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

Optimal. Leaf size=150 \[ -\frac{5 a^4 A \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{128 c^{3/2}}-\frac{5 a^3 A x \sqrt{a+c x^2}}{128 c}-\frac{5 a^2 A x \left (a+c x^2\right )^{3/2}}{192 c}-\frac{\left (a+c x^2\right )^{7/2} (16 a B-63 A c x)}{504 c^2}-\frac{a A x \left (a+c x^2\right )^{5/2}}{48 c}+\frac{B x^2 \left (a+c x^2\right )^{7/2}}{9 c} \]

[Out]

(-5*a^3*A*x*Sqrt[a + c*x^2])/(128*c) - (5*a^2*A*x*(a + c*x^2)^(3/2))/(192*c) - (a*A*x*(a + c*x^2)^(5/2))/(48*c
) + (B*x^2*(a + c*x^2)^(7/2))/(9*c) - ((16*a*B - 63*A*c*x)*(a + c*x^2)^(7/2))/(504*c^2) - (5*a^4*A*ArcTanh[(Sq
rt[c]*x)/Sqrt[a + c*x^2]])/(128*c^(3/2))

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Rubi [A]  time = 0.0661778, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {833, 780, 195, 217, 206} \[ -\frac{5 a^4 A \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{128 c^{3/2}}-\frac{5 a^3 A x \sqrt{a+c x^2}}{128 c}-\frac{5 a^2 A x \left (a+c x^2\right )^{3/2}}{192 c}-\frac{\left (a+c x^2\right )^{7/2} (16 a B-63 A c x)}{504 c^2}-\frac{a A x \left (a+c x^2\right )^{5/2}}{48 c}+\frac{B x^2 \left (a+c x^2\right )^{7/2}}{9 c} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(A + B*x)*(a + c*x^2)^(5/2),x]

[Out]

(-5*a^3*A*x*Sqrt[a + c*x^2])/(128*c) - (5*a^2*A*x*(a + c*x^2)^(3/2))/(192*c) - (a*A*x*(a + c*x^2)^(5/2))/(48*c
) + (B*x^2*(a + c*x^2)^(7/2))/(9*c) - ((16*a*B - 63*A*c*x)*(a + c*x^2)^(7/2))/(504*c^2) - (5*a^4*A*ArcTanh[(Sq
rt[c]*x)/Sqrt[a + c*x^2]])/(128*c^(3/2))

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])

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^2 (A+B x) \left (a+c x^2\right )^{5/2} \, dx &=\frac{B x^2 \left (a+c x^2\right )^{7/2}}{9 c}+\frac{\int x (-2 a B+9 A c x) \left (a+c x^2\right )^{5/2} \, dx}{9 c}\\ &=\frac{B x^2 \left (a+c x^2\right )^{7/2}}{9 c}-\frac{(16 a B-63 A c x) \left (a+c x^2\right )^{7/2}}{504 c^2}-\frac{(a A) \int \left (a+c x^2\right )^{5/2} \, dx}{8 c}\\ &=-\frac{a A x \left (a+c x^2\right )^{5/2}}{48 c}+\frac{B x^2 \left (a+c x^2\right )^{7/2}}{9 c}-\frac{(16 a B-63 A c x) \left (a+c x^2\right )^{7/2}}{504 c^2}-\frac{\left (5 a^2 A\right ) \int \left (a+c x^2\right )^{3/2} \, dx}{48 c}\\ &=-\frac{5 a^2 A x \left (a+c x^2\right )^{3/2}}{192 c}-\frac{a A x \left (a+c x^2\right )^{5/2}}{48 c}+\frac{B x^2 \left (a+c x^2\right )^{7/2}}{9 c}-\frac{(16 a B-63 A c x) \left (a+c x^2\right )^{7/2}}{504 c^2}-\frac{\left (5 a^3 A\right ) \int \sqrt{a+c x^2} \, dx}{64 c}\\ &=-\frac{5 a^3 A x \sqrt{a+c x^2}}{128 c}-\frac{5 a^2 A x \left (a+c x^2\right )^{3/2}}{192 c}-\frac{a A x \left (a+c x^2\right )^{5/2}}{48 c}+\frac{B x^2 \left (a+c x^2\right )^{7/2}}{9 c}-\frac{(16 a B-63 A c x) \left (a+c x^2\right )^{7/2}}{504 c^2}-\frac{\left (5 a^4 A\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{128 c}\\ &=-\frac{5 a^3 A x \sqrt{a+c x^2}}{128 c}-\frac{5 a^2 A x \left (a+c x^2\right )^{3/2}}{192 c}-\frac{a A x \left (a+c x^2\right )^{5/2}}{48 c}+\frac{B x^2 \left (a+c x^2\right )^{7/2}}{9 c}-\frac{(16 a B-63 A c x) \left (a+c x^2\right )^{7/2}}{504 c^2}-\frac{\left (5 a^4 A\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{128 c}\\ &=-\frac{5 a^3 A x \sqrt{a+c x^2}}{128 c}-\frac{5 a^2 A x \left (a+c x^2\right )^{3/2}}{192 c}-\frac{a A x \left (a+c x^2\right )^{5/2}}{48 c}+\frac{B x^2 \left (a+c x^2\right )^{7/2}}{9 c}-\frac{(16 a B-63 A c x) \left (a+c x^2\right )^{7/2}}{504 c^2}-\frac{5 a^4 A \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{128 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.260715, size = 131, normalized size = 0.87 \[ \frac{\sqrt{a+c x^2} \left (6 a^2 c^2 x^3 (413 A+320 B x)+a^3 c x (315 A+128 B x)-\frac{315 a^{7/2} A \sqrt{c} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{\frac{c x^2}{a}+1}}-256 a^4 B+8 a c^3 x^5 (357 A+304 B x)+112 c^4 x^7 (9 A+8 B x)\right )}{8064 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(-256*a^4*B + 112*c^4*x^7*(9*A + 8*B*x) + a^3*c*x*(315*A + 128*B*x) + 8*a*c^3*x^5*(357*A + 30
4*B*x) + 6*a^2*c^2*x^3*(413*A + 320*B*x) - (315*a^(7/2)*A*Sqrt[c]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/Sqrt[1 + (c*x^
2)/a]))/(8064*c^2)

