3.21.23 \(\int \frac {(d+e x)^{5/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\) [2023]

Optimal. Leaf size=146 \[ -\frac {\sqrt {d+e x}}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2}+\frac {3 e \sqrt {d+e x}}{4 \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 \sqrt {c} \sqrt {d} \left (c d^2-a e^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {640, 44, 65, 214} \begin {gather*} \frac {3 e \sqrt {d+e x}}{4 \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac {\sqrt {d+e x}}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 \sqrt {c} \sqrt {d} \left (c d^2-a e^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.31, size = 125, normalized size = 0.86 \begin {gather*} \frac {1}{4} \left (\frac {\sqrt {d+e x} \left (5 a e^2+c d (-2 d+3 e x)\right )}{\left (c d^2-a e^2\right )^2 (a e+c d x)^2}+\frac {3 e^2 \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{\sqrt {c} \sqrt {d} \left (-c d^2+a e^2\right )^{5/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d + e*x]*(5*a*e^2 + c*d*(-2*d + 3*e*x)))/((c*d^2 - a*e^2)^2*(a*e + c*d*x)^2) + (3*e^2*ArcTan[(Sqrt[c]*S
qrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]])/(Sqrt[c]*Sqrt[d]*(-(c*d^2) + a*e^2)^(5/2)))/4

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Maple [A]
time = 0.86, size = 175, normalized size = 1.20

method	result	size
derivativedivides	\(2 e^{2} \left (\frac {\sqrt {e x +d}}{4 \left (e^{2} a -c \,d^{2}\right ) \left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {\frac {3 \sqrt {e x +d}}{8 \left (e^{2} a -c \,d^{2}\right ) \left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )}+\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \left (e^{2} a -c \,d^{2}\right ) \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}}{e^{2} a -c \,d^{2}}\right )\)	\(175\)
default	\(2 e^{2} \left (\frac {\sqrt {e x +d}}{4 \left (e^{2} a -c \,d^{2}\right ) \left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {\frac {3 \sqrt {e x +d}}{8 \left (e^{2} a -c \,d^{2}\right ) \left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )}+\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \left (e^{2} a -c \,d^{2}\right ) \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}}{e^{2} a -c \,d^{2}}\right )\)	\(175\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x,method=_RETURNVERBOSE)

[Out]

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

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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)^(5/2)/(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. 313 vs. \(2 (123) = 246\).
time = 2.32, size = 645, normalized size = 4.42 \begin {gather*} \left [\frac {3 \, {\left (c^{2} d^{2} x^{2} e^{2} + 2 \, a c d x e^{3} + a^{2} e^{4}\right )} \sqrt {c^{2} d^{3} - a c d e^{2}} \log \left (\frac {c d x e + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {c^{2} d^{3} - a c d e^{2}} \sqrt {x e + d}}{c d x + a e}\right ) + 2 \, {\left (3 \, c^{3} d^{4} x e - 2 \, c^{3} d^{5} - 3 \, a c^{2} d^{2} x e^{3} + 7 \, a c^{2} d^{3} e^{2} - 5 \, a^{2} c d e^{4}\right )} \sqrt {x e + d}}{8 \, {\left (c^{6} d^{9} x^{2} + 2 \, a c^{5} d^{8} x e - 6 \, a^{2} c^{4} d^{6} x e^{3} + 6 \, a^{3} c^{3} d^{4} x e^{5} - 2 \, a^{4} c^{2} d^{2} x e^{7} - a^{5} c d e^{8} - {\left (a^{3} c^{3} d^{3} x^{2} - 3 \, a^{4} c^{2} d^{3}\right )} e^{6} + 3 \, {\left (a^{2} c^{4} d^{5} x^{2} - a^{3} c^{3} d^{5}\right )} e^{4} - {\left (3 \, a c^{5} d^{7} x^{2} - a^{2} c^{4} d^{7}\right )} e^{2}\right )}}, \frac {3 \, {\left (c^{2} d^{2} x^{2} e^{2} + 2 \, a c d x e^{3} + a^{2} e^{4}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}} \arctan \left (\frac {\sqrt {-c^{2} d^{3} + a c d e^{2}} \sqrt {x e + d}}{c d x e + c d^{2}}\right ) + {\left (3 \, c^{3} d^{4} x e - 2 \, c^{3} d^{5} - 3 \, a c^{2} d^{2} x e^{3} + 7 \, a c^{2} d^{3} e^{2} - 5 \, a^{2} c d e^{4}\right )} \sqrt {x e + d}}{4 \, {\left (c^{6} d^{9} x^{2} + 2 \, a c^{5} d^{8} x e - 6 \, a^{2} c^{4} d^{6} x e^{3} + 6 \, a^{3} c^{3} d^{4} x e^{5} - 2 \, a^{4} c^{2} d^{2} x e^{7} - a^{5} c d e^{8} - {\left (a^{3} c^{3} d^{3} x^{2} - 3 \, a^{4} c^{2} d^{3}\right )} e^{6} + 3 \, {\left (a^{2} c^{4} d^{5} x^{2} - a^{3} c^{3} d^{5}\right )} e^{4} - {\left (3 \, a c^{5} d^{7} x^{2} - a^{2} c^{4} d^{7}\right )} e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="fricas")

