3.522 \(\int x^4 (a+b x^2)^{3/2} (A+B x^2) \, dx\)

Optimal. Leaf size=188 \[ \frac{a^2 x^3 \sqrt{a+b x^2} (2 A b-a B)}{128 b^2}-\frac{3 a^3 x \sqrt{a+b x^2} (2 A b-a B)}{256 b^3}+\frac{3 a^4 (2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{256 b^{7/2}}+\frac{x^5 \left (a+b x^2\right )^{3/2} (2 A b-a B)}{16 b}+\frac{a x^5 \sqrt{a+b x^2} (2 A b-a B)}{32 b}+\frac{B x^5 \left (a+b x^2\right )^{5/2}}{10 b} \]

[Out]

(-3*a^3*(2*A*b - a*B)*x*Sqrt[a + b*x^2])/(256*b^3) + (a^2*(2*A*b - a*B)*x^3*Sqrt[a + b*x^2])/(128*b^2) + (a*(2
*A*b - a*B)*x^5*Sqrt[a + b*x^2])/(32*b) + ((2*A*b - a*B)*x^5*(a + b*x^2)^(3/2))/(16*b) + (B*x^5*(a + b*x^2)^(5
/2))/(10*b) + (3*a^4*(2*A*b - a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(256*b^(7/2))

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Rubi [A]  time = 0.0963296, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {459, 279, 321, 217, 206} \[ \frac{a^2 x^3 \sqrt{a+b x^2} (2 A b-a B)}{128 b^2}-\frac{3 a^3 x \sqrt{a+b x^2} (2 A b-a B)}{256 b^3}+\frac{3 a^4 (2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{256 b^{7/2}}+\frac{x^5 \left (a+b x^2\right )^{3/2} (2 A b-a B)}{16 b}+\frac{a x^5 \sqrt{a+b x^2} (2 A b-a B)}{32 b}+\frac{B x^5 \left (a+b x^2\right )^{5/2}}{10 b} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*x^2)^(3/2)*(A + B*x^2),x]

[Out]

(-3*a^3*(2*A*b - a*B)*x*Sqrt[a + b*x^2])/(256*b^3) + (a^2*(2*A*b - a*B)*x^3*Sqrt[a + b*x^2])/(128*b^2) + (a*(2
*A*b - a*B)*x^5*Sqrt[a + b*x^2])/(32*b) + ((2*A*b - a*B)*x^5*(a + b*x^2)^(3/2))/(16*b) + (B*x^5*(a + b*x^2)^(5
/2))/(10*b) + (3*a^4*(2*A*b - a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(256*b^(7/2))

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 x^4 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx &=\frac{B x^5 \left (a+b x^2\right )^{5/2}}{10 b}-\frac{(-10 A b+5 a B) \int x^4 \left (a+b x^2\right )^{3/2} \, dx}{10 b}\\ &=\frac{(2 A b-a B) x^5 \left (a+b x^2\right )^{3/2}}{16 b}+\frac{B x^5 \left (a+b x^2\right )^{5/2}}{10 b}+\frac{(3 a (2 A b-a B)) \int x^4 \sqrt{a+b x^2} \, dx}{16 b}\\ &=\frac{a (2 A b-a B) x^5 \sqrt{a+b x^2}}{32 b}+\frac{(2 A b-a B) x^5 \left (a+b x^2\right )^{3/2}}{16 b}+\frac{B x^5 \left (a+b x^2\right )^{5/2}}{10 b}+\frac{\left (a^2 (2 A b-a B)\right ) \int \frac{x^4}{\sqrt{a+b x^2}} \, dx}{32 b}\\ &=\frac{a^2 (2 A b-a B) x^3 \sqrt{a+b x^2}}{128 b^2}+\frac{a (2 A b-a B) x^5 \sqrt{a+b x^2}}{32 b}+\frac{(2 A b-a B) x^5 \left (a+b x^2\right )^{3/2}}{16 b}+\frac{B x^5 \left (a+b x^2\right )^{5/2}}{10 b}-\frac{\left (3 a^3 (2 A b-a B)\right ) \int \frac{x^2}{\sqrt{a+b x^2}} \, dx}{128 b^2}\\ &=-\frac{3 a^3 (2 A b-a B) x \sqrt{a+b x^2}}{256 b^3}+\frac{a^2 (2 A b-a B) x^3 \sqrt{a+b x^2}}{128 b^2}+\frac{a (2 A b-a B) x^5 \sqrt{a+b x^2}}{32 b}+\frac{(2 A b-a B) x^5 \left (a+b x^2\right )^{3/2}}{16 b}+\frac{B x^5 \left (a+b x^2\right )^{5/2}}{10 b}+\frac{\left (3 a^4 (2 A b-a B)\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{256 b^3}\\ &=-\frac{3 a^3 (2 A b-a B) x \sqrt{a+b x^2}}{256 b^3}+\frac{a^2 (2 A b-a B) x^3 \sqrt{a+b x^2}}{128 b^2}+\frac{a (2 A b-a B) x^5 \sqrt{a+b x^2}}{32 b}+\frac{(2 A b-a B) x^5 \left (a+b x^2\right )^{3/2}}{16 b}+\frac{B x^5 \left (a+b x^2\right )^{5/2}}{10 b}+\frac{\left (3 a^4 (2 A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{256 b^3}\\ &=-\frac{3 a^3 (2 A b-a B) x \sqrt{a+b x^2}}{256 b^3}+\frac{a^2 (2 A b-a B) x^3 \sqrt{a+b x^2}}{128 b^2}+\frac{a (2 A b-a B) x^5 \sqrt{a+b x^2}}{32 b}+\frac{(2 A b-a B) x^5 \left (a+b x^2\right )^{3/2}}{16 b}+\frac{B x^5 \left (a+b x^2\right )^{5/2}}{10 b}+\frac{3 a^4 (2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{256 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.306261, size = 150, normalized size = 0.8 \[ \frac{\sqrt{a+b x^2} \left (\sqrt{b} x \left (4 a^2 b^2 x^2 \left (5 A+2 B x^2\right )-10 a^3 b \left (3 A+B x^2\right )+15 a^4 B+16 a b^3 x^4 \left (15 A+11 B x^2\right )+32 b^4 x^6 \left (5 A+4 B x^2\right )\right )-\frac{15 a^{7/2} (a B-2 A b) \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{\frac{b x^2}{a}+1}}\right )}{1280 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*x^2)^(3/2)*(A + B*x^2),x]

