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

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

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Rubi [A]  time = 0.07, 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 {5 a^4 A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{3/2}}-\frac {5 a^3 A x \sqrt {a+b x^2}}{128 b}-\frac {5 a^2 A x \left (a+b x^2\right )^{3/2}}{192 b}-\frac {\left (a+b x^2\right )^{7/2} (16 a B-63 A b x)}{504 b^2}-\frac {a A x \left (a+b x^2\right )^{5/2}}{48 b}+\frac {B x^2 \left (a+b x^2\right )^{7/2}}{9 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.41, size = 131, normalized size = 0.87 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-\frac {315 a^{7/2} A \sqrt {b} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}-256 a^4 B+a^3 b x (315 A+128 B x)+6 a^2 b^2 x^3 (413 A+320 B x)+8 a b^3 x^5 (357 A+304 B x)+112 b^4 x^7 (9 A+8 B x)\right )}{8064 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.43, size = 140, normalized size = 0.93 \begin {gather*} \frac {5 a^4 A \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{128 b^{3/2}}+\frac {\sqrt {a+b x^2} \left (-256 a^4 B+315 a^3 A b x+128 a^3 b B x^2+2478 a^2 A b^2 x^3+1920 a^2 b^2 B x^4+2856 a A b^3 x^5+2432 a b^3 B x^6+1008 A b^4 x^7+896 b^4 B x^8\right )}{8064 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(-256*a^4*B + 315*a^3*A*b*x + 128*a^3*b*B*x^2 + 2478*a^2*A*b^2*x^3 + 1920*a^2*b^2*B*x^4 + 285
6*a*A*b^3*x^5 + 2432*a*b^3*B*x^6 + 1008*A*b^4*x^7 + 896*b^4*B*x^8))/(8064*b^2) + (5*a^4*A*Log[-(Sqrt[b]*x) + S
qrt[a + b*x^2]])/(128*b^(3/2))

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fricas [A]  time = 0.80, size = 271, normalized size = 1.81 \begin {gather*} \left [\frac {315 \, A a^{4} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (896 \, B b^{4} x^{8} + 1008 \, A b^{4} x^{7} + 2432 \, B a b^{3} x^{6} + 2856 \, A a b^{3} x^{5} + 1920 \, B a^{2} b^{2} x^{4} + 2478 \, A a^{2} b^{2} x^{3} + 128 \, B a^{3} b x^{2} + 315 \, A a^{3} b x - 256 \, B a^{4}\right )} \sqrt {b x^{2} + a}}{16128 \, b^{2}}, \frac {315 \, A a^{4} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (896 \, B b^{4} x^{8} + 1008 \, A b^{4} x^{7} + 2432 \, B a b^{3} x^{6} + 2856 \, A a b^{3} x^{5} + 1920 \, B a^{2} b^{2} x^{4} + 2478 \, A a^{2} b^{2} x^{3} + 128 \, B a^{3} b x^{2} + 315 \, A a^{3} b x - 256 \, B a^{4}\right )} \sqrt {b x^{2} + a}}{8064 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.46, size = 128, normalized size = 0.85 \begin {gather*} \frac {5 \, A a^{4} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {3}{2}}} - \frac {1}{8064} \, {\left (\frac {256 \, B a^{4}}{b^{2}} - {\left (\frac {315 \, A a^{3}}{b} + 2 \, {\left (\frac {64 \, B a^{3}}{b} + {\left (1239 \, A a^{2} + 4 \, {\left (240 \, B a^{2} + {\left (357 \, A a b + 2 \, {\left (152 \, B a b + 7 \, {\left (8 \, B b^{2} x + 9 \, A b^{2}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {b x^{2} + a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 132, normalized size = 0.88 \begin {gather*} -\frac {5 A \,a^{4} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {3}{2}}}-\frac {5 \sqrt {b \,x^{2}+a}\, A \,a^{3} x}{128 b}-\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,a^{2} x}{192 b}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A a x}{48 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} B \,x^{2}}{9 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} A x}{8 b}-\frac {2 \left (b \,x^{2}+a \right )^{\frac {7}{2}} B a}{63 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.37, size = 124, normalized size = 0.83 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B x^{2}}{9 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A x}{8 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A a x}{48 \, b} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{2} x}{192 \, b} - \frac {5 \, \sqrt {b x^{2} + a} A a^{3} x}{128 \, b} - \frac {5 \, A a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {3}{2}}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a}{63 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,{\left (b\,x^2+a\right )}^{5/2}\,\left (A+B\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 61.22, size = 442, normalized size = 2.95 \begin {gather*} \frac {5 A a^{\frac {7}{2}} x}{128 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {133 A a^{\frac {5}{2}} x^{3}}{384 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {127 A a^{\frac {3}{2}} b x^{5}}{192 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {23 A \sqrt {a} b^{2} x^{7}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {5 A a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{128 b^{\frac {3}{2}}} + \frac {A b^{3} x^{9}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + B a^{2} \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 ) + 2 B 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 ) + B b^{2} \left (\begin {cases} - \frac {16 a^{4} \sqrt {a + b x^{2}}}{315 b^{4}} + \frac {8 a^{3} x^{2} \sqrt {a + b x^{2}}}{315 b^{3}} - \frac {2 a^{2} x^{4} \sqrt {a + b x^{2}}}{105 b^{2}} + \frac {a x^{6} \sqrt {a + b x^{2}}}{63 b} + \frac {x^{8} \sqrt {a + b x^{2}}}{9} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{8}}{8} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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