3.17.96 \(\int \frac {(d+e x)^{7/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\)

Optimal. Leaf size=112 \[ -\frac {3 e \sqrt {c d^2-a e^2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{5/2} d^{5/2}}-\frac {(d+e x)^{3/2}}{c d (a e+c d x)}+\frac {3 e \sqrt {d+e x}}{c^2 d^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {3 e \sqrt {c d^2-a e^2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{5/2} d^{5/2}}-\frac {(d+e x)^{3/2}}{c d (a e+c d x)}+\frac {3 e \sqrt {d+e x}}{c^2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 59, normalized size = 0.53 \begin {gather*} \frac {2 e (d+e x)^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};-\frac {c d (d+e x)}{a e^2-c d^2}\right )}{5 \left (a e^2-c d^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*(d + e*x)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, -((c*d*(d + e*x))/(-(c*d^2) + a*e^2))])/(5*(-(c*d^2) + a*e
^2)^2)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.38, size = 144, normalized size = 1.29 \begin {gather*} \frac {3 e \sqrt {a e^2-c d^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^{5/2} d^{5/2}}+\frac {\sqrt {d+e x} \left (-3 a e^3+3 c d^2 e-2 c d e (d+e x)\right )}{c^2 d^2 \left (-a e^2+c d^2-c d (d+e x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*(3*c*d^2*e - 3*a*e^3 - 2*c*d*e*(d + e*x)))/(c^2*d^2*(c*d^2 - a*e^2 - c*d*(d + e*x))) + (3*e*Sqr
t[-(c*d^2) + a*e^2]*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[-(c*d^2) + a*e^2]*Sqrt[d + e*x])/(c*d^2 - a*e^2)])/(c^(5/2)*d
^(5/2))

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 282, normalized size = 2.52 \begin {gather*} \left [\frac {3 \, {\left (c d e x + a e^{2}\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 d e x - c d^{2} + 3 \, a e^{2}\right )} \sqrt {e x + d}}{2 \, {\left (c^{3} d^{3} x + a c^{2} d^{2} e\right )}}, -\frac {3 \, {\left (c d e x + a e^{2}\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 d e x - c d^{2} + 3 \, a e^{2}\right )} \sqrt {e x + d}}{c^{3} d^{3} x + a c^{2} d^{2} e}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="fricas")

[Out]

[1/2*(3*(c*d*e*x + a*e^2)*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*sqr
t((c*d^2 - a*e^2)/(c*d)))/(c*d*x + a*e)) + 2*(2*c*d*e*x - c*d^2 + 3*a*e^2)*sqrt(e*x + d))/(c^3*d^3*x + a*c^2*d
^2*e), -(3*(c*d*e*x + a*e^2)*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*d*e*x - c*d^2 + 3*a*e^2)*sqrt(e*x + d))/(c^3*d^3*x + a*c^2*d^2*e)]

________________________________________________________________________________________

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)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.07, size = 183, normalized size = 1.63 \begin {gather*} -\frac {3 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^{2}}+\frac {3 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 \,e^{3}}{\left (c d e x +a \,e^{2}\right ) c^{2} d^{2}}-\frac {\sqrt {e x +d}\, e}{\left (c d e x +a \,e^{2}\right ) c}+\frac {2 \sqrt {e x +d}\, e}{c^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(7/2)/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^2,x)

[Out]

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

________________________________________________________________________________________

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)^(7/2)/(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?

________________________________________________________________________________________

mupad [B]  time = 0.13, size = 140, normalized size = 1.25 \begin {gather*} \frac {\left (a\,e^3-c\,d^2\,e\right )\,\sqrt {d+e\,x}}{c^3\,d^3\,\left (d+e\,x\right )-c^3\,d^4+a\,c^2\,d^2\,e^2}+\frac {2\,e\,\sqrt {d+e\,x}}{c^2\,d^2}-\frac {3\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,e\,\sqrt {a\,e^2-c\,d^2}\,\sqrt {d+e\,x}}{a\,e^3-c\,d^2\,e}\right )\,\sqrt {a\,e^2-c\,d^2}}{c^{5/2}\,d^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(7/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^2,x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________