3.4.14 \(\int x^4 (A+B x) \sqrt {a+c x^2} \, dx\)

Optimal. Leaf size=152 \[ \frac {a^3 A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{5/2}}+\frac {a^2 A x \sqrt {a+c x^2}}{16 c^2}+\frac {a \left (a+c x^2\right )^{3/2} (64 a B-105 A c x)}{840 c^3}+\frac {A x^3 \left (a+c x^2\right )^{3/2}}{6 c}-\frac {4 a B x^2 \left (a+c x^2\right )^{3/2}}{35 c^2}+\frac {B x^4 \left (a+c x^2\right )^{3/2}}{7 c} \]

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Rubi [A]  time = 0.13, antiderivative size = 152, 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 {a^2 A x \sqrt {a+c x^2}}{16 c^2}+\frac {a^3 A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{5/2}}+\frac {a \left (a+c x^2\right )^{3/2} (64 a B-105 A c x)}{840 c^3}+\frac {A x^3 \left (a+c x^2\right )^{3/2}}{6 c}-\frac {4 a B x^2 \left (a+c x^2\right )^{3/2}}{35 c^2}+\frac {B x^4 \left (a+c x^2\right )^{3/2}}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(A + B*x)*Sqrt[a + c*x^2],x]

[Out]

(a^2*A*x*Sqrt[a + c*x^2])/(16*c^2) - (4*a*B*x^2*(a + c*x^2)^(3/2))/(35*c^2) + (A*x^3*(a + c*x^2)^(3/2))/(6*c)
+ (B*x^4*(a + c*x^2)^(3/2))/(7*c) + (a*(64*a*B - 105*A*c*x)*(a + c*x^2)^(3/2))/(840*c^3) + (a^3*A*ArcTanh[(Sqr
t[c]*x)/Sqrt[a + c*x^2]])/(16*c^(5/2))

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^4 (A+B x) \sqrt {a+c x^2} \, dx &=\frac {B x^4 \left (a+c x^2\right )^{3/2}}{7 c}+\frac {\int x^3 (-4 a B+7 A c x) \sqrt {a+c x^2} \, dx}{7 c}\\ &=\frac {A x^3 \left (a+c x^2\right )^{3/2}}{6 c}+\frac {B x^4 \left (a+c x^2\right )^{3/2}}{7 c}+\frac {\int x^2 (-21 a A c-24 a B c x) \sqrt {a+c x^2} \, dx}{42 c^2}\\ &=-\frac {4 a B x^2 \left (a+c x^2\right )^{3/2}}{35 c^2}+\frac {A x^3 \left (a+c x^2\right )^{3/2}}{6 c}+\frac {B x^4 \left (a+c x^2\right )^{3/2}}{7 c}+\frac {\int x \left (48 a^2 B c-105 a A c^2 x\right ) \sqrt {a+c x^2} \, dx}{210 c^3}\\ &=-\frac {4 a B x^2 \left (a+c x^2\right )^{3/2}}{35 c^2}+\frac {A x^3 \left (a+c x^2\right )^{3/2}}{6 c}+\frac {B x^4 \left (a+c x^2\right )^{3/2}}{7 c}+\frac {a (64 a B-105 A c x) \left (a+c x^2\right )^{3/2}}{840 c^3}+\frac {\left (a^2 A\right ) \int \sqrt {a+c x^2} \, dx}{8 c^2}\\ &=\frac {a^2 A x \sqrt {a+c x^2}}{16 c^2}-\frac {4 a B x^2 \left (a+c x^2\right )^{3/2}}{35 c^2}+\frac {A x^3 \left (a+c x^2\right )^{3/2}}{6 c}+\frac {B x^4 \left (a+c x^2\right )^{3/2}}{7 c}+\frac {a (64 a B-105 A c x) \left (a+c x^2\right )^{3/2}}{840 c^3}+\frac {\left (a^3 A\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{16 c^2}\\ &=\frac {a^2 A x \sqrt {a+c x^2}}{16 c^2}-\frac {4 a B x^2 \left (a+c x^2\right )^{3/2}}{35 c^2}+\frac {A x^3 \left (a+c x^2\right )^{3/2}}{6 c}+\frac {B x^4 \left (a+c x^2\right )^{3/2}}{7 c}+\frac {a (64 a B-105 A c x) \left (a+c x^2\right )^{3/2}}{840 c^3}+\frac {\left (a^3 A\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{16 c^2}\\ &=\frac {a^2 A x \sqrt {a+c x^2}}{16 c^2}-\frac {4 a B x^2 \left (a+c x^2\right )^{3/2}}{35 c^2}+\frac {A x^3 \left (a+c x^2\right )^{3/2}}{6 c}+\frac {B x^4 \left (a+c x^2\right )^{3/2}}{7 c}+\frac {a (64 a B-105 A c x) \left (a+c x^2\right )^{3/2}}{840 c^3}+\frac {a^3 A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 113, normalized size = 0.74 \begin {gather*} \frac {\sqrt {a+c x^2} \left (\frac {105 a^{5/2} A \sqrt {c} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {\frac {c x^2}{a}+1}}+128 a^3 B-a^2 c x (105 A+64 B x)+2 a c^2 x^3 (35 A+24 B x)+40 c^3 x^5 (7 A+6 B x)\right )}{1680 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(128*a^3*B + 40*c^3*x^5*(7*A + 6*B*x) + 2*a*c^2*x^3*(35*A + 24*B*x) - a^2*c*x*(105*A + 64*B*x
) + (105*a^(5/2)*A*Sqrt[c]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/Sqrt[1 + (c*x^2)/a]))/(1680*c^3)

