+1239 d e^2 x^2+525 e^3 x^3\right )+c^4 d^4 \left (-10 d^4+45 d^3 e x+483 d^2 e^2 x^2+735 d e^3 x^3+315 e^4 x^4\right )}{20 \left (c d^2-a e^2\right )^5 (a e+c d x)^2 (d+e x)^{5/2}}-\frac {63 c^{5/2} d^{5/2} e^2 \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{4 \left (-c d^2+a e^2\right )^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^4*e^8 - 8*a^3*c*d*e^6*(7*d + 3*e*x) + 24*a^2*c^2*d^2*e^4*(12*d^2 + 17*d*e*x + 7*e^2*x^2) + a*c^3*d^3*e^2*
(85*d^3 + 831*d^2*e*x + 1239*d*e^2*x^2 + 525*e^3*x^3) + c^4*d^4*(-10*d^4 + 45*d^3*e*x + 483*d^2*e^2*x^2 + 735*
d*e^3*x^3 + 315*e^4*x^4))/(20*(c*d^2 - a*e^2)^5*(a*e + c*d*x)^2*(d + e*x)^(5/2)) - (63*c^(5/2)*d^(5/2)*e^2*Arc
Tan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]])/(4*(-(c*d^2) + a*e^2)^(11/2))

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Maple [A]
time = 0.80, size = 209, normalized size = 0.86

method	result	size
derivativedivides	\(2 e^{2} \left (-\frac {c^{3} d^{3} \left (\frac {\frac {15 c d \left (e x +d \right )^{\frac {3}{2}}}{8}+\left (\frac {17 e^{2} a}{8}-\frac {17 c \,d^{2}}{8}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {63 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right )^{5}}-\frac {1}{5 \left (e^{2} a -c \,d^{2}\right )^{3} \left (e x +d \right )^{\frac {5}{2}}}-\frac {6 c^{2} d^{2}}{\left (e^{2} a -c \,d^{2}\right )^{5} \sqrt {e x +d}}+\frac {c d}{\left (e^{2} a -c \,d^{2}\right )^{4} \left (e x +d \right )^{\frac {3}{2}}}\right )\)	\(209\)
default	\(2 e^{2} \left (-\frac {c^{3} d^{3} \left (\frac {\frac {15 c d \left (e x +d \right )^{\frac {3}{2}}}{8}+\left (\frac {17 e^{2} a}{8}-\frac {17 c \,d^{2}}{8}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {63 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right )^{5}}-\frac {1}{5 \left (e^{2} a -c \,d^{2}\right )^{3} \left (e x +d \right )^{\frac {5}{2}}}-\frac {6 c^{2} d^{2}}{\left (e^{2} a -c \,d^{2}\right )^{5} \sqrt {e x +d}}+\frac {c d}{\left (e^{2} a -c \,d^{2}\right )^{4} \left (e x +d \right )^{\frac {3}{2}}}\right )\)	\(209\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*e^2*(-1/(a*e^2-c*d^2)^5*c^3*d^3*((15/8*c*d*(e*x+d)^(3/2)+(17/8*e^2*a-17/8*c*d^2)*(e*x+d)^(1/2))/(c*d*(e*x+d)
+e^2*a-c*d^2)^2+63/8/((a*e^2-c*d^2)*c*d)^(1/2)*arctan(c*d*(e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2)))-1/5/(a*e^2
-c*d^2)^3/(e*x+d)^(5/2)-6/(a*e^2-c*d^2)^5*c^2*d^2/(e*x+d)^(1/2)+1/(a*e^2-c*d^2)^4*c*d/(e*x+d)^(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)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,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(c*d^2-%e^2*a>0)', see `assume?
` for more d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 998 vs. \(2 (206) = 412\).
