3.1.57 \(\int \frac {(a+b x^2) \sqrt {c+d x^2}}{e+f x^2} \, dx\) [57]

Optimal. Leaf size=128 \[ \frac {b x \sqrt {c+d x^2}}{2 f}-\frac {(2 b d e-b c f-2 a d f) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} f^2}+\frac {(b e-a f) \sqrt {d e-c f} \tanh ^{-1}\left (\frac {\sqrt {d e-c f} x}{\sqrt {e} \sqrt {c+d x^2}}\right )}{\sqrt {e} f^2} \]

[Out]

-1/2*(-2*a*d*f-b*c*f+2*b*d*e)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/f^2/d^(1/2)+(-a*f+b*e)*arctanh(x*(-c*f+d*e)^(
1/2)/e^(1/2)/(d*x^2+c)^(1/2))*(-c*f+d*e)^(1/2)/f^2/e^(1/2)+1/2*b*x*(d*x^2+c)^(1/2)/f

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Rubi [A]
time = 0.10, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {542, 537, 223, 212, 385, 214} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right ) (-2 a d f-b c f+2 b d e)}{2 \sqrt {d} f^2}+\frac {(b e-a f) \sqrt {d e-c f} \tanh ^{-1}\left (\frac {x \sqrt {d e-c f}}{\sqrt {e} \sqrt {c+d x^2}}\right )}{\sqrt {e} f^2}+\frac {b x \sqrt {c+d x^2}}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 212

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

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 537

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 542

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
f*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*(n*(p + q + 1) + 1))), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right ) \sqrt {c+d x^2}}{e+f x^2} \, dx &=\frac {b x \sqrt {c+d x^2}}{2 f}+\frac {\int \frac {-c (b e-2 a f)+(-2 b d e+b c f+2 a d f) x^2}{\sqrt {c+d x^2} \left (e+f x^2\right )} \, dx}{2 f}\\ &=\frac {b x \sqrt {c+d x^2}}{2 f}+\frac {((b e-a f) (d e-c f)) \int \frac {1}{\sqrt {c+d x^2} \left (e+f x^2\right )} \, dx}{f^2}-\frac {(2 b d e-b c f-2 a d f) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{2 f^2}\\ &=\frac {b x \sqrt {c+d x^2}}{2 f}+\frac {((b e-a f) (d e-c f)) \text {Subst}\left (\int \frac {1}{e-(d e-c f) x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{f^2}-\frac {(2 b d e-b c f-2 a d f) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{2 f^2}\\ &=\frac {b x \sqrt {c+d x^2}}{2 f}-\frac {(2 b d e-b c f-2 a d f) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} f^2}+\frac {(b e-a f) \sqrt {d e-c f} \tanh ^{-1}\left (\frac {\sqrt {d e-c f} x}{\sqrt {e} \sqrt {c+d x^2}}\right )}{\sqrt {e} f^2}\\ \end {align*}

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Mathematica [A]
time = 0.38, size = 141, normalized size = 1.10 \begin {gather*} \frac {b f x \sqrt {c+d x^2}+\frac {2 (b e-a f) \sqrt {-d e+c f} \tan ^{-1}\left (\frac {-f x \sqrt {c+d x^2}+\sqrt {d} \left (e+f x^2\right )}{\sqrt {e} \sqrt {-d e+c f}}\right )}{\sqrt {e}}+\frac {(2 b d e-b c f-2 a d f) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{\sqrt {d}}}{2 f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*f*x*Sqrt[c + d*x^2] + (2*(b*e - a*f)*Sqrt[-(d*e) + c*f]*ArcTan[(-(f*x*Sqrt[c + d*x^2]) + Sqrt[d]*(e + f*x^2
))/(Sqrt[e]*Sqrt[-(d*e) + c*f])])/Sqrt[e] + ((2*b*d*e - b*c*f - 2*a*d*f)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/
Sqrt[d])/(2*f^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(706\) vs. \(2(106)=212\).
time = 0.18, size = 707, normalized size = 5.52 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(d*x^2+c)^(1/2)/(f*x^2+e),x,method=_RETURNVERBOSE)

