3.17 \(\int x (A+B x) (a+b x^2)^{5/2} \, dx\)

Optimal. Leaf size=126 \[ -\frac{5 a^4 B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{128 b^{3/2}}-\frac{5 a^3 B x \sqrt{a+b x^2}}{128 b}-\frac{5 a^2 B x \left (a+b x^2\right )^{3/2}}{192 b}+\frac{\left (a+b x^2\right )^{7/2} (8 A+7 B x)}{56 b}-\frac{a B x \left (a+b x^2\right )^{5/2}}{48 b} \]

[Out]

(-5*a^3*B*x*Sqrt[a + b*x^2])/(128*b) - (5*a^2*B*x*(a + b*x^2)^(3/2))/(192*b) - (a*B*x*(a + b*x^2)^(5/2))/(48*b
) + ((8*A + 7*B*x)*(a + b*x^2)^(7/2))/(56*b) - (5*a^4*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(128*b^(3/2))

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Rubi [A]  time = 0.0435616, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {780, 195, 217, 206} \[ -\frac{5 a^4 B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{128 b^{3/2}}-\frac{5 a^3 B x \sqrt{a+b x^2}}{128 b}-\frac{5 a^2 B x \left (a+b x^2\right )^{3/2}}{192 b}+\frac{\left (a+b x^2\right )^{7/2} (8 A+7 B x)}{56 b}-\frac{a B x \left (a+b x^2\right )^{5/2}}{48 b} \]

Antiderivative was successfully verified.

[In]

Int[x*(A + B*x)*(a + b*x^2)^(5/2),x]

[Out]

(-5*a^3*B*x*Sqrt[a + b*x^2])/(128*b) - (5*a^2*B*x*(a + b*x^2)^(3/2))/(192*b) - (a*B*x*(a + b*x^2)^(5/2))/(48*b
) + ((8*A + 7*B*x)*(a + b*x^2)^(7/2))/(56*b) - (5*a^4*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(128*b^(3/2))

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 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 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 (A+B x) \left (a+b x^2\right )^{5/2} \, dx &=\frac{(8 A+7 B x) \left (a+b x^2\right )^{7/2}}{56 b}-\frac{(a B) \int \left (a+b x^2\right )^{5/2} \, dx}{8 b}\\ &=-\frac{a B x \left (a+b x^2\right )^{5/2}}{48 b}+\frac{(8 A+7 B x) \left (a+b x^2\right )^{7/2}}{56 b}-\frac{\left (5 a^2 B\right ) \int \left (a+b x^2\right )^{3/2} \, dx}{48 b}\\ &=-\frac{5 a^2 B x \left (a+b x^2\right )^{3/2}}{192 b}-\frac{a B x \left (a+b x^2\right )^{5/2}}{48 b}+\frac{(8 A+7 B x) \left (a+b x^2\right )^{7/2}}{56 b}-\frac{\left (5 a^3 B\right ) \int \sqrt{a+b x^2} \, dx}{64 b}\\ &=-\frac{5 a^3 B x \sqrt{a+b x^2}}{128 b}-\frac{5 a^2 B x \left (a+b x^2\right )^{3/2}}{192 b}-\frac{a B x \left (a+b x^2\right )^{5/2}}{48 b}+\frac{(8 A+7 B x) \left (a+b x^2\right )^{7/2}}{56 b}-\frac{\left (5 a^4 B\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{128 b}\\ &=-\frac{5 a^3 B x \sqrt{a+b x^2}}{128 b}-\frac{5 a^2 B x \left (a+b x^2\right )^{3/2}}{192 b}-\frac{a B x \left (a+b x^2\right )^{5/2}}{48 b}+\frac{(8 A+7 B x) \left (a+b x^2\right )^{7/2}}{56 b}-\frac{\left (5 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}\\ &=-\frac{5 a^3 B x \sqrt{a+b x^2}}{128 b}-\frac{5 a^2 B x \left (a+b x^2\right )^{3/2}}{192 b}-\frac{a B x \left (a+b x^2\right )^{5/2}}{48 b}+\frac{(8 A+7 B x) \left (a+b x^2\right )^{7/2}}{56 b}-\frac{5 a^4 B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{128 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.516493, size = 112, normalized size = 0.89 \[ \frac{\left (a+b x^2\right )^{7/2} \left (-\frac{7 a B x \left (\left (a+b x^2\right ) \left (33 a^2+26 a b x^2+8 b^2 x^4\right )+\frac{15 a^{7/2} \sqrt{\frac{b x^2}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{b} x}\right )}{\left (a+b x^2\right )^4}+384 A+336 B x\right )}{2688 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(A + B*x)*(a + b*x^2)^(5/2),x]

[Out]

((a + b*x^2)^(7/2)*(384*A + 336*B*x - (7*a*B*x*((a + b*x^2)*(33*a^2 + 26*a*b*x^2 + 8*b^2*x^4) + (15*a^(7/2)*Sq
rt[1 + (b*x^2)/a]*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[b]*x)))/(a + b*x^2)^4))/(2688*b)