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Maple [A]  time = 0.008, size = 132, normalized size = 0.9 \begin{align*}{\frac{B{x}^{2}}{9\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{2\,aB}{63\,{c}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{Ax}{8\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{aAx}{48\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{a}^{2}Ax}{192\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{a}^{3}Ax}{128\,c}\sqrt{c{x}^{2}+a}}-{\frac{5\,A{a}^{4}}{128}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*B*x^2*(c*x^2+a)^(7/2)/c-2/63*B*a/c^2*(c*x^2+a)^(7/2)+1/8*A*x*(c*x^2+a)^(7/2)/c-1/48*a*A*x*(c*x^2+a)^(5/2)/
c-5/192*a^2*A*x*(c*x^2+a)^(3/2)/c-5/128*a^3*A*x*(c*x^2+a)^(1/2)/c-5/128*A*a^4/c^(3/2)*ln(x*c^(1/2)+(c*x^2+a)^(
1/2))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.6965, size = 687, normalized size = 4.58 \begin{align*} \left [\frac{315 \, A a^{4} \sqrt{c} \log \left (-2 \, c x^{2} + 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 2 \,{\left (896 \, B c^{4} x^{8} + 1008 \, A c^{4} x^{7} + 2432 \, B a c^{3} x^{6} + 2856 \, A a c^{3} x^{5} + 1920 \, B a^{2} c^{2} x^{4} + 2478 \, A a^{2} c^{2} x^{3} + 128 \, B a^{3} c x^{2} + 315 \, A a^{3} c x - 256 \, B a^{4}\right )} \sqrt{c x^{2} + a}}{16128 \, c^{2}}, \frac{315 \, A a^{4} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) +{\left (896 \, B c^{4} x^{8} + 1008 \, A c^{4} x^{7} + 2432 \, B a c^{3} x^{6} + 2856 \, A a c^{3} x^{5} + 1920 \, B a^{2} c^{2} x^{4} + 2478 \, A a^{2} c^{2} x^{3} + 128 \, B a^{3} c x^{2} + 315 \, A a^{3} c x - 256 \, B a^{4}\right )} \sqrt{c x^{2} + a}}{8064 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16128*(315*A*a^4*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(896*B*c^4*x^8 + 1008*A*c^4*x^
7 + 2432*B*a*c^3*x^6 + 2856*A*a*c^3*x^5 + 1920*B*a^2*c^2*x^4 + 2478*A*a^2*c^2*x^3 + 128*B*a^3*c*x^2 + 315*A*a^
3*c*x - 256*B*a^4)*sqrt(c*x^2 + a))/c^2, 1/8064*(315*A*a^4*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (896*
B*c^4*x^8 + 1008*A*c^4*x^7 + 2432*B*a*c^3*x^6 + 2856*A*a*c^3*x^5 + 1920*B*a^2*c^2*x^4 + 2478*A*a^2*c^2*x^3 + 1
28*B*a^3*c*x^2 + 315*A*a^3*c*x - 256*B*a^4)*sqrt(c*x^2 + a))/c^2]

