3.6.22 \(\int x^4 (a+b x^2)^{3/2} (A+B x^2) \, dx\) [522]

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

[Out]

1/16*(2*A*b-B*a)*x^5*(b*x^2+a)^(3/2)/b+1/10*B*x^5*(b*x^2+a)^(5/2)/b+3/256*a^4*(2*A*b-B*a)*arctanh(x*b^(1/2)/(b
*x^2+a)^(1/2))/b^(7/2)-3/256*a^3*(2*A*b-B*a)*x*(b*x^2+a)^(1/2)/b^3+1/128*a^2*(2*A*b-B*a)*x^3*(b*x^2+a)^(1/2)/b
^2+1/32*a*(2*A*b-B*a)*x^5*(b*x^2+a)^(1/2)/b

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Rubi [A]
time = 0.06, antiderivative size = 188, normalized size of antiderivative = 1.00, 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 = {470, 285, 327, 223, 212} \begin {gather*} \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 {3 a^3 x \sqrt {a+b x^2} (2 A b-a B)}{256 b^3}+\frac {a^2 x^3 \sqrt {a+b x^2} (2 A b-a B)}{128 b^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} \end {gather*}

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 285

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 327

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 470

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]

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 ) \text {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*}

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Mathematica [A]
time = 0.18, size = 142, normalized size = 0.76 \begin {gather*} \frac {\sqrt {b} x \sqrt {a+b x^2} \left (15 a^4 B-10 a^3 b \left (3 A+B x^2\right )+4 a^2 b^2 x^2 \left (5 A+2 B x^2\right )+32 b^4 x^6 \left (5 A+4 B x^2\right )+16 a b^3 x^4 \left (15 A+11 B x^2\right )\right )+15 a^4 (-2 A b+a B) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{1280 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*x*Sqrt[a + b*x^2]*(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^4*(-2*A*b + a*B)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(1
280*b^(7/2))

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Maple [A]
time = 0.08, size = 224, normalized size = 1.19

method	result	size
risch	\(-\frac {x \left (-128 B \,b^{4} x^{8}-160 A \,b^{4} x^{6}-176 B a \,b^{3} x^{6}-240 A a \,b^{3} x^{4}-8 B \,a^{2} b^{2} x^{4}-20 a^{2} A \,b^{2} x^{2}+10 B \,a^{3} b \,x^{2}+30 A \,a^{3} b -15 B \,a^{4}\right ) \sqrt {b \,x^{2}+a}}{1280 b^{3}}+\frac {3 a^{4} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) A}{128 b^{\frac {5}{2}}}-\frac {3 a^{5} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) B}{256 b^{\frac {7}{2}}}\)	\(153\)
default	\(B \left (\frac {x^{5} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{10 b}-\frac {a \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\right )}{2 b}\right )+A \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\right )\)	\(224\)

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,method=_RETURNVERBOSE)

[Out]

B*(1/10*x^5*(b*x^2+a)^(5/2)/b-1/2*a/b*(1/8*x^3*(b*x^2+a)^(5/2)/b-3/8*a/b*(1/6*x*(b*x^2+a)^(5/2)/b-1/6*a/b*(1/4
*x*(b*x^2+a)^(3/2)+3/4*a*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2)))))))+A*(1/8*x^3*(b
*x^2+a)^(5/2)/b-3/8*a/b*(1/6*x*(b*x^2+a)^(5/2)/b-1/6*a/b*(1/4*x*(b*x^2+a)^(3/2)+3/4*a*(1/2*x*(b*x^2+a)^(1/2)+1
/2*a/b^(1/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))))))

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Maxima [A]
time = 0.28, size = 204, normalized size = 1.09 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x^{5}}{10 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a x^{3}}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A x^{3}}{8 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{2} x}{32 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{3} x}{128 \, b^{3}} - \frac {3 \, \sqrt {b x^{2} + a} B a^{4} x}{256 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A a x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{2} x}{64 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} A a^{3} x}{128 \, b^{2}} - \frac {3 \, B a^{5} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{256 \, b^{\frac {7}{2}}} + \frac {3 \, A a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {5}{2}}} \end {gather*}

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]

1/10*(b*x^2 + a)^(5/2)*B*x^5/b - 1/16*(b*x^2 + a)^(5/2)*B*a*x^3/b^2 + 1/8*(b*x^2 + a)^(5/2)*A*x^3/b + 1/32*(b*
x^2 + a)^(5/2)*B*a^2*x/b^3 - 1/128*(b*x^2 + a)^(3/2)*B*a^3*x/b^3 - 3/256*sqrt(b*x^2 + a)*B*a^4*x/b^3 - 1/16*(b
*x^2 + a)^(5/2)*A*a*x/b^2 + 1/64*(b*x^2 + a)^(3/2)*A*a^2*x/b^2 + 3/128*sqrt(b*x^2 + a)*A*a^3*x/b^2 - 3/256*B*a
^5*arcsinh(b*x/sqrt(a*b))/b^(7/2) + 3/128*A*a^4*arcsinh(b*x/sqrt(a*b))/b^(5/2)

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Fricas [A]
time = 1.02, size = 299, normalized size = 1.59 \begin {gather*} \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (170) = 340\).
time = 164.25, size = 345, normalized size = 1.84 \begin {gather*} - \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 {gather*}

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 = 2.35, size = 159, normalized size = 0.85 \begin {gather*} \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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^4\,\left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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