time = 2.96, size = 2011, normalized size = 8.24 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/40*(315*(c^4*d^7*x^2*e^2 + a^2*c^2*d^2*x^3*e^7 + (2*a*c^3*d^3*x^4 + 3*a^2*c^2*d^3*x^2)*e^6 + (c^4*d^4*x^5 +
6*a*c^3*d^4*x^3 + 3*a^2*c^2*d^4*x)*e^5 + (3*c^4*d^5*x^4 + 6*a*c^3*d^5*x^2 + a^2*c^2*d^5)*e^4 + (3*c^4*d^6*x^3
+ 2*a*c^3*d^6*x)*e^3)*sqrt(c*d/(c*d^2 - a*e^2))*log((c*d*x*e + 2*c*d^2 - 2*(c*d^2 - a*e^2)*sqrt(x*e + d)*sqrt
(c*d/(c*d^2 - a*e^2)) - a*e^2)/(c*d*x + a*e)) + 2*(45*c^4*d^7*x*e - 10*c^4*d^8 - 24*a^3*c*d*x*e^7 + 8*a^4*e^8
+ 56*(3*a^2*c^2*d^2*x^2 - a^3*c*d^2)*e^6 + 3*(175*a*c^3*d^3*x^3 + 136*a^2*c^2*d^3*x)*e^5 + 3*(105*c^4*d^4*x^4
+ 413*a*c^3*d^4*x^2 + 96*a^2*c^2*d^4)*e^4 + 3*(245*c^4*d^5*x^3 + 277*a*c^3*d^5*x)*e^3 + (483*c^4*d^6*x^2 + 85*
a*c^3*d^6)*e^2)*sqrt(x*e + d))/(c^7*d^15*x^2 - a^7*x^3*e^15 - (2*a^6*c*d*x^4 + 3*a^7*d*x^2)*e^14 - (a^5*c^2*d^
2*x^5 + a^6*c*d^2*x^3 + 3*a^7*d^2*x)*e^13 + (7*a^5*c^2*d^3*x^4 + 9*a^6*c*d^3*x^2 - a^7*d^3)*e^12 + (5*a^4*c^3*
d^4*x^5 + 17*a^5*c^2*d^4*x^3 + 13*a^6*c*d^4*x)*e^11 - (5*a^4*c^3*d^5*x^4 + a^5*c^2*d^5*x^2 - 5*a^6*c*d^5)*e^10
- 5*(2*a^3*c^4*d^6*x^5 + 7*a^4*c^3*d^6*x^3 + 4*a^5*c^2*d^6*x)*e^9 - 5*(2*a^3*c^4*d^7*x^4 + 5*a^4*c^3*d^7*x^2
+ 2*a^5*c^2*d^7)*e^8 + 5*(2*a^2*c^5*d^8*x^5 + 5*a^3*c^4*d^8*x^3 + 2*a^4*c^3*d^8*x)*e^7 + 5*(4*a^2*c^5*d^9*x^4
+ 7*a^3*c^4*d^9*x^2 + 2*a^4*c^3*d^9)*e^6 - (5*a*c^6*d^10*x^5 - a^2*c^5*d^10*x^3 - 5*a^3*c^4*d^10*x)*e^5 - (13*
a*c^6*d^11*x^4 + 17*a^2*c^5*d^11*x^2 + 5*a^3*c^4*d^11)*e^4 + (c^7*d^12*x^5 - 9*a*c^6*d^12*x^3 - 7*a^2*c^5*d^12
*x)*e^3 + (3*c^7*d^13*x^4 + a*c^6*d^13*x^2 + a^2*c^5*d^13)*e^2 + (3*c^7*d^14*x^3 + 2*a*c^6*d^14*x)*e), -1/20*(
315*(c^4*d^7*x^2*e^2 + a^2*c^2*d^2*x^3*e^7 + (2*a*c^3*d^3*x^4 + 3*a^2*c^2*d^3*x^2)*e^6 + (c^4*d^4*x^5 + 6*a*c^
3*d^4*x^3 + 3*a^2*c^2*d^4*x)*e^5 + (3*c^4*d^5*x^4 + 6*a*c^3*d^5*x^2 + a^2*c^2*d^5)*e^4 + (3*c^4*d^6*x^3 + 2*a*
c^3*d^6*x)*e^3)*sqrt(-c*d/(c*d^2 - a*e^2))*arctan(-(c*d^2 - a*e^2)*sqrt(x*e + d)*sqrt(-c*d/(c*d^2 - a*e^2))/(c
*d*x*e + c*d^2)) - (45*c^4*d^7*x*e - 10*c^4*d^8 - 24*a^3*c*d*x*e^7 + 8*a^4*e^8 + 56*(3*a^2*c^2*d^2*x^2 - a^3*c
*d^2)*e^6 + 3*(175*a*c^3*d^3*x^3 + 136*a^2*c^2*d^3*x)*e^5 + 3*(105*c^4*d^4*x^4 + 413*a*c^3*d^4*x^2 + 96*a^2*c^
2*d^4)*e^4 + 3*(245*c^4*d^5*x^3 + 277*a*c^3*d^5*x)*e^3 + (483*c^4*d^6*x^2 + 85*a*c^3*d^6)*e^2)*sqrt(x*e + d))/
(c^7*d^15*x^2 - a^7*x^3*e^15 - (2*a^6*c*d*x^4 + 3*a^7*d*x^2)*e^14 - (a^5*c^2*d^2*x^5 + a^6*c*d^2*x^3 + 3*a^7*d
