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

Optimal. Leaf size=150 \[ \frac {3 a^4 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{5/2}}+\frac {3 a^3 B x \sqrt {a+b x^2}}{128 b^2}+\frac {a^2 B x \left (a+b x^2\right )^{3/2}}{64 b^2}-\frac {a \left (a+b x^2\right )^{5/2} (32 A+35 B x)}{560 b^2}+\frac {A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b} \]

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Rubi [A]  time = 0.10, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {833, 780, 195, 217, 206} \begin {gather*} \frac {3 a^3 B x \sqrt {a+b x^2}}{128 b^2}+\frac {a^2 B x \left (a+b x^2\right )^{3/2}}{64 b^2}+\frac {3 a^4 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{5/2}}-\frac {a \left (a+b x^2\right )^{5/2} (32 A+35 B x)}{560 b^2}+\frac {A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^3*B*x*Sqrt[a + b*x^2])/(128*b^2) + (a^2*B*x*(a + b*x^2)^(3/2))/(64*b^2) + (A*x^2*(a + b*x^2)^(5/2))/(7*b)
+ (B*x^3*(a + b*x^2)^(5/2))/(8*b) - (a*(32*A + 35*B*x)*(a + b*x^2)^(5/2))/(560*b^2) + (3*a^4*B*ArcTanh[(Sqrt[
b]*x)/Sqrt[a + b*x^2]])/(128*b^(5/2))

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 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])

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 780

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]

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])

Rubi steps

\begin {align*} \int x^3 (A+B x) \left (a+b x^2\right )^{3/2} \, dx &=\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}+\frac {\int x^2 (-3 a B+8 A b x) \left (a+b x^2\right )^{3/2} \, dx}{8 b}\\ &=\frac {A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}+\frac {\int x (-16 a A b-21 a b B x) \left (a+b x^2\right )^{3/2} \, dx}{56 b^2}\\ &=\frac {A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {a (32 A+35 B x) \left (a+b x^2\right )^{5/2}}{560 b^2}+\frac {\left (a^2 B\right ) \int \left (a+b x^2\right )^{3/2} \, dx}{16 b^2}\\ &=\frac {a^2 B x \left (a+b x^2\right )^{3/2}}{64 b^2}+\frac {A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {a (32 A+35 B x) \left (a+b x^2\right )^{5/2}}{560 b^2}+\frac {\left (3 a^3 B\right ) \int \sqrt {a+b x^2} \, dx}{64 b^2}\\ &=\frac {3 a^3 B x \sqrt {a+b x^2}}{128 b^2}+\frac {a^2 B x \left (a+b x^2\right )^{3/2}}{64 b^2}+\frac {A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {a (32 A+35 B x) \left (a+b x^2\right )^{5/2}}{560 b^2}+\frac {\left (3 a^4 B\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{128 b^2}\\ &=\frac {3 a^3 B x \sqrt {a+b x^2}}{128 b^2}+\frac {a^2 B x \left (a+b x^2\right )^{3/2}}{64 b^2}+\frac {A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {a (32 A+35 B x) \left (a+b x^2\right )^{5/2}}{560 b^2}+\frac {\left (3 a^4 B\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{128 b^2}\\ &=\frac {3 a^3 B x \sqrt {a+b x^2}}{128 b^2}+\frac {a^2 B x \left (a+b x^2\right )^{3/2}}{64 b^2}+\frac {A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {a (32 A+35 B x) \left (a+b x^2\right )^{5/2}}{560 b^2}+\frac {3 a^4 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 126, normalized size = 0.84 \begin {gather*} \frac {\sqrt {a+b x^2} \left (\frac {105 a^{7/2} B \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}+\sqrt {b} \left (-a^3 (256 A+105 B x)+2 a^2 b x^2 (64 A+35 B x)+8 a b^2 x^4 (128 A+105 B x)+80 b^3 x^6 (8 A+7 B x)\right )\right )}{4480 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(Sqrt[b]*(80*b^3*x^6*(8*A + 7*B*x) + 2*a^2*b*x^2*(64*A + 35*B*x) + 8*a*b^2*x^4*(128*A + 105*B
*x) - a^3*(256*A + 105*B*x)) + (105*a^(7/2)*B*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[1 + (b*x^2)/a]))/(4480*b^(5/2
))

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IntegrateAlgebraic [A]  time = 0.35, size = 125, normalized size = 0.83 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-256 a^3 A-105 a^3 B x+128 a^2 A b x^2+70 a^2 b B x^3+1024 a A b^2 x^4+840 a b^2 B x^5+640 A b^3 x^6+560 b^3 B x^7\right )}{4480 b^2}-\frac {3 a^4 B \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{128 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(-256*a^3*A - 105*a^3*B*x + 128*a^2*A*b*x^2 + 70*a^2*b*B*x^3 + 1024*a*A*b^2*x^4 + 840*a*b^2*B
*x^5 + 640*A*b^3*x^6 + 560*b^3*B*x^7))/(4480*b^2) - (3*a^4*B*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(128*b^(5/2)
)

