3.269 \(\int \sqrt{c+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4} \, dx\)

Optimal. Leaf size=178 \[ -\frac{c \sqrt{a^2+2 a b x^2+b^2 x^4} (b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 d^{3/2} \left (a+b x^2\right )}+\frac{b x \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2\right )^{3/2}}{4 d \left (a+b x^2\right )}-\frac{x \sqrt{a^2+2 a b x^2+b^2 x^4} \sqrt{c+d x^2} (b c-4 a d)}{8 d \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0762214, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {1148, 388, 195, 217, 206} \[ -\frac{c \sqrt{a^2+2 a b x^2+b^2 x^4} (b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 d^{3/2} \left (a+b x^2\right )}+\frac{b x \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2\right )^{3/2}}{4 d \left (a+b x^2\right )}-\frac{x \sqrt{a^2+2 a b x^2+b^2 x^4} \sqrt{c+d x^2} (b c-4 a d)}{8 d \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

Rule 1148

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

Rule 388

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]

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

Mathematica [A]  time = 0.111615, size = 121, normalized size = 0.68 \[ \frac{\sqrt{\left (a+b x^2\right )^2} \sqrt{c+d x^2} \left (\sqrt{d} x \sqrt{\frac{d x^2}{c}+1} \left (4 a d+b \left (c+2 d x^2\right )\right )-\sqrt{c} (b c-4 a d) \sinh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )\right )}{8 d^{3/2} \left (a+b x^2\right ) \sqrt{\frac{d x^2}{c}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

(Sqrt[(a + b*x^2)^2]*Sqrt[c + d*x^2]*(Sqrt[d]*x*Sqrt[1 + (d*x^2)/c]*(4*a*d + b*(c + 2*d*x^2)) - Sqrt[c]*(b*c -
4*a*d)*ArcSinh[(Sqrt[d]*x)/Sqrt[c]]))/(8*d^(3/2)*(a + b*x^2)*Sqrt[1 + (d*x^2)/c])

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Maple [A]  time = 0.007, size = 119, normalized size = 0.7 \begin{align*}{\frac{1}{8\,b{x}^{2}+8\,a}\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}} \left ( 2\,\sqrt{d} \left ( d{x}^{2}+c \right ) ^{3/2}xb+4\,{d}^{3/2}\sqrt{d{x}^{2}+c}xa-\sqrt{d}\sqrt{d{x}^{2}+c}xbc+4\,\ln \left ( \sqrt{d}x+\sqrt{d{x}^{2}+c} \right ) acd-\ln \left ( \sqrt{d}x+\sqrt{d{x}^{2}+c} \right ) b{c}^{2} \right ){d}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((b*x^2+a)^2)^(1/2)*(2*d^(1/2)*(d*x^2+c)^(3/2)*x*b+4*d^(3/2)*(d*x^2+c)^(1/2)*x*a-d^(1/2)*(d*x^2+c)^(1/2)*x
*b*c+4*ln(d^(1/2)*x+(d*x^2+c)^(1/2))*a*c*d-ln(d^(1/2)*x+(d*x^2+c)^(1/2))*b*c^2)/(b*x^2+a)/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^{2} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.82027, size = 370, normalized size = 2.08 \begin{align*} \left [-\frac{{\left (b c^{2} - 4 \, a c d\right )} \sqrt{d} \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) - 2 \,{\left (2 \, b d^{2} x^{3} +{\left (b c d + 4 \, a d^{2}\right )} x\right )} \sqrt{d x^{2} + c}}{16 \, d^{2}}, \frac{{\left (b c^{2} - 4 \, a c d\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) +{\left (2 \, b d^{2} x^{3} +{\left (b c d + 4 \, a d^{2}\right )} x\right )} \sqrt{d x^{2} + c}}{8 \, d^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{2}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1184, size = 147, normalized size = 0.83 \begin{align*} \frac{1}{8} \,{\left (2 \, b x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{b c d \mathrm{sgn}\left (b x^{2} + a\right ) + 4 \, a d^{2} \mathrm{sgn}\left (b x^{2} + a\right )}{d^{2}}\right )} \sqrt{d x^{2} + c} x + \frac{{\left (b c^{2} \mathrm{sgn}\left (b x^{2} + a\right ) - 4 \, a c d \mathrm{sgn}\left (b x^{2} + a\right )\right )} \log \left ({\left | -\sqrt{d} x + \sqrt{d x^{2} + c} \right |}\right )}{8 \, d^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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