3.1.47 \(\int \sqrt {a+b x^2} (c+d x^2) \, dx\) [47]

Optimal. Leaf size=87 \[ \frac {(4 b c-a d) x \sqrt {a+b x^2}}{8 b}+\frac {d x \left (a+b x^2\right )^{3/2}}{4 b}+\frac {a (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}} \]

[Out]

1/4*d*x*(b*x^2+a)^(3/2)/b+1/8*a*(-a*d+4*b*c)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(3/2)+1/8*(-a*d+4*b*c)*x*(b*
x^2+a)^(1/2)/b

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Rubi [A]
time = 0.02, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {396, 201, 223, 212} \begin {gather*} \frac {a (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}}+\frac {x \sqrt {a+b x^2} (4 b c-a d)}{8 b}+\frac {d x \left (a+b x^2\right )^{3/2}}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2]*(c + d*x^2),x]

[Out]

((4*b*c - a*d)*x*Sqrt[a + b*x^2])/(8*b) + (d*x*(a + b*x^2)^(3/2))/(4*b) + (a*(4*b*c - a*d)*ArcTanh[(Sqrt[b]*x)
/Sqrt[a + b*x^2]])/(8*b^(3/2))

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 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 396

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

Rubi steps

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

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Mathematica [A]
time = 0.08, size = 74, normalized size = 0.85 \begin {gather*} \frac {x \sqrt {a+b x^2} \left (4 b c+a d+2 b d x^2\right )}{8 b}+\frac {a (-4 b c+a d) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[a + b*x^2]*(4*b*c + a*d + 2*b*d*x^2))/(8*b) + (a*(-4*b*c + a*d)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(
8*b^(3/2))

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Maple [A]
time = 0.06, size = 98, normalized size = 1.13

method	result	size
risch	\(\frac {x \left (2 b d \,x^{2}+a d +4 b c \right ) \sqrt {b \,x^{2}+a}}{8 b}-\frac {a^{2} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) d}{8 b^{\frac {3}{2}}}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) c}{2 \sqrt {b}}\)	\(80\)
default	\(d \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )+c \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*(1/4*x*(b*x^2+a)^(3/2)/b-1/4*a/b*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))))+c*(1/2
*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2)))

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Maxima [A]
time = 0.30, size = 81, normalized size = 0.93 \begin {gather*} \frac {1}{2} \, \sqrt {b x^{2} + a} c x + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} d x}{4 \, b} - \frac {\sqrt {b x^{2} + a} a d x}{8 \, b} + \frac {a c \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} - \frac {a^{2} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(b*x^2 + a)*c*x + 1/4*(b*x^2 + a)^(3/2)*d*x/b - 1/8*sqrt(b*x^2 + a)*a*d*x/b + 1/2*a*c*arcsinh(b*x/sqrt
(a*b))/sqrt(b) - 1/8*a^2*d*arcsinh(b*x/sqrt(a*b))/b^(3/2)

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Fricas [A]
time = 2.68, size = 158, normalized size = 1.82 \begin {gather*} \left [-\frac {{\left (4 \, a b c - a^{2} d\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (2 \, b^{2} d x^{3} + {\left (4 \, b^{2} c + a b d\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, b^{2}}, -\frac {{\left (4 \, a b c - a^{2} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, b^{2} d x^{3} + {\left (4 \, b^{2} c + a b d\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*((4*a*b*c - a^2*d)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(2*b^2*d*x^3 + (4*b^2*c
+ a*b*d)*x)*sqrt(b*x^2 + a))/b^2, -1/8*((4*a*b*c - a^2*d)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (2*b^2
*d*x^3 + (4*b^2*c + a*b*d)*x)*sqrt(b*x^2 + a))/b^2]

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Sympy [A]
time = 3.30, size = 144, normalized size = 1.66 \begin {gather*} \frac {a^{\frac {3}{2}} d x}{8 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {\sqrt {a} c x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {3 \sqrt {a} d x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {a^{2} d \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 b^{\frac {3}{2}}} + \frac {a c \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 \sqrt {b}} + \frac {b d x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(3/2)*d*x/(8*b*sqrt(1 + b*x**2/a)) + sqrt(a)*c*x*sqrt(1 + b*x**2/a)/2 + 3*sqrt(a)*d*x**3/(8*sqrt(1 + b*x**2
/a)) - a**2*d*asinh(sqrt(b)*x/sqrt(a))/(8*b**(3/2)) + a*c*asinh(sqrt(b)*x/sqrt(a))/(2*sqrt(b)) + b*d*x**5/(4*s
qrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]
time = 0.68, size = 70, normalized size = 0.80 \begin {gather*} \frac {1}{8} \, \sqrt {b x^{2} + a} {\left (2 \, d x^{2} + \frac {4 \, b^{2} c + a b d}{b^{2}}\right )} x - \frac {{\left (4 \, a b c - a^{2} d\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(b*x^2 + a)*(2*d*x^2 + (4*b^2*c + a*b*d)/b^2)*x - 1/8*(4*a*b*c - a^2*d)*log(abs(-sqrt(b)*x + sqrt(b*x^
2 + a)))/b^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/2)*(c + d*x^2),x)

[Out]

int((a + b*x^2)^(1/2)*(c + d*x^2), x)

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