3.4.74 \(\int \frac {x^4 (A+B x)}{(a+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=99 \[ -\frac {x (3 A+4 B x)}{3 c^2 \sqrt {a+c x^2}}-\frac {x^3 (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}+\frac {A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{5/2}}+\frac {8 B \sqrt {a+c x^2}}{3 c^3} \]

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Rubi [A]  time = 0.06, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {819, 641, 217, 206} \begin {gather*} -\frac {x (3 A+4 B x)}{3 c^2 \sqrt {a+c x^2}}-\frac {x^3 (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}+\frac {A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{5/2}}+\frac {8 B \sqrt {a+c x^2}}{3 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x))/(a + c*x^2)^(5/2),x]

[Out]

-(x^3*(A + B*x))/(3*c*(a + c*x^2)^(3/2)) - (x*(3*A + 4*B*x))/(3*c^2*Sqrt[a + c*x^2]) + (8*B*Sqrt[a + c*x^2])/(
3*c^3) + (A*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/c^(5/2)

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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rubi steps

\begin {align*} \int \frac {x^4 (A+B x)}{\left (a+c x^2\right )^{5/2}} \, dx &=-\frac {x^3 (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}+\frac {\int \frac {x^2 (3 a A+4 a B x)}{\left (a+c x^2\right )^{3/2}} \, dx}{3 a c}\\ &=-\frac {x^3 (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac {x (3 A+4 B x)}{3 c^2 \sqrt {a+c x^2}}+\frac {\int \frac {3 a^2 A+8 a^2 B x}{\sqrt {a+c x^2}} \, dx}{3 a^2 c^2}\\ &=-\frac {x^3 (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac {x (3 A+4 B x)}{3 c^2 \sqrt {a+c x^2}}+\frac {8 B \sqrt {a+c x^2}}{3 c^3}+\frac {A \int \frac {1}{\sqrt {a+c x^2}} \, dx}{c^2}\\ &=-\frac {x^3 (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac {x (3 A+4 B x)}{3 c^2 \sqrt {a+c x^2}}+\frac {8 B \sqrt {a+c x^2}}{3 c^3}+\frac {A \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{c^2}\\ &=-\frac {x^3 (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}-\frac {x (3 A+4 B x)}{3 c^2 \sqrt {a+c x^2}}+\frac {8 B \sqrt {a+c x^2}}{3 c^3}+\frac {A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 89, normalized size = 0.90 \begin {gather*} \frac {8 a^2 B-3 a c x (A-4 B x)+3 A \sqrt {c} \left (a+c x^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )+c^2 x^3 (3 B x-4 A)}{3 c^3 \left (a+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x))/(a + c*x^2)^(5/2),x]

[Out]

(8*a^2*B - 3*a*c*x*(A - 4*B*x) + c^2*x^3*(-4*A + 3*B*x) + 3*A*Sqrt[c]*(a + c*x^2)^(3/2)*ArcTanh[(Sqrt[c]*x)/Sq
rt[a + c*x^2]])/(3*c^3*(a + c*x^2)^(3/2))

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IntegrateAlgebraic [A]  time = 0.52, size = 87, normalized size = 0.88 \begin {gather*} \frac {8 a^2 B-3 a A c x+12 a B c x^2-4 A c^2 x^3+3 B c^2 x^4}{3 c^3 \left (a+c x^2\right )^{3/2}}-\frac {A \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^4*(A + B*x))/(a + c*x^2)^(5/2),x]

[Out]

(8*a^2*B - 3*a*A*c*x + 12*a*B*c*x^2 - 4*A*c^2*x^3 + 3*B*c^2*x^4)/(3*c^3*(a + c*x^2)^(3/2)) - (A*Log[-(Sqrt[c]*
x) + Sqrt[a + c*x^2]])/c^(5/2)

