3.587 \(\int \frac{x^4 (A+B x^2)}{(a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=114 \[ -\frac{x (4 A b-7 a B)}{3 b^3 \sqrt{a+b x^2}}+\frac{a x (A b-a B)}{3 b^3 \left (a+b x^2\right )^{3/2}}+\frac{(2 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{7/2}}+\frac{B x \sqrt{a+b x^2}}{2 b^3} \]

[Out]

(a*(A*b - a*B)*x)/(3*b^3*(a + b*x^2)^(3/2)) - ((4*A*b - 7*a*B)*x)/(3*b^3*Sqrt[a + b*x^2]) + (B*x*Sqrt[a + b*x^
2])/(2*b^3) + ((2*A*b - 5*a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(7/2))

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Rubi [A]  time = 0.0900452, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {455, 1157, 388, 217, 206} \[ -\frac{x (4 A b-7 a B)}{3 b^3 \sqrt{a+b x^2}}+\frac{a x (A b-a B)}{3 b^3 \left (a+b x^2\right )^{3/2}}+\frac{(2 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{7/2}}+\frac{B x \sqrt{a+b x^2}}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x^2))/(a + b*x^2)^(5/2),x]

[Out]

(a*(A*b - a*B)*x)/(3*b^3*(a + b*x^2)^(3/2)) - ((4*A*b - 7*a*B)*x)/(3*b^3*Sqrt[a + b*x^2]) + (B*x*Sqrt[a + b*x^
2])/(2*b^3) + ((2*A*b - 5*a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(7/2))

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

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 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 \frac{x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx &=\frac{a (A b-a B) x}{3 b^3 \left (a+b x^2\right )^{3/2}}-\frac{\int \frac{a (A b-a B)-3 b (A b-a B) x^2-3 b^2 B x^4}{\left (a+b x^2\right )^{3/2}} \, dx}{3 b^3}\\ &=\frac{a (A b-a B) x}{3 b^3 \left (a+b x^2\right )^{3/2}}-\frac{(4 A b-7 a B) x}{3 b^3 \sqrt{a+b x^2}}+\frac{\int \frac{3 a (A b-2 a B)+3 a b B x^2}{\sqrt{a+b x^2}} \, dx}{3 a b^3}\\ &=\frac{a (A b-a B) x}{3 b^3 \left (a+b x^2\right )^{3/2}}-\frac{(4 A b-7 a B) x}{3 b^3 \sqrt{a+b x^2}}+\frac{B x \sqrt{a+b x^2}}{2 b^3}+\frac{(2 A b-5 a B) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{2 b^3}\\ &=\frac{a (A b-a B) x}{3 b^3 \left (a+b x^2\right )^{3/2}}-\frac{(4 A b-7 a B) x}{3 b^3 \sqrt{a+b x^2}}+\frac{B x \sqrt{a+b x^2}}{2 b^3}+\frac{(2 A b-5 a B) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{2 b^3}\\ &=\frac{a (A b-a B) x}{3 b^3 \left (a+b x^2\right )^{3/2}}-\frac{(4 A b-7 a B) x}{3 b^3 \sqrt{a+b x^2}}+\frac{B x \sqrt{a+b x^2}}{2 b^3}+\frac{(2 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.24877, size = 116, normalized size = 1.02 \[ \frac{\sqrt{b} x \left (15 a^2 B+a \left (20 b B x^2-6 A b\right )+b^2 x^2 \left (3 B x^2-8 A\right )\right )-3 \sqrt{a} \left (a+b x^2\right ) \sqrt{\frac{b x^2}{a}+1} (5 a B-2 A b) \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{6 b^{7/2} \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x^2))/(a + b*x^2)^(5/2),x]

[Out]

(Sqrt[b]*x*(15*a^2*B + b^2*x^2*(-8*A + 3*B*x^2) + a*(-6*A*b + 20*b*B*x^2)) - 3*Sqrt[a]*(-2*A*b + 5*a*B)*(a + b
*x^2)*Sqrt[1 + (b*x^2)/a]*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/(6*b^(7/2)*(a + b*x^2)^(3/2))

