3.6.89 \(\int \frac {a c+2 (b c+a d) x^2+3 b d x^4}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx\) [589]

Optimal. Leaf size=24 \[ x \sqrt {a+b x^2} \sqrt {c+d x^2} \]

[Out]

x*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {1604} \begin {gather*} x \sqrt {a+b x^2} \sqrt {c+d x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*c + 2*(b*c + a*d)*x^2 + 3*b*d*x^4)/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]),x]

[Out]

x*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]

Rule 1604

Int[(Pp_)*(Qq_)^(m_.)*(Rr_)^(n_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x], r = Expon[Rr, x]}, S
imp[Coeff[Pp, x, p]*x^(p - q - r + 1)*Qq^(m + 1)*(Rr^(n + 1)/((p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x,
r])), x] /; NeQ[p + m*q + n*r + 1, 0] && EqQ[(p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x, r]*Pp, Coeff[Pp
, x, p]*x^(p - q - r)*((p - q - r + 1)*Qq*Rr + (m + 1)*x*Rr*D[Qq, x] + (n + 1)*x*Qq*D[Rr, x])]] /; FreeQ[{m, n
}, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && PolyQ[Rr, x] && NeQ[m, -1] && NeQ[n, -1]

Rubi steps

\begin {align*} \int \frac {a c+2 (b c+a d) x^2+3 b d x^4}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx &=x \sqrt {a+b x^2} \sqrt {c+d x^2}\\ \end {align*}

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Mathematica [A]
time = 5.75, size = 24, normalized size = 1.00 \begin {gather*} x \sqrt {a+b x^2} \sqrt {c+d x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*c + 2*(b*c + a*d)*x^2 + 3*b*d*x^4)/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]),x]

[Out]

x*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]

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Maple [A]
time = 0.39, size = 21, normalized size = 0.88

method	result	size
gosper	\(x \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\)	\(21\)
default	\(x \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\)	\(21\)
risch	\(x \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\)	\(21\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, x \sqrt {b d \,x^{4}+a d \,x^{2}+c \,x^{2} b +a c}}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*c+2*(a*d+b*c)*x^2+3*b*d*x^4)/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

x*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)

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Maxima [A]
time = 0.33, size = 20, normalized size = 0.83 \begin {gather*} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+2*(a*d+b*c)*x^2+3*b*d*x^4)/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*x

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Fricas [A]
time = 0.37, size = 20, normalized size = 0.83 \begin {gather*} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+2*(a*d+b*c)*x^2+3*b*d*x^4)/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*x

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a c + 2 a d x^{2} + 2 b c x^{2} + 3 b d x^{4}}{\sqrt {a + b x^{2}} \sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+2*(a*d+b*c)*x**2+3*b*d*x**4)/(b*x**2+a)**(1/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral((a*c + 2*a*d*x**2 + 2*b*c*x**2 + 3*b*d*x**4)/(sqrt(a + b*x**2)*sqrt(c + d*x**2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+2*(a*d+b*c)*x^2+3*b*d*x^4)/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate((3*b*d*x^4 + 2*(b*c + a*d)*x^2 + a*c)/(sqrt(b*x^2 + a)*sqrt(d*x^2 + c)), x)

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Mupad [B]
time = 5.59, size = 20, normalized size = 0.83 \begin {gather*} x\,\sqrt {b\,x^2+a}\,\sqrt {d\,x^2+c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*c + 2*x^2*(a*d + b*c) + 3*b*d*x^4)/((a + b*x^2)^(1/2)*(c + d*x^2)^(1/2)),x)

[Out]

x*(a + b*x^2)^(1/2)*(c + d*x^2)^(1/2)

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