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fricas [A]  time = 0.45, size = 258, normalized size = 2.61 \begin {gather*} \left [\frac {3 \, {\left (A c^{2} x^{4} + 2 \, A a c x^{2} + A a^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (3 \, B c^{2} x^{4} - 4 \, A c^{2} x^{3} + 12 \, B a c x^{2} - 3 \, A a c x + 8 \, B a^{2}\right )} \sqrt {c x^{2} + a}}{6 \, {\left (c^{5} x^{4} + 2 \, a c^{4} x^{2} + a^{2} c^{3}\right )}}, -\frac {3 \, {\left (A c^{2} x^{4} + 2 \, A a c x^{2} + A a^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (3 \, B c^{2} x^{4} - 4 \, A c^{2} x^{3} + 12 \, B a c x^{2} - 3 \, A a c x + 8 \, B a^{2}\right )} \sqrt {c x^{2} + a}}{3 \, {\left (c^{5} x^{4} + 2 \, a c^{4} x^{2} + a^{2} c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(c*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(A*c^2*x^4 + 2*A*a*c*x^2 + A*a^2)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(3*B*c^2
*x^4 - 4*A*c^2*x^3 + 12*B*a*c*x^2 - 3*A*a*c*x + 8*B*a^2)*sqrt(c*x^2 + a))/(c^5*x^4 + 2*a*c^4*x^2 + a^2*c^3), -
1/3*(3*(A*c^2*x^4 + 2*A*a*c*x^2 + A*a^2)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (3*B*c^2*x^4 - 4*A*c^2*
x^3 + 12*B*a*c*x^2 - 3*A*a*c*x + 8*B*a^2)*sqrt(c*x^2 + a))/(c^5*x^4 + 2*a*c^4*x^2 + a^2*c^3)]

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giac [A]  time = 0.21, size = 82, normalized size = 0.83 \begin {gather*} \frac {{\left ({\left ({\left (\frac {3 \, B x}{c} - \frac {4 \, A}{c}\right )} x + \frac {12 \, B a}{c^{2}}\right )} x - \frac {3 \, A a}{c^{2}}\right )} x + \frac {8 \, B a^{2}}{c^{3}}}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}}} - \frac {A \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(c*x^2+a)^(5/2),x, algorithm="giac")

[Out]

1/3*((((3*B*x/c - 4*A/c)*x + 12*B*a/c^2)*x - 3*A*a/c^2)*x + 8*B*a^2/c^3)/(c*x^2 + a)^(3/2) - A*log(abs(-sqrt(c
)*x + sqrt(c*x^2 + a)))/c^(5/2)

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maple [A]  time = 0.05, size = 111, normalized size = 1.12 \begin {gather*} \frac {B \,x^{4}}{\left (c \,x^{2}+a \right )^{\frac {3}{2}} c}-\frac {A \,x^{3}}{3 \left (c \,x^{2}+a \right )^{\frac {3}{2}} c}+\frac {4 B a \,x^{2}}{\left (c \,x^{2}+a \right )^{\frac {3}{2}} c^{2}}-\frac {A x}{\sqrt {c \,x^{2}+a}\, c^{2}}+\frac {8 B \,a^{2}}{3 \left (c \,x^{2}+a \right )^{\frac {3}{2}} c^{3}}+\frac {A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x+A)/(c*x^2+a)^(5/2),x)

[Out]

B*x^4/c/(c*x^2+a)^(3/2)+4*B*a/c^2*x^2/(c*x^2+a)^(3/2)+8/3*B*a^2/c^3/(c*x^2+a)^(3/2)-1/3*A*x^3/c/(c*x^2+a)^(3/2
)-A/c^2*x/(c*x^2+a)^(1/2)+A/c^(5/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))

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maxima [A]  time = 0.62, size = 122, normalized size = 1.23 \begin {gather*} -\frac {1}{3} \, A x {\left (\frac {3 \, x^{2}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c} + \frac {2 \, a}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c^{2}}\right )} + \frac {B x^{4}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c} + \frac {4 \, B a x^{2}}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} c^{2}} - \frac {A x}{3 \, \sqrt {c x^{2} + a} c^{2}} + \frac {A \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{c^{\frac {5}{2}}} + \frac {8 \, B a^{2}}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(c*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