[Out]

(Sqrt[a + b*x^2]*(Sqrt[b]*x*(15*a^4*B - 10*a^3*b*(3*A + B*x^2) + 4*a^2*b^2*x^2*(5*A + 2*B*x^2) + 32*b^4*x^6*(5
*A + 4*B*x^2) + 16*a*b^3*x^4*(15*A + 11*B*x^2)) - (15*a^(7/2)*(-2*A*b + a*B)*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/Sqr
t[1 + (b*x^2)/a]))/(1280*b^(7/2))

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Maple [A]  time = 0.01, size = 219, normalized size = 1.2 \begin{align*}{\frac{B{x}^{5}}{10\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{Ba{x}^{3}}{16\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{{a}^{2}Bx}{32\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{B{a}^{3}x}{128\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{3\,B{a}^{4}x}{256\,{b}^{3}}\sqrt{b{x}^{2}+a}}-{\frac{3\,B{a}^{5}}{256}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{7}{2}}}}+{\frac{A{x}^{3}}{8\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{aAx}{16\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{{a}^{2}Ax}{64\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,A{a}^{3}x}{128\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{3\,A{a}^{4}}{128}\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)^(3/2)*(B*x^2+A),x)

[Out]

1/10*B*x^5*(b*x^2+a)^(5/2)/b-1/16*B/b^2*a*x^3*(b*x^2+a)^(5/2)+1/32*B/b^3*a^2*x*(b*x^2+a)^(5/2)-1/128*B/b^3*a^3
*x*(b*x^2+a)^(3/2)-3/256*B/b^3*a^4*x*(b*x^2+a)^(1/2)-3/256*B/b^(7/2)*a^5*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+1/8*A*x
^3*(b*x^2+a)^(5/2)/b-1/16*A/b^2*a*x*(b*x^2+a)^(5/2)+1/64*A/b^2*a^2*x*(b*x^2+a)^(3/2)+3/128*A/b^2*a^3*x*(b*x^2+
a)^(1/2)+3/128*A/b^(5/2)*a^4*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)^(3/2)*(B*x^2+A),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.01368, size = 693, normalized size = 3.69 \begin{align*} \left [-\frac{15 \,{\left (B a^{5} - 2 \, A a^{4} b\right )} \sqrt{b} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) - 2 \,{\left (128 \, B b^{5} x^{9} + 16 \,{\left (11 \, B a b^{4} + 10 \, A b^{5}\right )} x^{7} + 8 \,{\left (B a^{2} b^{3} + 30 \, A a b^{4}\right )} x^{5} - 10 \,{\left (B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x^{3} + 15 \,{\left (B a^{4} b - 2 \, A a^{3} b^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{2560 \, b^{4}}, \frac{15 \,{\left (B a^{5} - 2 \, A a^{4} b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) +{\left (128 \, B b^{5} x^{9} + 16 \,{\left (11 \, B a b^{4} + 10 \, A b^{5}\right )} x^{7} + 8 \,{\left (B a^{2} b^{3} + 30 \, A a b^{4}\right )} x^{5} - 10 \,{\left (B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x^{3} + 15 \,{\left (B a^{4} b - 2 \, A a^{3} b^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{1280 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^(3/2)*(B*x^2+A),x, algorithm="fricas")

