3.1.10 \(\int x (A+B x) (a+b x^2)^{3/2} \, dx\) [10]

Optimal. Leaf size=103 \[ -\frac {a^2 B x \sqrt {a+b x^2}}{16 b}-\frac {a B x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {(6 A+5 B x) \left (a+b x^2\right )^{5/2}}{30 b}-\frac {a^3 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}} \]

[Out]

-1/24*a*B*x*(b*x^2+a)^(3/2)/b+1/30*(5*B*x+6*A)*(b*x^2+a)^(5/2)/b-1/16*a^3*B*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))
/b^(3/2)-1/16*a^2*B*x*(b*x^2+a)^(1/2)/b

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Rubi [A]
time = 0.02, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {794, 201, 223, 212} \begin {gather*} -\frac {a^3 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}}-\frac {a^2 B x \sqrt {a+b x^2}}{16 b}+\frac {\left (a+b x^2\right )^{5/2} (6 A+5 B x)}{30 b}-\frac {a B x \left (a+b x^2\right )^{3/2}}{24 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(A + B*x)*(a + b*x^2)^(3/2),x]

[Out]

-1/16*(a^2*B*x*Sqrt[a + b*x^2])/b - (a*B*x*(a + b*x^2)^(3/2))/(24*b) + ((6*A + 5*B*x)*(a + b*x^2)^(5/2))/(30*b
) - (a^3*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*b^(3/2))

Rule 201

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 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 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int x (A+B x) \left (a+b x^2\right )^{3/2} \, dx &=\frac {(6 A+5 B x) \left (a+b x^2\right )^{5/2}}{30 b}-\frac {(a B) \int \left (a+b x^2\right )^{3/2} \, dx}{6 b}\\ &=-\frac {a B x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {(6 A+5 B x) \left (a+b x^2\right )^{5/2}}{30 b}-\frac {\left (a^2 B\right ) \int \sqrt {a+b x^2} \, dx}{8 b}\\ &=-\frac {a^2 B x \sqrt {a+b x^2}}{16 b}-\frac {a B x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {(6 A+5 B x) \left (a+b x^2\right )^{5/2}}{30 b}-\frac {\left (a^3 B\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{16 b}\\ &=-\frac {a^2 B x \sqrt {a+b x^2}}{16 b}-\frac {a B x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {(6 A+5 B x) \left (a+b x^2\right )^{5/2}}{30 b}-\frac {\left (a^3 B\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{16 b}\\ &=-\frac {a^2 B x \sqrt {a+b x^2}}{16 b}-\frac {a B x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {(6 A+5 B x) \left (a+b x^2\right )^{5/2}}{30 b}-\frac {a^3 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 101, normalized size = 0.98 \begin {gather*} \frac {\sqrt {a+b x^2} \left (48 a^2 A+15 a^2 B x+96 a A b x^2+70 a b B x^3+48 A b^2 x^4+40 b^2 B x^5\right )}{240 b}+\frac {a^3 B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(48*a^2*A + 15*a^2*B*x + 96*a*A*b*x^2 + 70*a*b*B*x^3 + 48*A*b^2*x^4 + 40*b^2*B*x^5))/(240*b)
+ (a^3*B*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(16*b^(3/2))

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Maple [A]
time = 0.12, size = 92, normalized size = 0.89

method	result	size
risch	\(\frac {\left (40 b^{2} B \,x^{5}+48 A \,b^{2} x^{4}+70 B a b \,x^{3}+96 a A b \,x^{2}+15 a^{2} B x +48 a^{2} A \right ) \sqrt {b \,x^{2}+a}}{240 b}-\frac {B \,a^{3} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {3}{2}}}\)	\(89\)
default	\(B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )+\frac {A \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5 b}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(B*x+A)*(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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)))))+1/5*A/b*(b*x^2+a)^(5/2)