-1/3*A*x*(3*x^2/((c*x^2 + a)^(3/2)*c) + 2*a/((c*x^2 + a)^(3/2)*c^2)) + B*x^4/((c*x^2 + a)^(3/2)*c) + 4*B*a*x^2
/((c*x^2 + a)^(3/2)*c^2) - 1/3*A*x/(sqrt(c*x^2 + a)*c^2) + A*arcsinh(c*x/sqrt(a*c))/c^(5/2) + 8/3*B*a^2/((c*x^
2 + a)^(3/2)*c^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4\,\left (A+B\,x\right )}{{\left (c\,x^2+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(A + B*x))/(a + c*x^2)^(5/2),x)

[Out]

int((x^4*(A + B*x))/(a + c*x^2)^(5/2), x)

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sympy [B]  time = 29.11, size = 445, normalized size = 4.49 \begin {gather*} A \left (\frac {3 a^{\frac {39}{2}} c^{11} \sqrt {1 + \frac {c x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} c^{\frac {27}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 3 a^{\frac {37}{2}} c^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {3 a^{\frac {37}{2}} c^{12} x^{2} \sqrt {1 + \frac {c x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} c^{\frac {27}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 3 a^{\frac {37}{2}} c^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {3 a^{19} c^{\frac {23}{2}} x}{3 a^{\frac {39}{2}} c^{\frac {27}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 3 a^{\frac {37}{2}} c^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {4 a^{18} c^{\frac {25}{2}} x^{3}}{3 a^{\frac {39}{2}} c^{\frac {27}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 3 a^{\frac {37}{2}} c^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {c x^{2}}{a}}}\right ) + B \left (\begin {cases} \frac {8 a^{2}}{3 a c^{3} \sqrt {a + c x^{2}} + 3 c^{4} x^{2} \sqrt {a + c x^{2}}} + \frac {12 a c x^{2}}{3 a c^{3} \sqrt {a + c x^{2}} + 3 c^{4} x^{2} \sqrt {a + c x^{2}}} + \frac {3 c^{2} x^{4}}{3 a c^{3} \sqrt {a + c x^{2}} + 3 c^{4} x^{2} \sqrt {a + c x^{2}}} & \text {for}\: c \neq 0 \\\frac {x^{6}}{6 a^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(B*x+A)/(c*x**2+a)**(5/2),x)

[Out]

A*(3*a**(39/2)*c**11*sqrt(1 + c*x**2/a)*asinh(sqrt(c)*x/sqrt(a))/(3*a**(39/2)*c**(27/2)*sqrt(1 + c*x**2/a) + 3
*a**(37/2)*c**(29/2)*x**2*sqrt(1 + c*x**2/a)) + 3*a**(37/2)*c**12*x**2*sqrt(1 + c*x**2/a)*asinh(sqrt(c)*x/sqrt
(a))/(3*a**(39/2)*c**(27/2)*sqrt(1 + c*x**2/a) + 3*a**(37/2)*c**(29/2)*x**2*sqrt(1 + c*x**2/a)) - 3*a**19*c**(
23/2)*x/(3*a**(39/2)*c**(27/2)*sqrt(1 + c*x**2/a) + 3*a**(37/2)*c**(29/2)*x**2*sqrt(1 + c*x**2/a)) - 4*a**18*c
**(25/2)*x**3/(3*a**(39/2)*c**(27/2)*sqrt(1 + c*x**2/a) + 3*a**(37/2)*c**(29/2)*x**2*sqrt(1 + c*x**2/a))) + B*
Piecewise((8*a**2/(3*a*c**3*sqrt(a + c*x**2) + 3*c**4*x**2*sqrt(a + c*x**2)) + 12*a*c*x**2/(3*a*c**3*sqrt(a +
c*x**2) + 3*c**4*x**2*sqrt(a + c*x**2)) + 3*c**2*x**4/(3*a*c**3*sqrt(a + c*x**2) + 3*c**4*x**2*sqrt(a + c*x**2
)), Ne(c, 0)), (x**6/(6*a**(5/2)), True))

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