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Sympy [A]  time = 24.4336, size = 442, normalized size = 2.95 \begin{align*} \frac{5 A a^{\frac{7}{2}} x}{128 c \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{133 A a^{\frac{5}{2}} x^{3}}{384 \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{127 A a^{\frac{3}{2}} c x^{5}}{192 \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{23 A \sqrt{a} c^{2} x^{7}}{48 \sqrt{1 + \frac{c x^{2}}{a}}} - \frac{5 A a^{4} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )}}{128 c^{\frac{3}{2}}} + \frac{A c^{3} x^{9}}{8 \sqrt{a} \sqrt{1 + \frac{c x^{2}}{a}}} + B a^{2} \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 ) + 2 B a c \left (\begin{cases} \frac{8 a^{3} \sqrt{a + c x^{2}}}{105 c^{3}} - \frac{4 a^{2} x^{2} \sqrt{a + c x^{2}}}{105 c^{2}} + \frac{a x^{4} \sqrt{a + c x^{2}}}{35 c} + \frac{x^{6} \sqrt{a + c x^{2}}}{7} & \text{for}\: c \neq 0 \\\frac{\sqrt{a} x^{6}}{6} & \text{otherwise} \end{cases}\right ) + B c^{2} \left (\begin{cases} - \frac{16 a^{4} \sqrt{a + c x^{2}}}{315 c^{4}} + \frac{8 a^{3} x^{2} \sqrt{a + c x^{2}}}{315 c^{3}} - \frac{2 a^{2} x^{4} \sqrt{a + c x^{2}}}{105 c^{2}} + \frac{a x^{6} \sqrt{a + c x^{2}}}{63 c} + \frac{x^{8} \sqrt{a + c x^{2}}}{9} & \text{for}\: c \neq 0 \\\frac{\sqrt{a} x^{8}}{8} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*A*a**(7/2)*x/(128*c*sqrt(1 + c*x**2/a)) + 133*A*a**(5/2)*x**3/(384*sqrt(1 + c*x**2/a)) + 127*A*a**(3/2)*c*x*
*5/(192*sqrt(1 + c*x**2/a)) + 23*A*sqrt(a)*c**2*x**7/(48*sqrt(1 + c*x**2/a)) - 5*A*a**4*asinh(sqrt(c)*x/sqrt(a
))/(128*c**(3/2)) + A*c**3*x**9/(8*sqrt(a)*sqrt(1 + c*x**2/a)) + B*a**2*Piecewise((-2*a**2*sqrt(a + c*x**2)/(1
5*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, True)) + 2*B*a
*c*Piecewise((8*a**3*sqrt(a + c*x**2)/(105*c**3) - 4*a**2*x**2*sqrt(a + c*x**2)/(105*c**2) + a*x**4*sqrt(a + c
*x**2)/(35*c) + x**6*sqrt(a + c*x**2)/7, Ne(c, 0)), (sqrt(a)*x**6/6, True)) + B*c**2*Piecewise((-16*a**4*sqrt(
a + c*x**2)/(315*c**4) + 8*a**3*x**2*sqrt(a + c*x**2)/(315*c**3) - 2*a**2*x**4*sqrt(a + c*x**2)/(105*c**2) + a
*x**6*sqrt(a + c*x**2)/(63*c) + x**8*sqrt(a + c*x**2)/9, Ne(c, 0)), (sqrt(a)*x**8/8, True))

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Giac [A]  time = 1.19179, size = 173, normalized size = 1.15 \begin{align*} \frac{5 \, A a^{4} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{128 \, c^{\frac{3}{2}}} - \frac{1}{8064} \,{\left (\frac{256 \, B a^{4}}{c^{2}} -{\left (\frac{315 \, A a^{3}}{c} + 2 \,{\left (\frac{64 \, B a^{3}}{c} +{\left (1239 \, A a^{2} + 4 \,{\left (240 \, B a^{2} +{\left (357 \, A a c + 2 \,{\left (152 \, B a c + 7 \,{\left (8 \, B c^{2} x + 9 \, A c^{2}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt{c x^{2} + a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/128*A*a^4*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(3/2) - 1/8064*(256*B*a^4/c^2 - (315*A*a^3/c + 2*(64*B*a^
3/c + (1239*A*a^2 + 4*(240*B*a^2 + (357*A*a*c + 2*(152*B*a*c + 7*(8*B*c^2*x + 9*A*c^2)*x)*x)*x)*x)*x)*x)*x)*sq
rt(c*x^2 + a)