3.465 \(\int \frac {1}{(d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}})^{5/2}} \, dx\)

Optimal. Leaf size=199 \[ \frac {5 a f^2 \tanh ^{-1}\left (\frac {\sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {d}}\right )}{2 d^{7/2} e}-\frac {a f^2 \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{2 d^3 e \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )}-\frac {2 a f^2}{d^3 e \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}-\frac {\frac {a f^2}{d^2}+1}{3 e \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )^{3/2}} \]

[Out]

5/2*a*f^2*arctanh((d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2)/d^(1/2))/d^(7/2)/e+1/3*(-1-a*f^2/d^2)/e/(d+e*x+f*(a+e^
2*x^2/f^2)^(1/2))^(3/2)-2*a*f^2/d^3/e/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2)-1/2*a*f^2*(d+e*x+f*(a+e^2*x^2/f^2)
^(1/2))^(1/2)/d^3/e/(e*x+f*(a+e^2*x^2/f^2)^(1/2))

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Rubi [A]  time = 0.18, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2117, 897, 1259, 1261, 206} \[ -\frac {a f^2 \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{2 d^3 e \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x\right )}-\frac {2 a f^2}{d^3 e \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}-\frac {\frac {a f^2}{d^2}+1}{3 e \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )^{3/2}}+\frac {5 a f^2 \tanh ^{-1}\left (\frac {\sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {d}}\right )}{2 d^{7/2} e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(-5/2),x]

[Out]

-(1 + (a*f^2)/d^2)/(3*e*(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(3/2)) - (2*a*f^2)/(d^3*e*Sqrt[d + e*x + f*Sqrt[
a + (e^2*x^2)/f^2]]) - (a*f^2*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/(2*d^3*e*(e*x + f*Sqrt[a + (e^2*x^2)/
f^2])) + (5*a*f^2*ArcTanh[Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]]/Sqrt[d]])/(2*d^(7/2)*e)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 2117

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[((g + h*x^n)^p*(d^2 + a*f^2 - 2*d*x + x^2))/(d - x)^2, x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.60, size = 186, normalized size = 0.93 \[ -\frac {\frac {2 d \left (a f^2+d^2\right )}{3 \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )^{3/2}}+\frac {a f^2 \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{f \sqrt {a+\frac {e^2 x^2}{f^2}}+e x}+\frac {4 a f^2}{\sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}-\frac {5 a f^2 \tanh ^{-1}\left (\frac {\sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}}{2 d^3 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(-5/2),x]

[Out]

-1/2*((2*d*(d^2 + a*f^2))/(3*(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(3/2)) + (4*a*f^2)/Sqrt[d + e*x + f*Sqrt[a
+ (e^2*x^2)/f^2]] + (a*f^2*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/(e*x + f*Sqrt[a + (e^2*x^2)/f^2]) - (5*a
*f^2*ArcTanh[Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]]/Sqrt[d]])/Sqrt[d])/(d^3*e)

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fricas [B]  time = 0.60, size = 812, normalized size = 4.08 \[ \left [\frac {15 \, {\left (a^{3} f^{6} + 4 \, a d^{2} e^{2} f^{2} x^{2} - 2 \, a^{2} d^{2} f^{4} + a d^{4} f^{2} - 4 \, {\left (a^{2} d e f^{4} - a d^{3} e f^{2}\right )} x\right )} \sqrt {d} \log \left (a f^{2} - 2 \, d e x + 2 \, d f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} - 2 \, {\left (\sqrt {d} e x - \sqrt {d} f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d}\right ) + 2 \, {\left (12 \, d^{3} e^{3} x^{3} + 10 \, a^{2} d^{2} f^{4} - 16 \, a d^{4} f^{2} - 2 \, d^{6} - 8 \, {\left (5 \, a d^{2} e^{2} f^{2} - d^{4} e^{2}\right )} x^{2} + {\left (15 \, a^{2} d e f^{4} - 46 \, a d^{3} e f^{2} - d^{5} e\right )} x - {\left (15 \, a^{2} d f^{5} + 12 \, d^{3} e^{2} f x^{2} - 22 \, a d^{3} f^{3} - d^{5} f - 8 \, {\left (5 \, a d^{2} e f^{3} - d^{4} e f\right )} x\right )} \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{12 \, {\left (a^{2} d^{4} e f^{4} + 4 \, d^{6} e^{3} x^{2} - 2 \, a d^{6} e f^{2} + d^{8} e - 4 \, {\left (a d^{5} e^{2} f^{2} - d^{7} e^{2}\right )} x\right )}}, -\frac {15 \, {\left (a^{3} f^{6} + 4 \, a d^{2} e^{2} f^{2} x^{2} - 2 \, a^{2} d^{2} f^{4} + a d^{4} f^{2} - 4 \, {\left (a^{2} d e f^{4} - a d^{3} e f^{2}\right )} x\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d} \sqrt {-d}}{d}\right ) - {\left (12 \, d^{3} e^{3} x^{3} + 10 \, a^{2} d^{2} f^{4} - 16 \, a d^{4} f^{2} - 2 \, d^{6} - 8 \, {\left (5 \, a d^{2} e^{2} f^{2} - d^{4} e^{2}\right )} x^{2} + {\left (15 \, a^{2} d e f^{4} - 46 \, a d^{3} e f^{2} - d^{5} e\right )} x - {\left (15 \, a^{2} d f^{5} + 12 \, d^{3} e^{2} f x^{2} - 22 \, a d^{3} f^{3} - d^{5} f - 8 \, {\left (5 \, a d^{2} e f^{3} - d^{4} e f\right )} x\right )} \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{6 \, {\left (a^{2} d^{4} e f^{4} + 4 \, d^{6} e^{3} x^{2} - 2 \, a d^{6} e f^{2} + d^{8} e - 4 \, {\left (a d^{5} e^{2} f^{2} - d^{7} e^{2}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(5/2),x, algorithm="fricas")