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Maple [A]  time = 0.006, size = 113, normalized size = 0.9 \begin{align*}{\frac{Bx}{8\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{Bax}{48\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{a}^{2}Bx}{192\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{a}^{3}Bx}{128\,b}\sqrt{b{x}^{2}+a}}-{\frac{5\,B{a}^{4}}{128}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{3}{2}}}}+{\frac{A}{7\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(B*x+A)*(b*x^2+a)^(5/2),x)

[Out]

1/8*B*x*(b*x^2+a)^(7/2)/b-1/48*B/b*a*x*(b*x^2+a)^(5/2)-5/192*B/b*a^2*x*(b*x^2+a)^(3/2)-5/128*B/b*a^3*x*(b*x^2+
a)^(1/2)-5/128*B/b^(3/2)*a^4*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+1/7*A/b*(b*x^2+a)^(7/2)

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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*(B*x+A)*(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.66048, size = 635, normalized size = 5.04 \begin{align*} \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 (336 \, B b^{4} x^{7} + 384 \, A b^{4} x^{6} + 952 \, B a b^{3} x^{5} + 1152 \, A a b^{3} x^{4} + 826 \, B a^{2} b^{2} x^{3} + 1152 \, A a^{2} b^{2} x^{2} + 105 \, B a^{3} b x + 384 \, A a^{3} b\right )} \sqrt{b x^{2} + a}}{5376 \, b^{2}}, \frac{105 \, B a^{4} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) +{\left (336 \, B b^{4} x^{7} + 384 \, A b^{4} x^{6} + 952 \, B a b^{3} x^{5} + 1152 \, A a b^{3} x^{4} + 826 \, B a^{2} b^{2} x^{3} + 1152 \, A a^{2} b^{2} x^{2} + 105 \, B a^{3} b x + 384 \, A a^{3} b\right )} \sqrt{b x^{2} + a}}{2688 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

[1/5376*(105*B*a^4*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(336*B*b^4*x^7 + 384*A*b^4*x^6
+ 952*B*a*b^3*x^5 + 1152*A*a*b^3*x^4 + 826*B*a^2*b^2*x^3 + 1152*A*a^2*b^2*x^2 + 105*B*a^3*b*x + 384*A*a^3*b)*s
qrt(b*x^2 + a))/b^2, 1/2688*(105*B*a^4*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (336*B*b^4*x^7 + 384*A*b^
4*x^6 + 952*B*a*b^3*x^5 + 1152*A*a*b^3*x^4 + 826*B*a^2*b^2*x^3 + 1152*A*a^2*b^2*x^2 + 105*B*a^3*b*x + 384*A*a^
3*b)*sqrt(b*x^2 + a))/b^2]

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Sympy [A]  time = 19.8831, size = 354, normalized size = 2.81 \begin{align*} A a^{2} \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 ) + 2 A 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 ) + A b^{2} \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{5 B a^{\frac{7}{2}} x}{128 b \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{133 B a^{\frac{5}{2}} x^{3}}{384 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{127 B a^{\frac{3}{2}} b x^{5}}{192 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{23 B \sqrt{a} b^{2} x^{7}}{48 \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{5 B a^{4} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{128 b^{\frac{3}{2}}} + \frac{B b^{3} x^{9}}{8 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(b*x**2+a)**(5/2),x)

[Out]

A*a**2*Piecewise((sqrt(a)*x**2/2, Eq(b, 0)), ((a + b*x**2)**(3/2)/(3*b), True)) + 2*A*a*b*Piecewise((-2*a**2*s
qrt(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**2*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, True)) + 5*B*a**(7/2)*x/
(128*b*sqrt(1 + b*x**2/a)) + 133*B*a**(5/2)*x**3/(384*sqrt(1 + b*x**2/a)) + 127*B*a**(3/2)*b*x**5/(192*sqrt(1
+ b*x**2/a)) + 23*B*sqrt(a)*b**2*x**7/(48*sqrt(1 + b*x**2/a)) - 5*B*a**4*asinh(sqrt(b)*x/sqrt(a))/(128*b**(3/2
)) + B*b**3*x**9/(8*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]  time = 1.22827, size = 154, normalized size = 1.22 \begin{align*} \frac{5 \, B a^{4} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{128 \, b^{\frac{3}{2}}} + \frac{1}{2688} \,{\left (\frac{384 \, A a^{3}}{b} +{\left (\frac{105 \, B a^{3}}{b} + 2 \,{\left (576 \, A a^{2} +{\left (413 \, B a^{2} + 4 \,{\left (144 \, A a b +{\left (119 \, B a b + 6 \,{\left (7 \, B b^{2} x + 8 \, A b^{2}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt{b x^{2} + a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)*(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

5/128*B*a^4*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2) + 1/2688*(384*A*a^3/b + (105*B*a^3/b + 2*(576*A*a^2
+ (413*B*a^2 + 4*(144*A*a*b + (119*B*a*b + 6*(7*B*b^2*x + 8*A*b^2)*x)*x)*x)*x)*x)*x)*sqrt(b*x^2 + a)