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

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

[Out]

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

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Rubi [A]  time = 0.0186641, antiderivative size = 67, normalized size of antiderivative = 1., 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} \[ \frac{1}{2} A x \sqrt{a+c x^2}+\frac{a A \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 \sqrt{c}}+\frac{B \left (a+c x^2\right )^{3/2}}{3 c} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*Sqrt[a + c*x^2],x]

[Out]

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

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]

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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.0526464, size = 67, normalized size = 1. \[ \frac{\sqrt{a+c x^2} (2 a B+c x (3 A+2 B x))+3 a A \sqrt{c} \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right )}{6 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 53, normalized size = 0.8 \begin{align*}{\frac{B}{3\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{Ax}{2}\sqrt{c{x}^{2}+a}}+{\frac{aA}{2}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.64579, size = 316, normalized size = 4.72 \begin{align*} \left [\frac{3 \, A a \sqrt{c} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 2 \,{\left (2 \, B c x^{2} + 3 \, A c x + 2 \, B a\right )} \sqrt{c x^{2} + a}}{12 \, c}, -\frac{3 \, A a \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) -{\left (2 \, B c x^{2} + 3 \, A c x + 2 \, B a\right )} \sqrt{c x^{2} + a}}{6 \, c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.21722, size = 70, normalized size = 1.04 \begin{align*} \frac{A \sqrt{a} x \sqrt{1 + \frac{c x^{2}}{a}}}{2} + \frac{A a \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )}}{2 \sqrt{c}} + B \left (\begin{cases} \frac{\sqrt{a} x^{2}}{2} & \text{for}\: c = 0 \\\frac{\left (a + c x^{2}\right )^{\frac{3}{2}}}{3 c} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10577, size = 74, normalized size = 1.1 \begin{align*} -\frac{A a \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{2 \, \sqrt{c}} + \frac{1}{6} \, \sqrt{c x^{2} + a}{\left ({\left (2 \, B x + 3 \, A\right )} x + \frac{2 \, B a}{c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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