3.2034 \(\int \frac {\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=199 \[ \frac {c^2 d^2 \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d^2-a e^2}}\right )}{4 e^{3/2} \left (c d^2-a e^2\right )^{3/2}}+\frac {c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 e (d+e x)^{3/2} \left (c d^2-a e^2\right )}-\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{2 e (d+e x)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {662, 672, 660, 205} \[ \frac {c^2 d^2 \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt {d+e x} \sqrt {c d^2-a e^2}}\right )}{4 e^{3/2} \left (c d^2-a e^2\right )^{3/2}}+\frac {c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 e (d+e x)^{3/2} \left (c d^2-a e^2\right )}-\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{2 e (d+e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

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

Rule 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*(m + 2*p + 2))/((m + p + 1)*(2*c*d - b*e)), I
nt[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ
[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{(d+e x)^{7/2}} \, dx &=-\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 e (d+e x)^{5/2}}+\frac {(c d) \int \frac {1}{(d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{4 e}\\ &=-\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 e (d+e x)^{5/2}}+\frac {c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e \left (c d^2-a e^2\right ) (d+e x)^{3/2}}+\frac {\left (c^2 d^2\right ) \int \frac {1}{\sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{8 e \left (c d^2-a e^2\right )}\\ &=-\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 e (d+e x)^{5/2}}+\frac {c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e \left (c d^2-a e^2\right ) (d+e x)^{3/2}}+\frac {\left (c^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{2 c d^2 e-e \left (c d^2+a e^2\right )+e^2 x^2} \, dx,x,\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {d+e x}}\right )}{4 \left (c d^2-a e^2\right )}\\ &=-\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 e (d+e x)^{5/2}}+\frac {c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 e \left (c d^2-a e^2\right ) (d+e x)^{3/2}}+\frac {c^2 d^2 \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {c d^2-a e^2} \sqrt {d+e x}}\right )}{4 e^{3/2} \left (c d^2-a e^2\right )^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 83, normalized size = 0.42 \[ \frac {2 c^2 d^2 ((d+e x) (a e+c d x))^{3/2} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};\frac {e (a e+c d x)}{a e^2-c d^2}\right )}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.09, size = 749, normalized size = 3.76 \[ \left [-\frac {{\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\right )} \sqrt {-c d^{2} e + a e^{3}} \log \left (-\frac {c d e^{2} x^{2} + 2 \, a e^{3} x - c d^{3} + 2 \, a d e^{2} - 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {-c d^{2} e + a e^{3}} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, {\left (c^{2} d^{4} e - 3 \, a c d^{2} e^{3} + 2 \, a^{2} e^{5} - {\left (c^{2} d^{3} e^{2} - a c d e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{8 \, {\left (c^{2} d^{7} e^{2} - 2 \, a c d^{5} e^{4} + a^{2} d^{3} e^{6} + {\left (c^{2} d^{4} e^{5} - 2 \, a c d^{2} e^{7} + a^{2} e^{9}\right )} x^{3} + 3 \, {\left (c^{2} d^{5} e^{4} - 2 \, a c d^{3} e^{6} + a^{2} d e^{8}\right )} x^{2} + 3 \, {\left (c^{2} d^{6} e^{3} - 2 \, a c d^{4} e^{5} + a^{2} d^{2} e^{7}\right )} x\right )}}, -\frac {{\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\right )} \sqrt {c d^{2} e - a e^{3}} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {c d^{2} e - a e^{3}} \sqrt {e x + d}}{c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x}\right ) + {\left (c^{2} d^{4} e - 3 \, a c d^{2} e^{3} + 2 \, a^{2} e^{5} - {\left (c^{2} d^{3} e^{2} - a c d e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{4 \, {\left (c^{2} d^{7} e^{2} - 2 \, a c d^{5} e^{4} + a^{2} d^{3} e^{6} + {\left (c^{2} d^{4} e^{5} - 2 \, a c d^{2} e^{7} + a^{2} e^{9}\right )} x^{3} + 3 \, {\left (c^{2} d^{5} e^{4} - 2 \, a c d^{3} e^{6} + a^{2} d e^{8}\right )} x^{2} + 3 \, {\left (c^{2} d^{6} e^{3} - 2 \, a c d^{4} e^{5} + a^{2} d^{2} e^{7}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{{\left (e x + d\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.09, size = 292, normalized size = 1.47 \[ \frac {\sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}\, \left (c^{2} d^{2} e^{2} x^{2} \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )+2 c^{2} d^{3} e x \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )+c^{2} d^{4} \arctanh \left (\frac {\sqrt {c d x +a e}\, e}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}}\right )-\sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, c d e x -2 \sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, a \,e^{2}+\sqrt {c d x +a e}\, \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, c \,d^{2}\right )}{4 \left (e x +d \right )^{\frac {5}{2}} \sqrt {c d x +a e}\, \left (a \,e^{2}-c \,d^{2}\right ) \sqrt {\left (a \,e^{2}-c \,d^{2}\right ) e}\, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)/(e*x+d)^(7/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{{\left (e x + d\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{{\left (d+e\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)/(d + e*x)^(7/2),x)

[Out]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)/(d + e*x)^(7/2), x)

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

[Out]

Timed out

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