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

Optimal. Leaf size=126 \[ -\frac {5 a^4 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{128 c^{3/2}}-\frac {5 a^3 B x \sqrt {a+c x^2}}{128 c}-\frac {5 a^2 B x \left (a+c x^2\right )^{3/2}}{192 c}+\frac {\left (a+c x^2\right )^{7/2} (8 A+7 B x)}{56 c}-\frac {a B x \left (a+c x^2\right )^{5/2}}{48 c} \]

[Out]

-5/192*a^2*B*x*(c*x^2+a)^(3/2)/c-1/48*a*B*x*(c*x^2+a)^(5/2)/c+1/56*(7*B*x+8*A)*(c*x^2+a)^(7/2)/c-5/128*a^4*B*a
rctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/c^(3/2)-5/128*a^3*B*x*(c*x^2+a)^(1/2)/c

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Rubi [A]  time = 0.04, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {780, 195, 217, 206} \[ -\frac {5 a^4 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{128 c^{3/2}}-\frac {5 a^3 B x \sqrt {a+c x^2}}{128 c}-\frac {5 a^2 B x \left (a+c x^2\right )^{3/2}}{192 c}+\frac {\left (a+c x^2\right )^{7/2} (8 A+7 B x)}{56 c}-\frac {a B x \left (a+c x^2\right )^{5/2}}{48 c} \]

Antiderivative was successfully verified.

[In]

Int[x*(A + B*x)*(a + c*x^2)^(5/2),x]

[Out]

(-5*a^3*B*x*Sqrt[a + c*x^2])/(128*c) - (5*a^2*B*x*(a + c*x^2)^(3/2))/(192*c) - (a*B*x*(a + c*x^2)^(5/2))/(48*c
) + ((8*A + 7*B*x)*(a + c*x^2)^(7/2))/(56*c) - (5*a^4*B*ArcTanh[(Sqrt[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]

Rubi steps

\begin {align*} \int x (A+B x) \left (a+c x^2\right )^{5/2} \, dx &=\frac {(8 A+7 B x) \left (a+c x^2\right )^{7/2}}{56 c}-\frac {(a B) \int \left (a+c x^2\right )^{5/2} \, dx}{8 c}\\ &=-\frac {a B x \left (a+c x^2\right )^{5/2}}{48 c}+\frac {(8 A+7 B x) \left (a+c x^2\right )^{7/2}}{56 c}-\frac {\left (5 a^2 B\right ) \int \left (a+c x^2\right )^{3/2} \, dx}{48 c}\\ &=-\frac {5 a^2 B x \left (a+c x^2\right )^{3/2}}{192 c}-\frac {a B x \left (a+c x^2\right )^{5/2}}{48 c}+\frac {(8 A+7 B x) \left (a+c x^2\right )^{7/2}}{56 c}-\frac {\left (5 a^3 B\right ) \int \sqrt {a+c x^2} \, dx}{64 c}\\ &=-\frac {5 a^3 B x \sqrt {a+c x^2}}{128 c}-\frac {5 a^2 B x \left (a+c x^2\right )^{3/2}}{192 c}-\frac {a B x \left (a+c x^2\right )^{5/2}}{48 c}+\frac {(8 A+7 B x) \left (a+c x^2\right )^{7/2}}{56 c}-\frac {\left (5 a^4 B\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{128 c}\\ &=-\frac {5 a^3 B x \sqrt {a+c x^2}}{128 c}-\frac {5 a^2 B x \left (a+c x^2\right )^{3/2}}{192 c}-\frac {a B x \left (a+c x^2\right )^{5/2}}{48 c}+\frac {(8 A+7 B x) \left (a+c x^2\right )^{7/2}}{56 c}-\frac {\left (5 a^4 B\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 B x \sqrt {a+c x^2}}{128 c}-\frac {5 a^2 B x \left (a+c x^2\right )^{3/2}}{192 c}-\frac {a B x \left (a+c x^2\right )^{5/2}}{48 c}+\frac {(8 A+7 B x) \left (a+c x^2\right )^{7/2}}{56 c}-\frac {5 a^4 B \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 = 0.49, size = 112, normalized size = 0.89 \[ \frac {\left (a+c x^2\right )^{7/2} \left (-\frac {7 a B x \left (\frac {15 a^{7/2} \sqrt {\frac {c x^2}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {c} x}+\left (a+c x^2\right ) \left (33 a^2+26 a c x^2+8 c^2 x^4\right )\right )}{\left (a+c x^2\right )^4}+384 A+336 B x\right )}{2688 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + c*x^2)^(7/2)*(384*A + 336*B*x - (7*a*B*x*((a + c*x^2)*(33*a^2 + 26*a*c*x^2 + 8*c^2*x^4) + (15*a^(7/2)*Sq
rt[1 + (c*x^2)/a]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[c]*x)))/(a + c*x^2)^4))/(2688*c)