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fricas [A]  time = 0.73, size = 254, normalized size = 1.69 \begin {gather*} \left [\frac {105 \, B a^{4} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (560 \, B b^{4} x^{7} + 640 \, A b^{4} x^{6} + 840 \, B a b^{3} x^{5} + 1024 \, A a b^{3} x^{4} + 70 \, B a^{2} b^{2} x^{3} + 128 \, A a^{2} b^{2} x^{2} - 105 \, B a^{3} b x - 256 \, A a^{3} b\right )} \sqrt {b x^{2} + a}}{8960 \, b^{3}}, -\frac {105 \, B a^{4} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (560 \, B b^{4} x^{7} + 640 \, A b^{4} x^{6} + 840 \, B a b^{3} x^{5} + 1024 \, A a b^{3} x^{4} + 70 \, B a^{2} b^{2} x^{3} + 128 \, A a^{2} b^{2} x^{2} - 105 \, B a^{3} b x - 256 \, A a^{3} b\right )} \sqrt {b x^{2} + a}}{4480 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8960*(105*B*a^4*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(560*B*b^4*x^7 + 640*A*b^4*x^6
+ 840*B*a*b^3*x^5 + 1024*A*a*b^3*x^4 + 70*B*a^2*b^2*x^3 + 128*A*a^2*b^2*x^2 - 105*B*a^3*b*x - 256*A*a^3*b)*sqr
t(b*x^2 + a))/b^3, -1/4480*(105*B*a^4*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (560*B*b^4*x^7 + 640*A*b^4
*x^6 + 840*B*a*b^3*x^5 + 1024*A*a*b^3*x^4 + 70*B*a^2*b^2*x^3 + 128*A*a^2*b^2*x^2 - 105*B*a^3*b*x - 256*A*a^3*b
)*sqrt(b*x^2 + a))/b^3]

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giac [A]  time = 0.44, size = 115, normalized size = 0.77 \begin {gather*} -\frac {3 \, B a^{4} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {5}{2}}} - \frac {1}{4480} \, \sqrt {b x^{2} + a} {\left (\frac {256 \, A a^{3}}{b^{2}} + {\left (\frac {105 \, B a^{3}}{b^{2}} - 2 \, {\left (\frac {64 \, A a^{2}}{b} + {\left (\frac {35 \, B a^{2}}{b} + 4 \, {\left (128 \, A a + 5 \, {\left (21 \, B a + 2 \, {\left (7 \, B b x + 8 \, A b\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/128*B*a^4*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2) - 1/4480*sqrt(b*x^2 + a)*(256*A*a^3/b^2 + (105*B*a
^3/b^2 - 2*(64*A*a^2/b + (35*B*a^2/b + 4*(128*A*a + 5*(21*B*a + 2*(7*B*b*x + 8*A*b)*x)*x)*x)*x)*x)*x)

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maple [A]  time = 0.01, size = 134, normalized size = 0.89 \begin {gather*} \frac {3 B \,a^{4} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {5}{2}}}+\frac {3 \sqrt {b \,x^{2}+a}\, B \,a^{3} x}{128 b^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B \,x^{3}}{8 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A \,x^{2}}{7 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,a^{2} x}{64 b^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B a x}{16 b^{2}}-\frac {2 \left (b \,x^{2}+a \right )^{\frac {5}{2}} A a}{35 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*B*x^3*(b*x^2+a)^(5/2)/b-1/16*B*a/b^2*x*(b*x^2+a)^(5/2)+1/64*a^2*B*x*(b*x^2+a)^(3/2)/b^2+3/128*a^3*B*x*(b*x
^2+a)^(1/2)/b^2+3/128*B*a^4/b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))+1/7*A*x^2*(b*x^2+a)^(5/2)/b-2/35*A*a/b^2*(b*
x^2+a)^(5/2)

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maxima [A]  time = 1.39, size = 126, normalized size = 0.84 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x^{3}}{8 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A x^{2}}{7 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} x}{64 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} B a^{3} x}{128 \, b^{2}} + \frac {3 \, B a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {5}{2}}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A a}{35 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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sympy [A]  time = 20.79, size = 318, normalized size = 2.12 \begin {gather*} A a \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 ) + A b \left (\begin {cases} \frac {8 a^{3} \sqrt {a + b x^{2}}}{105 b^{3}} - \frac {4 a^{2} x^{2} \sqrt {a + b x^{2}}}{105 b^{2}} + \frac {a x^{4} \sqrt {a + b x^{2}}}{35 b} + \frac {x^{6} \sqrt {a + b x^{2}}}{7} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{6}}{6} & \text {otherwise} \end {cases}\right ) - \frac {3 B a^{\frac {7}{2}} x}{128 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{\frac {5}{2}} x^{3}}{128 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {13 B a^{\frac {3}{2}} x^{5}}{64 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 B \sqrt {a} b x^{7}}{16 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 B a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{128 b^{\frac {5}{2}}} + \frac {B b^{2} x^{9}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a*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, True)) + A*b*Piecewise((8*a**3*sqrt(a + b*x**2)/(105*b**3) - 4*a**2*x**2*sqrt(a +
b*x**2)/(105*b**2) + a*x**4*sqrt(a + b*x**2)/(35*b) + x**6*sqrt(a + b*x**2)/7, Ne(b, 0)), (sqrt(a)*x**6/6, Tru
e)) - 3*B*a**(7/2)*x/(128*b**2*sqrt(1 + b*x**2/a)) - B*a**(5/2)*x**3/(128*b*sqrt(1 + b*x**2/a)) + 13*B*a**(3/2
)*x**5/(64*sqrt(1 + b*x**2/a)) + 5*B*sqrt(a)*b*x**7/(16*sqrt(1 + b*x**2/a)) + 3*B*a**4*asinh(sqrt(b)*x/sqrt(a)
)/(128*b**(5/2)) + B*b**2*x**9/(8*sqrt(a)*sqrt(1 + b*x**2/a))

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