3.55 \(\int (a+b x^2)^{3/2} (c+d x^2) \, dx\)

Optimal. Leaf size=118 \[ \frac{a^2 (6 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{3/2}}+\frac{x \left (a+b x^2\right )^{3/2} (6 b c-a d)}{24 b}+\frac{a x \sqrt{a+b x^2} (6 b c-a d)}{16 b}+\frac{d x \left (a+b x^2\right )^{5/2}}{6 b} \]

[Out]

(a*(6*b*c - a*d)*x*Sqrt[a + b*x^2])/(16*b) + ((6*b*c - a*d)*x*(a + b*x^2)^(3/2))/(24*b) + (d*x*(a + b*x^2)^(5/
2))/(6*b) + (a^2*(6*b*c - a*d)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*b^(3/2))

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Rubi [A]  time = 0.0400311, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {388, 195, 217, 206} \[ \frac{a^2 (6 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{3/2}}+\frac{x \left (a+b x^2\right )^{3/2} (6 b c-a d)}{24 b}+\frac{a x \sqrt{a+b x^2} (6 b c-a d)}{16 b}+\frac{d x \left (a+b x^2\right )^{5/2}}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)*(c + d*x^2),x]

[Out]

(a*(6*b*c - a*d)*x*Sqrt[a + b*x^2])/(16*b) + ((6*b*c - a*d)*x*(a + b*x^2)^(3/2))/(24*b) + (d*x*(a + b*x^2)^(5/
2))/(6*b) + (a^2*(6*b*c - a*d)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*b^(3/2))

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 \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \, dx &=\frac{d x \left (a+b x^2\right )^{5/2}}{6 b}-\frac{(-6 b c+a d) \int \left (a+b x^2\right )^{3/2} \, dx}{6 b}\\ &=\frac{(6 b c-a d) x \left (a+b x^2\right )^{3/2}}{24 b}+\frac{d x \left (a+b x^2\right )^{5/2}}{6 b}+\frac{(a (6 b c-a d)) \int \sqrt{a+b x^2} \, dx}{8 b}\\ &=\frac{a (6 b c-a d) x \sqrt{a+b x^2}}{16 b}+\frac{(6 b c-a d) x \left (a+b x^2\right )^{3/2}}{24 b}+\frac{d x \left (a+b x^2\right )^{5/2}}{6 b}+\frac{\left (a^2 (6 b c-a d)\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{16 b}\\ &=\frac{a (6 b c-a d) x \sqrt{a+b x^2}}{16 b}+\frac{(6 b c-a d) x \left (a+b x^2\right )^{3/2}}{24 b}+\frac{d x \left (a+b x^2\right )^{5/2}}{6 b}+\frac{\left (a^2 (6 b c-a d)\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{16 b}\\ &=\frac{a (6 b c-a d) x \sqrt{a+b x^2}}{16 b}+\frac{(6 b c-a d) x \left (a+b x^2\right )^{3/2}}{24 b}+\frac{d x \left (a+b x^2\right )^{5/2}}{6 b}+\frac{a^2 (6 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.199388, size = 109, normalized size = 0.92 \[ \frac{\sqrt{a+b x^2} \left (\sqrt{b} x \left (3 a^2 d+2 a b \left (15 c+7 d x^2\right )+4 b^2 x^2 \left (3 c+2 d x^2\right )\right )-\frac{3 a^{3/2} (a d-6 b c) \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{\frac{b x^2}{a}+1}}\right )}{48 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)*(c + d*x^2),x]

[Out]

(Sqrt[a + b*x^2]*(Sqrt[b]*x*(3*a^2*d + 4*b^2*x^2*(3*c + 2*d*x^2) + 2*a*b*(15*c + 7*d*x^2)) - (3*a^(3/2)*(-6*b*
c + a*d)*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[1 + (b*x^2)/a]))/(48*b^(3/2))

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Maple [A]  time = 0.003, size = 131, normalized size = 1.1 \begin{align*}{\frac{dx}{6\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{adx}{24\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{d{a}^{2}x}{16\,b}\sqrt{b{x}^{2}+a}}-{\frac{d{a}^{3}}{16}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{3}{2}}}}+{\frac{cx}{4} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,acx}{8}\sqrt{b{x}^{2}+a}}+{\frac{3\,{a}^{2}c}{8}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)*(d*x^2+c),x)

[Out]