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IntegrateAlgebraic [A]  time = 0.31, size = 116, normalized size = 0.76 \begin {gather*} \frac {\sqrt {a+c x^2} \left (128 a^3 B-105 a^2 A c x-64 a^2 B c x^2+70 a A c^2 x^3+48 a B c^2 x^4+280 A c^3 x^5+240 B c^3 x^6\right )}{1680 c^3}-\frac {a^3 A \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{16 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(128*a^3*B - 105*a^2*A*c*x - 64*a^2*B*c*x^2 + 70*a*A*c^2*x^3 + 48*a*B*c^2*x^4 + 280*A*c^3*x^5
+ 240*B*c^3*x^6))/(1680*c^3) - (a^3*A*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(16*c^(5/2))

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fricas [A]  time = 0.49, size = 224, normalized size = 1.47 \begin {gather*} \left [\frac {105 \, A a^{3} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (240 \, B c^{3} x^{6} + 280 \, A c^{3} x^{5} + 48 \, B a c^{2} x^{4} + 70 \, A a c^{2} x^{3} - 64 \, B a^{2} c x^{2} - 105 \, A a^{2} c x + 128 \, B a^{3}\right )} \sqrt {c x^{2} + a}}{3360 \, c^{3}}, -\frac {105 \, A a^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (240 \, B c^{3} x^{6} + 280 \, A c^{3} x^{5} + 48 \, B a c^{2} x^{4} + 70 \, A a c^{2} x^{3} - 64 \, B a^{2} c x^{2} - 105 \, A a^{2} c x + 128 \, B a^{3}\right )} \sqrt {c x^{2} + a}}{1680 \, c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)*(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/3360*(105*A*a^3*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(240*B*c^3*x^6 + 280*A*c^3*x^5
+ 48*B*a*c^2*x^4 + 70*A*a*c^2*x^3 - 64*B*a^2*c*x^2 - 105*A*a^2*c*x + 128*B*a^3)*sqrt(c*x^2 + a))/c^3, -1/1680*
(105*A*a^3*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (240*B*c^3*x^6 + 280*A*c^3*x^5 + 48*B*a*c^2*x^4 + 70*
A*a*c^2*x^3 - 64*B*a^2*c*x^2 - 105*A*a^2*c*x + 128*B*a^3)*sqrt(c*x^2 + a))/c^3]

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giac [A]  time = 0.19, size = 106, normalized size = 0.70 \begin {gather*} -\frac {A a^{3} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{16 \, c^{\frac {5}{2}}} + \frac {1}{1680} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, {\left (6 \, B x + 7 \, A\right )} x + \frac {6 \, B a}{c}\right )} x + \frac {35 \, A a}{c}\right )} x - \frac {32 \, B a^{2}}{c^{2}}\right )} x - \frac {105 \, A a^{2}}{c^{2}}\right )} x + \frac {128 \, B a^{3}}{c^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)*(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