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Maple [A]  time = 0.007, size = 134, normalized size = 1.2 \begin{align*}{\frac{{x}^{5}B}{2\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{5\,Ba{x}^{3}}{6\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{5\,Bax}{2\,{b}^{3}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{5\,Ba}{2}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{7}{2}}}}-{\frac{A{x}^{3}}{3\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{Ax}{{b}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{A\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x^2+A)/(b*x^2+a)^(5/2),x)

[Out]

1/2*B*x^5/b/(b*x^2+a)^(3/2)+5/6*B/b^2*a*x^3/(b*x^2+a)^(3/2)+5/2*B/b^3*a*x/(b*x^2+a)^(1/2)-5/2*B/b^(7/2)*a*ln(x
*b^(1/2)+(b*x^2+a)^(1/2))-1/3*A*x^3/b/(b*x^2+a)^(3/2)-A/b^2*x/(b*x^2+a)^(1/2)+A/b^(5/2)*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(x^4*(B*x^2+A)/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.97516, size = 732, normalized size = 6.42 \begin{align*} \left [-\frac{3 \,{\left ({\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{4} + 5 \, B a^{3} - 2 \, A a^{2} b + 2 \,{\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x^{2}\right )} \sqrt{b} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) - 2 \,{\left (3 \, B b^{3} x^{5} + 4 \,{\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{3} + 3 \,{\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{12 \,{\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}, \frac{3 \,{\left ({\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{4} + 5 \, B a^{3} - 2 \, A a^{2} b + 2 \,{\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) +{\left (3 \, B b^{3} x^{5} + 4 \,{\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} x^{3} + 3 \,{\left (5 \, B a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{6 \,{\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^2+A)/(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*((5*B*a*b^2 - 2*A*b^3)*x^4 + 5*B*a^3 - 2*A*a^2*b + 2*(5*B*a^2*b - 2*A*a*b^2)*x^2)*sqrt(b)*log(-2*b*x
^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(3*B*b^3*x^5 + 4*(5*B*a*b^2 - 2*A*b^3)*x^3 + 3*(5*B*a^2*b - 2*A*a*b^
2)*x)*sqrt(b*x^2 + a))/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b^4), 1/6*(3*((5*B*a*b^2 - 2*A*b^3)*x^4 + 5*B*a^3 - 2*A*a^
2*b + 2*(5*B*a^2*b - 2*A*a*b^2)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (3*B*b^3*x^5 + 4*(5*B*a*b^2
- 2*A*b^3)*x^3 + 3*(5*B*a^2*b - 2*A*a*b^2)*x)*sqrt(b*x^2 + a))/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b^4)]

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Sympy [B]  time = 17.8795, size = 675, normalized size = 5.92 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(B*x**2+A)/(b*x**2+a)**(5/2),x)

[Out]

A*(3*a**(39/2)*b**11*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3
*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) + 3*a**(37/2)*b**12*x**2*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt
(a))/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) - 3*a**19*b**(
23/2)*x/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) - 4*a**18*b
**(25/2)*x**3/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a))) + B*
(-15*a**(81/2)*b**22*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6
*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)) - 15*a**(79/2)*b**23*x**2*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqr
t(a))/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)) + 15*a**40*b*
*(45/2)*x/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)) + 20*a**3
9*b**(47/2)*x**3/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)) +
3*a**38*b**(49/2)*x**5/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/
a)))

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Giac [A]  time = 1.13172, size = 151, normalized size = 1.32 \begin{align*} \frac{{\left ({\left (\frac{3 \, B x^{2}}{b} + \frac{4 \,{\left (5 \, B a^{2} b^{3} - 2 \, A a b^{4}\right )}}{a b^{5}}\right )} x^{2} + \frac{3 \,{\left (5 \, B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )}}{a b^{5}}\right )} x}{6 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}}} + \frac{{\left (5 \, B a - 2 \, A b\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{2 \, b^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^2+A)/(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

1/6*((3*B*x^2/b + 4*(5*B*a^2*b^3 - 2*A*a*b^4)/(a*b^5))*x^2 + 3*(5*B*a^3*b^2 - 2*A*a^2*b^3)/(a*b^5))*x/(b*x^2 +
a)^(3/2) + 1/2*(5*B*a - 2*A*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)