1/6*d*x*(b*x^2+a)^(5/2)/b-1/24*d/b*a*x*(b*x^2+a)^(3/2)-1/16*d/b*a^2*x*(b*x^2+a)^(1/2)-1/16*d/b^(3/2)*a^3*ln(x*
b^(1/2)+(b*x^2+a)^(1/2))+1/4*c*x*(b*x^2+a)^(3/2)+3/8*c*a*x*(b*x^2+a)^(1/2)+3/8*c*a^2/b^(1/2)*ln(x*b^(1/2)+(b*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^2+a)^(3/2)*(d*x^2+c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72869, size = 483, normalized size = 4.09 \begin{align*} \left [-\frac{3 \,{\left (6 \, a^{2} b c - a^{3} d\right )} \sqrt{b} \log \left (-2 \, b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) - 2 \,{\left (8 \, b^{3} d x^{5} + 2 \,{\left (6 \, b^{3} c + 7 \, a b^{2} d\right )} x^{3} + 3 \,{\left (10 \, a b^{2} c + a^{2} b d\right )} x\right )} \sqrt{b x^{2} + a}}{96 \, b^{2}}, -\frac{3 \,{\left (6 \, a^{2} b c - a^{3} d\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (8 \, b^{3} d x^{5} + 2 \,{\left (6 \, b^{3} c + 7 \, a b^{2} d\right )} x^{3} + 3 \,{\left (10 \, a b^{2} c + a^{2} b d\right )} x\right )} \sqrt{b x^{2} + a}}{48 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(d*x^2+c),x, algorithm="fricas")

[Out]

[-1/96*(3*(6*a^2*b*c - a^3*d)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(8*b^3*d*x^5 + 2*(6*
b^3*c + 7*a*b^2*d)*x^3 + 3*(10*a*b^2*c + a^2*b*d)*x)*sqrt(b*x^2 + a))/b^2, -1/48*(3*(6*a^2*b*c - a^3*d)*sqrt(-
b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8*b^3*d*x^5 + 2*(6*b^3*c + 7*a*b^2*d)*x^3 + 3*(10*a*b^2*c + a^2*b*d)*
x)*sqrt(b*x^2 + a))/b^2]

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Sympy [B]  time = 13.2377, size = 253, normalized size = 2.14 \begin{align*} \frac{a^{\frac{5}{2}} d x}{16 b \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{a^{\frac{3}{2}} c x \sqrt{1 + \frac{b x^{2}}{a}}}{2} + \frac{a^{\frac{3}{2}} c x}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{17 a^{\frac{3}{2}} d x^{3}}{48 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 \sqrt{a} b c x^{3}}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{11 \sqrt{a} b d x^{5}}{24 \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{a^{3} d \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{16 b^{\frac{3}{2}}} + \frac{3 a^{2} c \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{8 \sqrt{b}} + \frac{b^{2} c x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{b^{2} d x^{7}}{6 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)*(d*x**2+c),x)

[Out]

a**(5/2)*d*x/(16*b*sqrt(1 + b*x**2/a)) + a**(3/2)*c*x*sqrt(1 + b*x**2/a)/2 + a**(3/2)*c*x/(8*sqrt(1 + b*x**2/a
)) + 17*a**(3/2)*d*x**3/(48*sqrt(1 + b*x**2/a)) + 3*sqrt(a)*b*c*x**3/(8*sqrt(1 + b*x**2/a)) + 11*sqrt(a)*b*d*x
**5/(24*sqrt(1 + b*x**2/a)) - a**3*d*asinh(sqrt(b)*x/sqrt(a))/(16*b**(3/2)) + 3*a**2*c*asinh(sqrt(b)*x/sqrt(a)
)/(8*sqrt(b)) + b**2*c*x**5/(4*sqrt(a)*sqrt(1 + b*x**2/a)) + b**2*d*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]  time = 1.10687, size = 139, normalized size = 1.18 \begin{align*} \frac{1}{48} \,{\left (2 \,{\left (4 \, b d x^{2} + \frac{6 \, b^{5} c + 7 \, a b^{4} d}{b^{4}}\right )} x^{2} + \frac{3 \,{\left (10 \, a b^{4} c + a^{2} b^{3} d\right )}}{b^{4}}\right )} \sqrt{b x^{2} + a} x - \frac{{\left (6 \, a^{2} b c - a^{3} d\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{16 \, b^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(d*x^2+c),x, algorithm="giac")

[Out]

1/48*(2*(4*b*d*x^2 + (6*b^5*c + 7*a*b^4*d)/b^4)*x^2 + 3*(10*a*b^4*c + a^2*b^3*d)/b^4)*sqrt(b*x^2 + a)*x - 1/16
*(6*a^2*b*c - a^3*d)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2)