-1/16*A*a^3*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(5/2) + 1/1680*sqrt(c*x^2 + a)*((2*((4*(5*(6*B*x + 7*A)*x
+ 6*B*a/c)*x + 35*A*a/c)*x - 32*B*a^2/c^2)*x - 105*A*a^2/c^2)*x + 128*B*a^3/c^3)

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maple [A]  time = 0.06, size = 136, normalized size = 0.89 \begin {gather*} \frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} B \,x^{4}}{7 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} A \,x^{3}}{6 c}+\frac {A \,a^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{16 c^{\frac {5}{2}}}+\frac {\sqrt {c \,x^{2}+a}\, A \,a^{2} x}{16 c^{2}}-\frac {4 \left (c \,x^{2}+a \right )^{\frac {3}{2}} B a \,x^{2}}{35 c^{2}}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} A a x}{8 c^{2}}+\frac {8 \left (c \,x^{2}+a \right )^{\frac {3}{2}} B \,a^{2}}{105 c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x+A)*(c*x^2+a)^(1/2),x)

[Out]

1/7*B*x^4*(c*x^2+a)^(3/2)/c-4/35*a*B*x^2*(c*x^2+a)^(3/2)/c^2+8/105*B*a^2/c^3*(c*x^2+a)^(3/2)+1/6*A*x^3*(c*x^2+
a)^(3/2)/c-1/8*A*a/c^2*x*(c*x^2+a)^(3/2)+1/16*a^2*A*x*(c*x^2+a)^(1/2)/c^2+1/16*A*a^3/c^(5/2)*ln(c^(1/2)*x+(c*x
^2+a)^(1/2))

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maxima [A]  time = 0.56, size = 128, normalized size = 0.84 \begin {gather*} \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} B x^{4}}{7 \, c} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} A x^{3}}{6 \, c} - \frac {4 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} B a x^{2}}{35 \, c^{2}} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} A a x}{8 \, c^{2}} + \frac {\sqrt {c x^{2} + a} A a^{2} x}{16 \, c^{2}} + \frac {A a^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{16 \, c^{\frac {5}{2}}} + \frac {8 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} B a^{2}}{105 \, 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)^(1/2),x, algorithm="maxima")

[Out]

1/7*(c*x^2 + a)^(3/2)*B*x^4/c + 1/6*(c*x^2 + a)^(3/2)*A*x^3/c - 4/35*(c*x^2 + a)^(3/2)*B*a*x^2/c^2 - 1/8*(c*x^
2 + a)^(3/2)*A*a*x/c^2 + 1/16*sqrt(c*x^2 + a)*A*a^2*x/c^2 + 1/16*A*a^3*arcsinh(c*x/sqrt(a*c))/c^(5/2) + 8/105*
(c*x^2 + a)^(3/2)*B*a^2/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + c*x^2)^(1/2)*(A + B*x),x)

[Out]

int(x^4*(a + c*x^2)^(1/2)*(A + B*x), x)

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sympy [A]  time = 8.36, size = 216, normalized size = 1.42 \begin {gather*} - \frac {A a^{\frac {5}{2}} x}{16 c^{2} \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {A a^{\frac {3}{2}} x^{3}}{48 c \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {5 A \sqrt {a} x^{5}}{24 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {A a^{3} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{16 c^{\frac {5}{2}}} + \frac {A c x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} + B \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(B*x+A)*(c*x**2+a)**(1/2),x)

[Out]

-A*a**(5/2)*x/(16*c**2*sqrt(1 + c*x**2/a)) - A*a**(3/2)*x**3/(48*c*sqrt(1 + c*x**2/a)) + 5*A*sqrt(a)*x**5/(24*
sqrt(1 + c*x**2/a)) + A*a**3*asinh(sqrt(c)*x/sqrt(a))/(16*c**(5/2)) + A*c*x**7/(6*sqrt(a)*sqrt(1 + c*x**2/a))
+ B*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))

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