[Out]

[1/12*(15*(a^3*f^6 + 4*a*d^2*e^2*f^2*x^2 - 2*a^2*d^2*f^4 + a*d^4*f^2 - 4*(a^2*d*e*f^4 - a*d^3*e*f^2)*x)*sqrt(d
)*log(a*f^2 - 2*d*e*x + 2*d*f*sqrt((e^2*x^2 + a*f^2)/f^2) - 2*(sqrt(d)*e*x - sqrt(d)*f*sqrt((e^2*x^2 + a*f^2)/
f^2))*sqrt(e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) + d)) + 2*(12*d^3*e^3*x^3 + 10*a^2*d^2*f^4 - 16*a*d^4*f^2 - 2*d
^6 - 8*(5*a*d^2*e^2*f^2 - d^4*e^2)*x^2 + (15*a^2*d*e*f^4 - 46*a*d^3*e*f^2 - d^5*e)*x - (15*a^2*d*f^5 + 12*d^3*
e^2*f*x^2 - 22*a*d^3*f^3 - d^5*f - 8*(5*a*d^2*e*f^3 - d^4*e*f)*x)*sqrt((e^2*x^2 + a*f^2)/f^2))*sqrt(e*x + f*sq
rt((e^2*x^2 + a*f^2)/f^2) + d))/(a^2*d^4*e*f^4 + 4*d^6*e^3*x^2 - 2*a*d^6*e*f^2 + d^8*e - 4*(a*d^5*e^2*f^2 - d^
7*e^2)*x), -1/6*(15*(a^3*f^6 + 4*a*d^2*e^2*f^2*x^2 - 2*a^2*d^2*f^4 + a*d^4*f^2 - 4*(a^2*d*e*f^4 - a*d^3*e*f^2)
*x)*sqrt(-d)*arctan(sqrt(e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) + d)*sqrt(-d)/d) - (12*d^3*e^3*x^3 + 10*a^2*d^2*f
^4 - 16*a*d^4*f^2 - 2*d^6 - 8*(5*a*d^2*e^2*f^2 - d^4*e^2)*x^2 + (15*a^2*d*e*f^4 - 46*a*d^3*e*f^2 - d^5*e)*x -
(15*a^2*d*f^5 + 12*d^3*e^2*f*x^2 - 22*a*d^3*f^3 - d^5*f - 8*(5*a*d^2*e*f^3 - d^4*e*f)*x)*sqrt((e^2*x^2 + a*f^2
)/f^2))*sqrt(e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) + d))/(a^2*d^4*e*f^4 + 4*d^6*e^3*x^2 - 2*a*d^6*e*f^2 + d^8*e
- 4*(a*d^5*e^2*f^2 - d^7*e^2)*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(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e x +d +\sqrt {\frac {e^{2} x^{2}}{f^{2}}+a}\, f \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d+(e^2/f^2*x^2+a)^(1/2)*f)^(5/2),x)

[Out]

int(1/(e*x+d+(e^2/f^2*x^2+a)^(1/2)*f)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(5/2),x, algorithm="maxima")

[Out]

integrate((e*x + sqrt(e^2*x^2/f^2 + a)*f + d)^(-5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (d+e\,x+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d + e*x + f*(a + (e^2*x^2)/f^2)^(1/2))^(5/2),x)

[Out]

int(1/(d + e*x + f*(a + (e^2*x^2)/f^2)^(1/2))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d + e x + f \sqrt {a + \frac {e^{2} x^{2}}{f^{2}}}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x+f*(a+e**2*x**2/f**2)**(1/2))**(5/2),x)

[Out]

Integral((d + e*x + f*sqrt(a + e**2*x**2/f**2))**(-5/2), x)

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