^2*x)*e^13 + (7*a^5*c^2*d^3*x^4 + 9*a^6*c*d^3*x^2 - a^7*d^3)*e^12 + (5*a^4*c^3*d^4*x^5 + 17*a^5*c^2*d^4*x^3 +
13*a^6*c*d^4*x)*e^11 - (5*a^4*c^3*d^5*x^4 + a^5*c^2*d^5*x^2 - 5*a^6*c*d^5)*e^10 - 5*(2*a^3*c^4*d^6*x^5 + 7*a^4
*c^3*d^6*x^3 + 4*a^5*c^2*d^6*x)*e^9 - 5*(2*a^3*c^4*d^7*x^4 + 5*a^4*c^3*d^7*x^2 + 2*a^5*c^2*d^7)*e^8 + 5*(2*a^2
*c^5*d^8*x^5 + 5*a^3*c^4*d^8*x^3 + 2*a^4*c^3*d^8*x)*e^7 + 5*(4*a^2*c^5*d^9*x^4 + 7*a^3*c^4*d^9*x^2 + 2*a^4*c^3
*d^9)*e^6 - (5*a*c^6*d^10*x^5 - a^2*c^5*d^10*x^3 - 5*a^3*c^4*d^10*x)*e^5 - (13*a*c^6*d^11*x^4 + 17*a^2*c^5*d^1
1*x^2 + 5*a^3*c^4*d^11)*e^4 + (c^7*d^12*x^5 - 9*a*c^6*d^12*x^3 - 7*a^2*c^5*d^12*x)*e^3 + (3*c^7*d^13*x^4 + a*c
^6*d^13*x^2 + a^2*c^5*d^13)*e^2 + (3*c^7*d^14*x^3 + 2*a*c^6*d^14*x)*e)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (d + e x\right )^{\frac {7}{2}} \left (a e + c d x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((d + e*x)**(7/2)*(a*e + c*d*x)**3), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 414 vs. \(2 (206) = 412\).
time = 2.46, size = 414, normalized size = 1.70 \begin {gather*} \frac {63 \, c^{3} d^{3} \arctan \left (\frac {\sqrt {x e + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right ) e^{2}}{4 \, {\left (c^{5} d^{10} - 5 \, a c^{4} d^{8} e^{2} + 10 \, a^{2} c^{3} d^{6} e^{4} - 10 \, a^{3} c^{2} d^{4} e^{6} + 5 \, a^{4} c d^{2} e^{8} - a^{5} e^{10}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} + \frac {15 \, {\left (x e + d\right )}^{\frac {3}{2}} c^{4} d^{4} e^{2} - 17 \, \sqrt {x e + d} c^{4} d^{5} e^{2} + 17 \, \sqrt {x e + d} a c^{3} d^{3} e^{4}}{4 \, {\left (c^{5} d^{10} - 5 \, a c^{4} d^{8} e^{2} + 10 \, a^{2} c^{3} d^{6} e^{4} - 10 \, a^{3} c^{2} d^{4} e^{6} + 5 \, a^{4} c d^{2} e^{8} - a^{5} e^{10}\right )} {\left ({\left (x e + d\right )} c d - c d^{2} + a e^{2}\right )}^{2}} + \frac {2 \, {\left (30 \, {\left (x e + d\right )}^{2} c^{2} d^{2} e^{2} + 5 \, {\left (x e + d\right )} c^{2} d^{3} e^{2} + c^{2} d^{4} e^{2} - 5 \, {\left (x e + d\right )} a c d e^{4} - 2 \, a c d^{2} e^{4} + a^{2} e^{6}\right )}}{5 \, {\left (c^{5} d^{10} - 5 \, a c^{4} d^{8} e^{2} + 10 \, a^{2} c^{3} d^{6} e^{4} - 10 \, a^{3} c^{2} d^{4} e^{6} + 5 \, a^{4} c d^{2} e^{8} - a^{5} e^{10}\right )} {\left (x e + d\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