[Out]

[-1/2560*(15*(B*a^5 - 2*A*a^4*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(128*B*b^5*x^9 +
16*(11*B*a*b^4 + 10*A*b^5)*x^7 + 8*(B*a^2*b^3 + 30*A*a*b^4)*x^5 - 10*(B*a^3*b^2 - 2*A*a^2*b^3)*x^3 + 15*(B*a^4
*b - 2*A*a^3*b^2)*x)*sqrt(b*x^2 + a))/b^4, 1/1280*(15*(B*a^5 - 2*A*a^4*b)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^
2 + a)) + (128*B*b^5*x^9 + 16*(11*B*a*b^4 + 10*A*b^5)*x^7 + 8*(B*a^2*b^3 + 30*A*a*b^4)*x^5 - 10*(B*a^3*b^2 - 2
*A*a^2*b^3)*x^3 + 15*(B*a^4*b - 2*A*a^3*b^2)*x)*sqrt(b*x^2 + a))/b^4]

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Sympy [B]  time = 34.0349, size = 345, normalized size = 1.84 \begin{align*} - \frac{3 A a^{\frac{7}{2}} x}{128 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{A a^{\frac{5}{2}} x^{3}}{128 b \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{13 A a^{\frac{3}{2}} x^{5}}{64 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{5 A \sqrt{a} b x^{7}}{16 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 A a^{4} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{128 b^{\frac{5}{2}}} + \frac{A b^{2} x^{9}}{8 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 B a^{\frac{9}{2}} x}{256 b^{3} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{B a^{\frac{7}{2}} x^{3}}{256 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{B a^{\frac{5}{2}} x^{5}}{640 b \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{23 B a^{\frac{3}{2}} x^{7}}{160 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{19 B \sqrt{a} b x^{9}}{80 \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{3 B a^{5} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{256 b^{\frac{7}{2}}} + \frac{B b^{2} x^{11}}{10 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b*x**2+a)**(3/2)*(B*x**2+A),x)

[Out]

-3*A*a**(7/2)*x/(128*b**2*sqrt(1 + b*x**2/a)) - A*a**(5/2)*x**3/(128*b*sqrt(1 + b*x**2/a)) + 13*A*a**(3/2)*x**
5/(64*sqrt(1 + b*x**2/a)) + 5*A*sqrt(a)*b*x**7/(16*sqrt(1 + b*x**2/a)) + 3*A*a**4*asinh(sqrt(b)*x/sqrt(a))/(12
8*b**(5/2)) + A*b**2*x**9/(8*sqrt(a)*sqrt(1 + b*x**2/a)) + 3*B*a**(9/2)*x/(256*b**3*sqrt(1 + b*x**2/a)) + B*a*
*(7/2)*x**3/(256*b**2*sqrt(1 + b*x**2/a)) - B*a**(5/2)*x**5/(640*b*sqrt(1 + b*x**2/a)) + 23*B*a**(3/2)*x**7/(1
60*sqrt(1 + b*x**2/a)) + 19*B*sqrt(a)*b*x**9/(80*sqrt(1 + b*x**2/a)) - 3*B*a**5*asinh(sqrt(b)*x/sqrt(a))/(256*
b**(7/2)) + B*b**2*x**11/(10*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]  time = 1.12732, size = 215, normalized size = 1.14 \begin{align*} \frac{1}{1280} \,{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \, B b x^{2} + \frac{11 \, B a b^{8} + 10 \, A b^{9}}{b^{8}}\right )} x^{2} + \frac{B a^{2} b^{7} + 30 \, A a b^{8}}{b^{8}}\right )} x^{2} - \frac{5 \,{\left (B a^{3} b^{6} - 2 \, A a^{2} b^{7}\right )}}{b^{8}}\right )} x^{2} + \frac{15 \,{\left (B a^{4} b^{5} - 2 \, A a^{3} b^{6}\right )}}{b^{8}}\right )} \sqrt{b x^{2} + a} x + \frac{3 \,{\left (B a^{5} - 2 \, A a^{4} b\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{256 \, b^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^(3/2)*(B*x^2+A),x, algorithm="giac")

[Out]

1/1280*(2*(4*(2*(8*B*b*x^2 + (11*B*a*b^8 + 10*A*b^9)/b^8)*x^2 + (B*a^2*b^7 + 30*A*a*b^8)/b^8)*x^2 - 5*(B*a^3*b
^6 - 2*A*a^2*b^7)/b^8)*x^2 + 15*(B*a^4*b^5 - 2*A*a^3*b^6)/b^8)*sqrt(b*x^2 + a)*x + 3/256*(B*a^5 - 2*A*a^4*b)*l
og(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)