[Out]

[1/8*(3*(c^2*d^2*x^2*e^2 + 2*a*c*d*x*e^3 + a^2*e^4)*sqrt(c^2*d^3 - a*c*d*e^2)*log((c*d*x*e + 2*c*d^2 - a*e^2 -
2*sqrt(c^2*d^3 - a*c*d*e^2)*sqrt(x*e + d))/(c*d*x + a*e)) + 2*(3*c^3*d^4*x*e - 2*c^3*d^5 - 3*a*c^2*d^2*x*e^3
+ 7*a*c^2*d^3*e^2 - 5*a^2*c*d*e^4)*sqrt(x*e + d))/(c^6*d^9*x^2 + 2*a*c^5*d^8*x*e - 6*a^2*c^4*d^6*x*e^3 + 6*a^3
*c^3*d^4*x*e^5 - 2*a^4*c^2*d^2*x*e^7 - a^5*c*d*e^8 - (a^3*c^3*d^3*x^2 - 3*a^4*c^2*d^3)*e^6 + 3*(a^2*c^4*d^5*x^
2 - a^3*c^3*d^5)*e^4 - (3*a*c^5*d^7*x^2 - a^2*c^4*d^7)*e^2), 1/4*(3*(c^2*d^2*x^2*e^2 + 2*a*c*d*x*e^3 + a^2*e^4
)*sqrt(-c^2*d^3 + a*c*d*e^2)*arctan(sqrt(-c^2*d^3 + a*c*d*e^2)*sqrt(x*e + d)/(c*d*x*e + c*d^2)) + (3*c^3*d^4*x
*e - 2*c^3*d^5 - 3*a*c^2*d^2*x*e^3 + 7*a*c^2*d^3*e^2 - 5*a^2*c*d*e^4)*sqrt(x*e + d))/(c^6*d^9*x^2 + 2*a*c^5*d^
8*x*e - 6*a^2*c^4*d^6*x*e^3 + 6*a^3*c^3*d^4*x*e^5 - 2*a^4*c^2*d^2*x*e^7 - a^5*c*d*e^8 - (a^3*c^3*d^3*x^2 - 3*a
^4*c^2*d^3)*e^6 + 3*(a^2*c^4*d^5*x^2 - a^3*c^3*d^5)*e^4 - (3*a*c^5*d^7*x^2 - a^2*c^4*d^7)*e^2)]

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Sympy [F(-1)] Timed out
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)**(5/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3,x)

[Out]

Timed out

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Giac [A]
time = 1.32, size = 169, normalized size = 1.16 \begin {gather*} \frac {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^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} + \frac {3 \, {\left (x e + d\right )}^{\frac {3}{2}} c d e^{2} - 5 \, \sqrt {x e + d} c d^{2} e^{2} + 5 \, \sqrt {x e + d} a e^{4}}{4 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} {\left ({\left (x e + d\right )} c d - c d^{2} + a e^{2}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.72, size = 177, normalized size = 1.21 \begin {gather*} \frac {\frac {5\,e^2\,\sqrt {d+e\,x}}{4\,\left (a\,e^2-c\,d^2\right )}+\frac {3\,c\,d\,e^2\,{\left (d+e\,x\right )}^{3/2}}{4\,{\left (a\,e^2-c\,d^2\right )}^2}}{a^2\,e^4+c^2\,d^4-\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )\,\left (d+e\,x\right )+c^2\,d^2\,{\left (d+e\,x\right )}^2-2\,a\,c\,d^2\,e^2}+\frac {3\,e^2\,\mathrm {atan}\left (\frac {c\,d\,\sqrt {d+e\,x}}{\sqrt {c\,d}\,\sqrt {a\,e^2-c\,d^2}}\right )}{4\,\sqrt {c\,d}\,{\left (a\,e^2-c\,d^2\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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