63/4*c^3*d^3*arctan(sqrt(x*e + d)*c*d/sqrt(-c^2*d^3 + a*c*d*e^2))*e^2/((c^5*d^10 - 5*a*c^4*d^8*e^2 + 10*a^2*c^
3*d^6*e^4 - 10*a^3*c^2*d^4*e^6 + 5*a^4*c*d^2*e^8 - a^5*e^10)*sqrt(-c^2*d^3 + a*c*d*e^2)) + 1/4*(15*(x*e + d)^(
3/2)*c^4*d^4*e^2 - 17*sqrt(x*e + d)*c^4*d^5*e^2 + 17*sqrt(x*e + d)*a*c^3*d^3*e^4)/((c^5*d^10 - 5*a*c^4*d^8*e^2
+ 10*a^2*c^3*d^6*e^4 - 10*a^3*c^2*d^4*e^6 + 5*a^4*c*d^2*e^8 - a^5*e^10)*((x*e + d)*c*d - c*d^2 + a*e^2)^2) +
2/5*(30*(x*e + d)^2*c^2*d^2*e^2 + 5*(x*e + d)*c^2*d^3*e^2 + c^2*d^4*e^2 - 5*(x*e + d)*a*c*d*e^4 - 2*a*c*d^2*e^
4 + a^2*e^6)/((c^5*d^10 - 5*a*c^4*d^8*e^2 + 10*a^2*c^3*d^6*e^4 - 10*a^3*c^2*d^4*e^6 + 5*a^4*c*d^2*e^8 - a^5*e^
10)*(x*e + d)^(5/2))

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Mupad [B]
time = 0.95, size = 344, normalized size = 1.41 \begin {gather*} -\frac {\frac {2\,e^2}{5\,\left (a\,e^2-c\,d^2\right )}-\frac {6\,c\,d\,e^2\,\left (d+e\,x\right )}{5\,{\left (a\,e^2-c\,d^2\right )}^2}+\frac {42\,c^2\,d^2\,e^2\,{\left (d+e\,x\right )}^2}{5\,{\left (a\,e^2-c\,d^2\right )}^3}+\frac {105\,c^3\,d^3\,e^2\,{\left (d+e\,x\right )}^3}{4\,{\left (a\,e^2-c\,d^2\right )}^4}+\frac {63\,c^4\,d^4\,e^2\,{\left (d+e\,x\right )}^4}{4\,{\left (a\,e^2-c\,d^2\right )}^5}}{{\left (d+e\,x\right )}^{5/2}\,\left (a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4\right )-\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )\,{\left (d+e\,x\right )}^{7/2}+c^2\,d^2\,{\left (d+e\,x\right )}^{9/2}}-\frac {63\,c^{5/2}\,d^{5/2}\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}\,\left (a^5\,e^{10}-5\,a^4\,c\,d^2\,e^8+10\,a^3\,c^2\,d^4\,e^6-10\,a^2\,c^3\,d^6\,e^4+5\,a\,c^4\,d^8\,e^2-c^5\,d^{10}\right )}{{\left (a\,e^2-c\,d^2\right )}^{11/2}}\right )}{4\,{\left (a\,e^2-c\,d^2\right )}^{11/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*e^2)/(5*(a*e^2 - c*d^2)) - (6*c*d*e^2*(d + e*x))/(5*(a*e^2 - c*d^2)^2) + (42*c^2*d^2*e^2*(d + e*x)^2)/(5
*(a*e^2 - c*d^2)^3) + (105*c^3*d^3*e^2*(d + e*x)^3)/(4*(a*e^2 - c*d^2)^4) + (63*c^4*d^4*e^2*(d + e*x)^4)/(4*(a
*e^2 - c*d^2)^5))/((d + e*x)^(5/2)*(a^2*e^4 + c^2*d^4 - 2*a*c*d^2*e^2) - (2*c^2*d^3 - 2*a*c*d*e^2)*(d + e*x)^(
7/2) + c^2*d^2*(d + e*x)^(9/2)) - (63*c^(5/2)*d^(5/2)*e^2*atan((c^(1/2)*d^(1/2)*(d + e*x)^(1/2)*(a^5*e^10 - c^
5*d^10 + 5*a*c^4*d^8*e^2 - 5*a^4*c*d^2*e^8 - 10*a^2*c^3*d^6*e^4 + 10*a^3*c^2*d^4*e^6))/(a*e^2 - c*d^2)^(11/2))
)/(4*(a*e^2 - c*d^2)^(11/2))

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