3.2022 \(\int \frac {(d+e x)^{7/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\)

Optimal. Leaf size=144 \[ \frac {e^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{3/2} d^{3/2} \left (c d^2-a e^2\right )^{3/2}}-\frac {e \sqrt {d+e x}}{4 c d \left (c d^2-a e^2\right ) (a e+c d x)}-\frac {\sqrt {d+e x}}{2 c d (a e+c d x)^2} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 144, 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, 51, 63, 208} \[ \frac {e^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{3/2} d^{3/2} \left (c d^2-a e^2\right )^{3/2}}-\frac {e \sqrt {d+e x}}{4 c d \left (c d^2-a e^2\right ) (a e+c d x)}-\frac {\sqrt {d+e x}}{2 c d (a e+c d x)^2} \]

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)^3,x]

[Out]

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

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 51

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 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 )^3} \, dx &=\int \frac {\sqrt {d+e x}}{(a e+c d x)^3} \, dx\\ &=-\frac {\sqrt {d+e x}}{2 c d (a e+c d x)^2}+\frac {e \int \frac {1}{(a e+c d x)^2 \sqrt {d+e x}} \, dx}{4 c d}\\ &=-\frac {\sqrt {d+e x}}{2 c d (a e+c d x)^2}-\frac {e \sqrt {d+e x}}{4 c d \left (c d^2-a e^2\right ) (a e+c d x)}-\frac {e^2 \int \frac {1}{(a e+c d x) \sqrt {d+e x}} \, dx}{8 c d \left (c d^2-a e^2\right )}\\ &=-\frac {\sqrt {d+e x}}{2 c d (a e+c d x)^2}-\frac {e \sqrt {d+e x}}{4 c d \left (c d^2-a e^2\right ) (a e+c d x)}-\frac {e \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 )}{4 c d \left (c d^2-a e^2\right )}\\ &=-\frac {\sqrt {d+e x}}{2 c d (a e+c d x)^2}-\frac {e \sqrt {d+e x}}{4 c d \left (c d^2-a e^2\right ) (a e+c d x)}+\frac {e^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{3/2} d^{3/2} \left (c d^2-a e^2\right )^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 61, normalized size = 0.42 \[ \frac {2 e^2 (d+e x)^{3/2} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};-\frac {c d (d+e x)}{a e^2-c d^2}\right )}{3 \left (a e^2-c d^2\right )^3} \]

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)^3,x]

[Out]

(2*e^2*(d + e*x)^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, -((c*d*(d + e*x))/(-(c*d^2) + a*e^2))])/(3*(-(c*d^2) + a
*e^2)^3)

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fricas [B]  time = 1.08, size = 565, normalized size = 3.92 \[ \left [\frac {{\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {c^{2} d^{3} - a c d e^{2}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} + 2 \, \sqrt {c^{2} d^{3} - a c d e^{2}} \sqrt {e x + d}}{c d x + a e}\right ) - 2 \, {\left (2 \, c^{3} d^{5} - 3 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4} + {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} c^{4} d^{6} e^{2} - 2 \, a^{3} c^{3} d^{4} e^{4} + a^{4} c^{2} d^{2} e^{6} + {\left (c^{6} d^{8} - 2 \, a c^{5} d^{6} e^{2} + a^{2} c^{4} d^{4} e^{4}\right )} x^{2} + 2 \, {\left (a c^{5} d^{7} e - 2 \, a^{2} c^{4} d^{5} e^{3} + a^{3} c^{3} d^{3} e^{5}\right )} x\right )}}, -\frac {{\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}} \arctan \left (\frac {\sqrt {-c^{2} d^{3} + a c d e^{2}} \sqrt {e x + d}}{c d e x + c d^{2}}\right ) + {\left (2 \, c^{3} d^{5} - 3 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4} + {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} c^{4} d^{6} e^{2} - 2 \, a^{3} c^{3} d^{4} e^{4} + a^{4} c^{2} d^{2} e^{6} + {\left (c^{6} d^{8} - 2 \, a c^{5} d^{6} e^{2} + a^{2} c^{4} d^{4} e^{4}\right )} x^{2} + 2 \, {\left (a c^{5} d^{7} e - 2 \, a^{2} c^{4} d^{5} e^{3} + a^{3} c^{3} d^{3} e^{5}\right )} x\right )}}\right ] \]

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)^3,x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)^3,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.07, size = 142, normalized size = 0.99 \[ \frac {\left (e x +d \right )^{\frac {3}{2}} e^{2}}{4 \left (c d e x +a \,e^{2}\right )^{2} \left (a \,e^{2}-c \,d^{2}\right )}+\frac {e^{2} \arctan \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}\right )}{4 \left (a \,e^{2}-c \,d^{2}\right ) \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}\, c d}-\frac {\sqrt {e x +d}\, e^{2}}{4 \left (c d e x +a \,e^{2}\right )^{2} c d} \]

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)^3,x)

[Out]

1/4*e^2/(c*d*e*x+a*e^2)^2/(a*e^2-c*d^2)*(e*x+d)^(3/2)-1/4*e^2/(c*d*e*x+a*e^2)^2/c/d*(e*x+d)^(1/2)+1/4*e^2/(a*e
^2-c*d^2)/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 \[ \text {Exception raised: ValueError} \]

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)^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(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 = 0.10, size = 166, normalized size = 1.15 \[ \frac {\frac {e^2\,{\left (d+e\,x\right )}^{3/2}}{4\,\left (a\,e^2-c\,d^2\right )}-\frac {e^2\,\sqrt {d+e\,x}}{4\,c\,d}}{a^2\,e^4+c^2\,d^4-\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )\,\left (d+e\,x\right )+c^2\,d^2\,{\left (d+e\,x\right )}^2-2\,a\,c\,d^2\,e^2}+\frac {e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}}{\sqrt {a\,e^2-c\,d^2}}\right )}{4\,c^{3/2}\,d^{3/2}\,{\left (a\,e^2-c\,d^2\right )}^{3/2}} \]

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)^3,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**3,x)

[Out]

Timed out

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