3.42 \(\int \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2} \, dx\)

Optimal. Leaf size=148 \[ \frac{b \sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{3 d (a+b x)}+\frac{a x \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{2 (a+b x)}+\frac{a c \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 \sqrt{d} (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0567116, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {970, 641, 195, 217, 206} \[ \frac{b \sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{3 d (a+b x)}+\frac{a x \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{2 (a+b x)}+\frac{a c \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 \sqrt{d} (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + d*x^2],x]

[Out]

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

Rule 970

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_.), x_Symbol] :> Dist[(a + b*x + c*x^2)^F
racPart[p]/((4*c)^IntPart[p]*(b + 2*c*x)^(2*FracPart[p])), Int[(b + 2*c*x)^(2*p)*(d + f*x^2)^q, x], x] /; Free
Q[{a, b, c, d, f, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

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

Mathematica [A]  time = 0.0588663, size = 85, normalized size = 0.57 \[ \frac{\sqrt{(a+b x)^2} \left (\sqrt{c+d x^2} \left (3 a d x+2 b \left (c+d x^2\right )\right )+3 a c \sqrt{d} \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )\right )}{6 d (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.216, size = 65, normalized size = 0.4 \begin{align*}{\frac{{\it csgn} \left ( bx+a \right ) }{6} \left ( 2\,b \left ( d{x}^{2}+c \right ) ^{3/2}\sqrt{d}+3\,{d}^{3/2}\sqrt{d{x}^{2}+c}xa+3\,\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ) acd \right ){d}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*csgn(b*x+a)*(2*b*(d*x^2+c)^(3/2)*d^(1/2)+3*d^(3/2)*(d*x^2+c)^(1/2)*x*a+3*ln(x*d^(1/2)+(d*x^2+c)^(1/2))*a*c
*d)/d^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d x^{2} + c} \sqrt{{\left (b x + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.87309, size = 316, normalized size = 2.14 \begin{align*} \left [\frac{3 \, a c \sqrt{d} \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) + 2 \,{\left (2 \, b d x^{2} + 3 \, a d x + 2 \, b c\right )} \sqrt{d x^{2} + c}}{12 \, d}, -\frac{3 \, a c \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) -{\left (2 \, b d x^{2} + 3 \, a d x + 2 \, b c\right )} \sqrt{d x^{2} + c}}{6 \, d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20071, size = 107, normalized size = 0.72 \begin{align*} -\frac{a c \log \left ({\left | -\sqrt{d} x + \sqrt{d x^{2} + c} \right |}\right ) \mathrm{sgn}\left (b x + a\right )}{2 \, \sqrt{d}} + \frac{1}{6} \, \sqrt{d x^{2} + c}{\left ({\left (2 \, b x \mathrm{sgn}\left (b x + a\right ) + 3 \, a \mathrm{sgn}\left (b x + a\right )\right )} x + \frac{2 \, b c \mathrm{sgn}\left (b x + a\right )}{d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*c*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))*sgn(b*x + a)/sqrt(d) + 1/6*sqrt(d*x^2 + c)*((2*b*x*sgn(b*x + a
) + 3*a*sgn(b*x + a))*x + 2*b*c*sgn(b*x + a)/d)