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

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

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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 {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} \end {gather*}

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 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+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*}

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Mathematica [A]  time = 0.26, size = 131, normalized size = 0.87 \begin {gather*} \frac {\sqrt {a+c x^2} \left (-\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+a^3 c x (315 A+128 B x)+6 a^2 c^2 x^3 (413 A+320 B x)+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} \end {gather*}

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 271, normalized size = 1.81 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.21, size = 128, normalized size = 0.85 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.06, size = 132, normalized size = 0.88 \begin {gather*} -\frac {5 A \,a^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{128 c^{\frac {3}{2}}}-\frac {5 \sqrt {c \,x^{2}+a}\, A \,a^{3} x}{128 c}-\frac {5 \left (c \,x^{2}+a \right )^{\frac {3}{2}} A \,a^{2} x}{192 c}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} A a x}{48 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} B \,x^{2}}{9 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} A x}{8 c}-\frac {2 \left (c \,x^{2}+a \right )^{\frac {7}{2}} B a}{63 c^{2}} \end {gather*}

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(c^(1/2)*x+(c*x^2+a)^(
1/2))

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maxima [A]  time = 0.67, size = 124, normalized size = 0.83 \begin {gather*} \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} B x^{2}}{9 \, c} + \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} A x}{8 \, c} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A a x}{48 \, c} - \frac {5 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} A a^{2} x}{192 \, c} - \frac {5 \, \sqrt {c x^{2} + a} A a^{3} x}{128 \, c} - \frac {5 \, A a^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{128 \, c^{\frac {3}{2}}} - \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {7}{2}} B a}{63 \, c^{2}} \end {gather*}

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]

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

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

[Out]

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

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sympy [A]  time = 24.06, size = 442, normalized size = 2.95 \begin {gather*} \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 {gather*}

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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