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Maxima [A]
time = 0.30, size = 86, normalized size = 0.83 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x}{24 \, b} - \frac {\sqrt {b x^{2} + a} B a^{2} x}{16 \, b} - \frac {B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{5 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(b*x^2 + a)^(5/2)*B*x/b - 1/24*(b*x^2 + a)^(3/2)*B*a*x/b - 1/16*sqrt(b*x^2 + a)*B*a^2*x/b - 1/16*B*a^3*arc
sinh(b*x/sqrt(a*b))/b^(3/2) + 1/5*(b*x^2 + a)^(5/2)*A/b

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Fricas [A]
time = 3.58, size = 205, normalized size = 1.99 \begin {gather*} \left [\frac {15 \, B a^{3} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 70 \, B a b^{2} x^{3} + 96 \, A a b^{2} x^{2} + 15 \, B a^{2} b x + 48 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{480 \, b^{2}}, \frac {15 \, B a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 70 \, B a b^{2} x^{3} + 96 \, A a b^{2} x^{2} + 15 \, B a^{2} b x + 48 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{240 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/480*(15*B*a^3*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(40*B*b^3*x^5 + 48*A*b^3*x^4 + 70
*B*a*b^2*x^3 + 96*A*a*b^2*x^2 + 15*B*a^2*b*x + 48*A*a^2*b)*sqrt(b*x^2 + a))/b^2, 1/240*(15*B*a^3*sqrt(-b)*arct
an(sqrt(-b)*x/sqrt(b*x^2 + a)) + (40*B*b^3*x^5 + 48*A*b^3*x^4 + 70*B*a*b^2*x^3 + 96*A*a*b^2*x^2 + 15*B*a^2*b*x
+ 48*A*a^2*b)*sqrt(b*x^2 + a))/b^2]

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Sympy [A]
time = 7.83, size = 223, normalized size = 2.17 \begin {gather*} A a \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\left (a + b x^{2}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) + A b \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {a x^{2} \sqrt {a + b x^{2}}}{15 b} + \frac {x^{4} \sqrt {a + b x^{2}}}{5} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + \frac {B a^{\frac {5}{2}} x}{16 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {17 B a^{\frac {3}{2}} x^{3}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {11 B \sqrt {a} b x^{5}}{24 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 b^{\frac {3}{2}}} + \frac {B b^{2} x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a*Piecewise((sqrt(a)*x**2/2, Eq(b, 0)), ((a + b*x**2)**(3/2)/(3*b), True)) + A*b*Piecewise((-2*a**2*sqrt(a +
b*x**2)/(15*b**2) + a*x**2*sqrt(a + b*x**2)/(15*b) + x**4*sqrt(a + b*x**2)/5, Ne(b, 0)), (sqrt(a)*x**4/4, Tru
e)) + B*a**(5/2)*x/(16*b*sqrt(1 + b*x**2/a)) + 17*B*a**(3/2)*x**3/(48*sqrt(1 + b*x**2/a)) + 11*B*sqrt(a)*b*x**
5/(24*sqrt(1 + b*x**2/a)) - B*a**3*asinh(sqrt(b)*x/sqrt(a))/(16*b**(3/2)) + B*b**2*x**7/(6*sqrt(a)*sqrt(1 + b*
x**2/a))

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Giac [A]
time = 0.92, size = 89, normalized size = 0.86 \begin {gather*} \frac {B a^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {3}{2}}} + \frac {1}{240} \, \sqrt {b x^{2} + a} {\left (\frac {48 \, A a^{2}}{b} + {\left (\frac {15 \, B a^{2}}{b} + 2 \, {\left (48 \, A a + {\left (35 \, B a + 4 \, {\left (5 \, B b x + 6 \, A b\right )} x\right )} x\right )} x\right )} x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*B*a^3*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2) + 1/240*sqrt(b*x^2 + a)*(48*A*a^2/b + (15*B*a^2/b +
2*(48*A*a + (35*B*a + 4*(5*B*b*x + 6*A*b)*x)*x)*x)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*x^2)^(3/2)*(A + B*x),x)

[Out]

int(x*(a + b*x^2)^(3/2)*(A + B*x), x)

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