3.1.4 \(\int (A+B x) \sqrt {a+b x^2} \, dx\)

Optimal. Leaf size=67 \[ \frac {1}{2} A x \sqrt {a+b x^2}+\frac {a A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {B \left (a+b x^2\right )^{3/2}}{3 b} \]

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Rubi [A]  time = 0.02, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {641, 195, 217, 206} \begin {gather*} \frac {1}{2} A x \sqrt {a+b x^2}+\frac {a A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {B \left (a+b x^2\right )^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*Sqrt[a + b*x^2],x]

[Out]

(A*x*Sqrt[a + b*x^2])/2 + (B*(a + b*x^2)^(3/2))/(3*b) + (a*A*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])

Rule 195

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 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 217

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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rubi steps

\begin {align*} \int (A+B x) \sqrt {a+b x^2} \, dx &=\frac {B \left (a+b x^2\right )^{3/2}}{3 b}+A \int \sqrt {a+b x^2} \, dx\\ &=\frac {1}{2} A x \sqrt {a+b x^2}+\frac {B \left (a+b x^2\right )^{3/2}}{3 b}+\frac {1}{2} (a A) \int \frac {1}{\sqrt {a+b x^2}} \, dx\\ &=\frac {1}{2} A x \sqrt {a+b x^2}+\frac {B \left (a+b x^2\right )^{3/2}}{3 b}+\frac {1}{2} (a A) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )\\ &=\frac {1}{2} A x \sqrt {a+b x^2}+\frac {B \left (a+b x^2\right )^{3/2}}{3 b}+\frac {a A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 67, normalized size = 1.00 \begin {gather*} \frac {\sqrt {a+b x^2} (2 a B+b x (3 A+2 B x))+3 a A \sqrt {b} \log \left (\sqrt {b} \sqrt {a+b x^2}+b x\right )}{6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)*Sqrt[a + b*x^2],x]

[Out]

(Sqrt[a + b*x^2]*(2*a*B + b*x*(3*A + 2*B*x)) + 3*a*A*Sqrt[b]*Log[b*x + Sqrt[b]*Sqrt[a + b*x^2]])/(6*b)

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IntegrateAlgebraic [A]  time = 0.23, size = 68, normalized size = 1.01 \begin {gather*} \frac {\sqrt {a+b x^2} \left (2 a B+3 A b x+2 b B x^2\right )}{6 b}-\frac {a A \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{2 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)*Sqrt[a + b*x^2],x]

[Out]

(Sqrt[a + b*x^2]*(2*a*B + 3*A*b*x + 2*b*B*x^2))/(6*b) - (a*A*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(2*Sqrt[b])

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fricas [A]  time = 0.74, size = 128, normalized size = 1.91 \begin {gather*} \left [\frac {3 \, A a \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, B b x^{2} + 3 \, A b x + 2 \, B a\right )} \sqrt {b x^{2} + a}}{12 \, b}, -\frac {3 \, A a \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, B b x^{2} + 3 \, A b x + 2 \, B a\right )} \sqrt {b x^{2} + a}}{6 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*A*a*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*B*b*x^2 + 3*A*b*x + 2*B*a)*sqrt(b*
x^2 + a))/b, -1/6*(3*A*a*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (2*B*b*x^2 + 3*A*b*x + 2*B*a)*sqrt(b*x^
2 + a))/b]

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giac [A]  time = 0.41, size = 55, normalized size = 0.82 \begin {gather*} -\frac {A a \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, \sqrt {b}} + \frac {1}{6} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, B x + 3 \, A\right )} x + \frac {2 \, B a}{b}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*A*a*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/sqrt(b) + 1/6*sqrt(b*x^2 + a)*((2*B*x + 3*A)*x + 2*B*a/b)

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maple [A]  time = 0.01, size = 53, normalized size = 0.79 \begin {gather*} \frac {A a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}+\frac {\sqrt {b \,x^{2}+a}\, A x}{2}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B}{3 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b*x^2+a)^(1/2),x)

[Out]

1/3*B*(b*x^2+a)^(3/2)/b+1/2*A*x*(b*x^2+a)^(1/2)+1/2*A*a/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.32, size = 45, normalized size = 0.67 \begin {gather*} \frac {1}{2} \, \sqrt {b x^{2} + a} A x + \frac {A a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(b*x^2 + a)*A*x + 1/2*A*a*arcsinh(b*x/sqrt(a*b))/sqrt(b) + 1/3*(b*x^2 + a)^(3/2)*B/b

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mupad [B]  time = 1.16, size = 52, normalized size = 0.78 \begin {gather*} \frac {B\,{\left (b\,x^2+a\right )}^{3/2}}{3\,b}+\frac {A\,x\,\sqrt {b\,x^2+a}}{2}+\frac {A\,a\,\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{2\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/2)*(A + B*x),x)

[Out]

(B*(a + b*x^2)^(3/2))/(3*b) + (A*x*(a + b*x^2)^(1/2))/2 + (A*a*log(b^(1/2)*x + (a + b*x^2)^(1/2)))/(2*b^(1/2))

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sympy [A]  time = 6.51, size = 70, normalized size = 1.04 \begin {gather*} \frac {A \sqrt {a} x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {A a \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 \sqrt {b}} + B \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\left (a + b x^{2}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x**2+a)**(1/2),x)

[Out]

A*sqrt(a)*x*sqrt(1 + b*x**2/a)/2 + A*a*asinh(sqrt(b)*x/sqrt(a))/(2*sqrt(b)) + B*Piecewise((sqrt(a)*x**2/2, Eq(
b, 0)), ((a + b*x**2)**(3/2)/(3*b), True))

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