[Out]

b/f*(1/2*x*(d*x^2+c)^(1/2)+1/2*c/d^(1/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2)))+1/2*(a*f-b*e)/(-f*e)^(1/2)/f*(((x-(-f*
e)^(1/2)/f)^2*d+2*d*(-f*e)^(1/2)/f*(x-(-f*e)^(1/2)/f)+(c*f-d*e)/f)^(1/2)+d^(1/2)*(-f*e)^(1/2)/f*ln((d*(-f*e)^(
1/2)/f+d*(x-(-f*e)^(1/2)/f))/d^(1/2)+((x-(-f*e)^(1/2)/f)^2*d+2*d*(-f*e)^(1/2)/f*(x-(-f*e)^(1/2)/f)+(c*f-d*e)/f
)^(1/2))-(c*f-d*e)/f/((c*f-d*e)/f)^(1/2)*ln((2*(c*f-d*e)/f+2*d*(-f*e)^(1/2)/f*(x-(-f*e)^(1/2)/f)+2*((c*f-d*e)/
f)^(1/2)*((x-(-f*e)^(1/2)/f)^2*d+2*d*(-f*e)^(1/2)/f*(x-(-f*e)^(1/2)/f)+(c*f-d*e)/f)^(1/2))/(x-(-f*e)^(1/2)/f))
)+1/2*(-a*f+b*e)/(-f*e)^(1/2)/f*(((x+(-f*e)^(1/2)/f)^2*d-2*d*(-f*e)^(1/2)/f*(x+(-f*e)^(1/2)/f)+(c*f-d*e)/f)^(1
/2)-d^(1/2)*(-f*e)^(1/2)/f*ln((-d*(-f*e)^(1/2)/f+d*(x+(-f*e)^(1/2)/f))/d^(1/2)+((x+(-f*e)^(1/2)/f)^2*d-2*d*(-f
*e)^(1/2)/f*(x+(-f*e)^(1/2)/f)+(c*f-d*e)/f)^(1/2))-(c*f-d*e)/f/((c*f-d*e)/f)^(1/2)*ln((2*(c*f-d*e)/f-2*d*(-f*e
)^(1/2)/f*(x+(-f*e)^(1/2)/f)+2*((c*f-d*e)/f)^(1/2)*((x+(-f*e)^(1/2)/f)^2*d-2*d*(-f*e)^(1/2)/f*(x+(-f*e)^(1/2)/
f)+(c*f-d*e)/f)^(1/2))/(x+(-f*e)^(1/2)/f)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.25, size = 786, normalized size = 6.14 \begin {gather*} \left [\frac {2 \, \sqrt {d x^{2} + c} b d f x + {\left (2 \, b d e - {\left (b c + 2 \, a d\right )} f\right )} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - {\left (a d f - b d e\right )} \sqrt {-{\left (c f - d e\right )} e^{\left (-1\right )}} \log \left (\frac {c^{2} f^{2} x^{4} - 4 \, {\left (c f x^{3} e - {\left (2 \, d x^{3} + c x\right )} e^{2}\right )} \sqrt {d x^{2} + c} \sqrt {-{\left (c f - d e\right )} e^{\left (-1\right )}} + {\left (8 \, d^{2} x^{4} + 8 \, c d x^{2} + c^{2}\right )} e^{2} - 2 \, {\left (4 \, c d f x^{4} + 3 \, c^{2} f x^{2}\right )} e}{f^{2} x^{4} + 2 \, f x^{2} e + e^{2}}\right )}{4 \, d f^{2}}, \frac {2 \, \sqrt {d x^{2} + c} b d f x + 2 \, {\left (2 \, b d e - {\left (b c + 2 \, a d\right )} f\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (a d f - b d e\right )} \sqrt {-{\left (c f - d e\right )} e^{\left (-1\right )}} \log \left (\frac {c^{2} f^{2} x^{4} - 4 \, {\left (c f x^{3} e - {\left (2 \, d x^{3} + c x\right )} e^{2}\right )} \sqrt {d x^{2} + c} \sqrt {-{\left (c f - d e\right )} e^{\left (-1\right )}} + {\left (8 \, d^{2} x^{4} + 8 \, c d x^{2} + c^{2}\right )} e^{2} - 2 \, {\left (4 \, c d f x^{4} + 3 \, c^{2} f x^{2}\right )} e}{f^{2} x^{4} + 2 \, f x^{2} e + e^{2}}\right )}{4 \, d f^{2}}, \frac {2 \, \sqrt {d x^{2} + c} b d f x + 2 \, {\left (a d f - b d e\right )} \sqrt {c f - d e} \arctan \left (\frac {{\left (c f x^{2} - {\left (2 \, d x^{2} + c\right )} e\right )} \sqrt {d x^{2} + c} \sqrt {c f - d e} e^{\left (-\frac {1}{2}\right )}}{2 \, {\left (c d f x^{3} + c^{2} f x - {\left (d^{2} x^{3} + c d x\right )} e\right )}}\right ) e^{\left (-\frac {1}{2}\right )} + {\left (2 \, b d e - {\left (b c + 2 \, a d\right )} f\right )} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right )}{4 \, d f^{2}}, \frac {\sqrt {d x^{2} + c} b d f x + {\left (a d f - b d e\right )} \sqrt {c f - d e} \arctan \left (\frac {{\left (c f x^{2} - {\left (2 \, d x^{2} + c\right )} e\right )} \sqrt {d x^{2} + c} \sqrt {c f - d e} e^{\left (-\frac {1}{2}\right )}}{2 \, {\left (c d f x^{3} + c^{2} f x - {\left (d^{2} x^{3} + c d x\right )} e\right )}}\right ) e^{\left (-\frac {1}{2}\right )} + {\left (2 \, b d e - {\left (b c + 2 \, a d\right )} f\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right )}{2 \, d f^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*sqrt(d*x^2 + c)*b*d*f*x + (2*b*d*e - (b*c + 2*a*d)*f)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)
*x - c) - (a*d*f - b*d*e)*sqrt(-(c*f - d*e)*e^(-1))*log((c^2*f^2*x^4 - 4*(c*f*x^3*e - (2*d*x^3 + c*x)*e^2)*sqr
t(d*x^2 + c)*sqrt(-(c*f - d*e)*e^(-1)) + (8*d^2*x^4 + 8*c*d*x^2 + c^2)*e^2 - 2*(4*c*d*f*x^4 + 3*c^2*f*x^2)*e)/
(f^2*x^4 + 2*f*x^2*e + e^2)))/(d*f^2), 1/4*(2*sqrt(d*x^2 + c)*b*d*f*x + 2*(2*b*d*e - (b*c + 2*a*d)*f)*sqrt(-d)
*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (a*d*f - b*d*e)*sqrt(-(c*f - d*e)*e^(-1))*log((c^2*f^2*x^4 - 4*(c*f*x^3*
e - (2*d*x^3 + c*x)*e^2)*sqrt(d*x^2 + c)*sqrt(-(c*f - d*e)*e^(-1)) + (8*d^2*x^4 + 8*c*d*x^2 + c^2)*e^2 - 2*(4*
c*d*f*x^4 + 3*c^2*f*x^2)*e)/(f^2*x^4 + 2*f*x^2*e + e^2)))/(d*f^2), 1/4*(2*sqrt(d*x^2 + c)*b*d*f*x + 2*(a*d*f -
b*d*e)*sqrt(c*f - d*e)*arctan(1/2*(c*f*x^2 - (2*d*x^2 + c)*e)*sqrt(d*x^2 + c)*sqrt(c*f - d*e)*e^(-1/2)/(c*d*f
*x^3 + c^2*f*x - (d^2*x^3 + c*d*x)*e))*e^(-1/2) + (2*b*d*e - (b*c + 2*a*d)*f)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*
x^2 + c)*sqrt(d)*x - c))/(d*f^2), 1/2*(sqrt(d*x^2 + c)*b*d*f*x + (a*d*f - b*d*e)*sqrt(c*f - d*e)*arctan(1/2*(c
*f*x^2 - (2*d*x^2 + c)*e)*sqrt(d*x^2 + c)*sqrt(c*f - d*e)*e^(-1/2)/(c*d*f*x^3 + c^2*f*x - (d^2*x^3 + c*d*x)*e)
)*e^(-1/2) + (2*b*d*e - (b*c + 2*a*d)*f)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)))/(d*f^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x^{2}\right ) \sqrt {c + d x^{2}}}{e + f x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x**2)*sqrt(c + d*x**2)/(e + f*x**2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (b\,x^2+a\right )\,\sqrt {d\,x^2+c}}{f\,x^2+e} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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