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

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

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Rubi [A]  time = 0.03, 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 = {780, 195, 217, 206} \begin {gather*} -\frac {a^3 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{3/2}}-\frac {a^2 B x \sqrt {a+c x^2}}{16 c}+\frac {\left (a+c x^2\right )^{5/2} (6 A+5 B x)}{30 c}-\frac {a B x \left (a+c x^2\right )^{3/2}}{24 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(A + B*x)*(a + c*x^2)^(3/2),x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 107, normalized size = 1.04 \begin {gather*} \frac {\sqrt {a+c x^2} \left (\sqrt {c} \left (3 a^2 (16 A+5 B x)+2 a c x^2 (48 A+35 B x)+8 c^2 x^4 (6 A+5 B x)\right )-\frac {15 a^{5/2} B \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {\frac {c x^2}{a}+1}}\right )}{240 c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(Sqrt[c]*(8*c^2*x^4*(6*A + 5*B*x) + 3*a^2*(16*A + 5*B*x) + 2*a*c*x^2*(48*A + 35*B*x)) - (15*a
^(5/2)*B*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/Sqrt[1 + (c*x^2)/a]))/(240*c^(3/2))

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IntegrateAlgebraic [A]  time = 0.36, size = 101, normalized size = 0.98 \begin {gather*} \frac {a^3 B \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{16 c^{3/2}}+\frac {\sqrt {a+c x^2} \left (48 a^2 A+15 a^2 B x+96 a A c x^2+70 a B c x^3+48 A c^2 x^4+40 B c^2 x^5\right )}{240 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 205, normalized size = 1.99 \begin {gather*} \left [\frac {15 \, B a^{3} \sqrt {c} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (40 \, B c^{3} x^{5} + 48 \, A c^{3} x^{4} + 70 \, B a c^{2} x^{3} + 96 \, A a c^{2} x^{2} + 15 \, B a^{2} c x + 48 \, A a^{2} c\right )} \sqrt {c x^{2} + a}}{480 \, c^{2}}, \frac {15 \, B a^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (40 \, B c^{3} x^{5} + 48 \, A c^{3} x^{4} + 70 \, B a c^{2} x^{3} + 96 \, A a c^{2} x^{2} + 15 \, B a^{2} c x + 48 \, A a^{2} c\right )} \sqrt {c x^{2} + a}}{240 \, c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.24, size = 89, normalized size = 0.86 \begin {gather*} \frac {B a^{3} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{16 \, c^{\frac {3}{2}}} + \frac {1}{240} \, \sqrt {c x^{2} + a} {\left (\frac {48 \, A a^{2}}{c} + {\left (\frac {15 \, B a^{2}}{c} + 2 \, {\left (48 \, A a + {\left (35 \, B a + 4 \, {\left (5 \, B c x + 6 \, A c\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)*(c*x^2+a)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 94, normalized size = 0.91 \begin {gather*} -\frac {B \,a^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{16 c^{\frac {3}{2}}}-\frac {\sqrt {c \,x^{2}+a}\, B \,a^{2} x}{16 c}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} B a x}{24 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B x}{6 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} A}{5 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*B*x*(c*x^2+a)^(5/2)/c-1/24*a*B*x*(c*x^2+a)^(3/2)/c-1/16*a^2*B*x*(c*x^2+a)^(1/2)/c-1/16*B*a^3/c^(3/2)*ln(c^
(1/2)*x+(c*x^2+a)^(1/2))+1/5*A*(c*x^2+a)^(5/2)/c

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maxima [A]  time = 0.49, size = 86, normalized size = 0.83 \begin {gather*} \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B x}{6 \, c} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} B a x}{24 \, c} - \frac {\sqrt {c x^{2} + a} B a^{2} x}{16 \, c} - \frac {B a^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{16 \, c^{\frac {3}{2}}} + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A}{5 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

int(x*(a + c*x^2)^(3/2)*(A + B*x), x)

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sympy [A]  time = 12.11, size = 223, normalized size = 2.17 \begin {gather*} A a \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: c = 0 \\\frac {\left (a + c x^{2}\right )^{\frac {3}{2}}}{3 c} & \text {otherwise} \end {cases}\right ) + A c \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + c x^{2}}}{15 c^{2}} + \frac {a x^{2} \sqrt {a + c x^{2}}}{15 c} + \frac {x^{4} \sqrt {a + c x^{2}}}{5} & \text {for}\: c \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + \frac {B a^{\frac {5}{2}} x}{16 c \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {17 B a^{\frac {3}{2}} x^{3}}{48 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {11 B \sqrt {a} c x^{5}}{24 \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {B a^{3} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{16 c^{\frac {3}{2}}} + \frac {B c^{2} x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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