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fricas [A]  time = 0.78, size = 253, normalized size = 2.01 \[ \left [\frac {105 \, B a^{4} \sqrt {c} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (336 \, B c^{4} x^{7} + 384 \, A c^{4} x^{6} + 952 \, B a c^{3} x^{5} + 1152 \, A a c^{3} x^{4} + 826 \, B a^{2} c^{2} x^{3} + 1152 \, A a^{2} c^{2} x^{2} + 105 \, B a^{3} c x + 384 \, A a^{3} c\right )} \sqrt {c x^{2} + a}}{5376 \, c^{2}}, \frac {105 \, B a^{4} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (336 \, B c^{4} x^{7} + 384 \, A c^{4} x^{6} + 952 \, B a c^{3} x^{5} + 1152 \, A a c^{3} x^{4} + 826 \, B a^{2} c^{2} x^{3} + 1152 \, A a^{2} c^{2} x^{2} + 105 \, B a^{3} c x + 384 \, A a^{3} c\right )} \sqrt {c x^{2} + a}}{2688 \, c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/5376*(105*B*a^4*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(336*B*c^4*x^7 + 384*A*c^4*x^6
+ 952*B*a*c^3*x^5 + 1152*A*a*c^3*x^4 + 826*B*a^2*c^2*x^3 + 1152*A*a^2*c^2*x^2 + 105*B*a^3*c*x + 384*A*a^3*c)*s
qrt(c*x^2 + a))/c^2, 1/2688*(105*B*a^4*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (336*B*c^4*x^7 + 384*A*c^
4*x^6 + 952*B*a*c^3*x^5 + 1152*A*a*c^3*x^4 + 826*B*a^2*c^2*x^3 + 1152*A*a^2*c^2*x^2 + 105*B*a^3*c*x + 384*A*a^
3*c)*sqrt(c*x^2 + a))/c^2]

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giac [A]  time = 0.20, size = 114, normalized size = 0.90 \[ \frac {5 \, B a^{4} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{128 \, c^{\frac {3}{2}}} + \frac {1}{2688} \, {\left (\frac {384 \, A a^{3}}{c} + {\left (\frac {105 \, B a^{3}}{c} + 2 \, {\left (576 \, A a^{2} + {\left (413 \, B a^{2} + 4 \, {\left (144 \, A a c + {\left (119 \, B a c + 6 \, {\left (7 \, B c^{2} x + 8 \, A c^{2}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {c x^{2} + a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/128*B*a^4*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(3/2) + 1/2688*(384*A*a^3/c + (105*B*a^3/c + 2*(576*A*a^2
+ (413*B*a^2 + 4*(144*A*a*c + (119*B*a*c + 6*(7*B*c^2*x + 8*A*c^2)*x)*x)*x)*x)*x)*x)*sqrt(c*x^2 + a)

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maple [A]  time = 0.05, size = 113, normalized size = 0.90 \[ -\frac {5 B \,a^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{128 c^{\frac {3}{2}}}-\frac {5 \sqrt {c \,x^{2}+a}\, B \,a^{3} x}{128 c}-\frac {5 \left (c \,x^{2}+a \right )^{\frac {3}{2}} B \,a^{2} x}{192 c}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B a x}{48 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} B x}{8 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} A}{7 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.56, size = 105, normalized size = 0.83 \[ \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} B x}{8 \, c} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B a x}{48 \, c} - \frac {5 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} B a^{2} x}{192 \, c} - \frac {5 \, \sqrt {c x^{2} + a} B a^{3} x}{128 \, c} - \frac {5 \, B a^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{128 \, c^{\frac {3}{2}}} + \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} A}{7 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + c*x^2)^(5/2)*(A + B*x),x)

[Out]

int(x*(a + c*x^2)^(5/2)*(A + B*x), x)

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sympy [A]  time = 22.22, size = 354, normalized size = 2.81 \[ A a^{2} \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: c = 0 \\\frac {\left (a + c x^{2}\right )^{\frac {3}{2}}}{3 c} & \text {otherwise} \end {cases}\right ) + 2 A a c \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + c x^{2}}}{15 c^{2}} + \frac {a x^{2} \sqrt {a + c x^{2}}}{15 c} + \frac {x^{4} \sqrt {a + c x^{2}}}{5} & \text {for}\: c \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + A c^{2} \left (\begin {cases} \frac {8 a^{3} \sqrt {a + c x^{2}}}{105 c^{3}} - \frac {4 a^{2} x^{2} \sqrt {a + c x^{2}}}{105 c^{2}} + \frac {a x^{4} \sqrt {a + c x^{2}}}{35 c} + \frac {x^{6} \sqrt {a + c x^{2}}}{7} & \text {for}\: c \neq 0 \\\frac {\sqrt {a} x^{6}}{6} & \text {otherwise} \end {cases}\right ) + \frac {5 B a^{\frac {7}{2}} x}{128 c \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {133 B a^{\frac {5}{2}} x^{3}}{384 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {127 B a^{\frac {3}{2}} c x^{5}}{192 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {23 B \sqrt {a} c^{2} x^{7}}{48 \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {5 B a^{4} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{128 c^{\frac {3}{2}}} + \frac {B c^{3} x^{9}}{8 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**2*Piecewise((sqrt(a)*x**2/2, Eq(c, 0)), ((a + c*x**2)**(3/2)/(3*c), True)) + 2*A*a*c*Piecewise((-2*a**2*s
qrt(a + c*x**2)/(15*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)) + A*c**2*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)) + 5*B*a**(7/2)*x/
(128*c*sqrt(1 + c*x**2/a)) + 133*B*a**(5/2)*x**3/(384*sqrt(1 + c*x**2/a)) + 127*B*a**(3/2)*c*x**5/(192*sqrt(1
+ c*x**2/a)) + 23*B*sqrt(a)*c**2*x**7/(48*sqrt(1 + c*x**2/a)) - 5*B*a**4*asinh(sqrt(c)*x/sqrt(a))/(128*c**(3/2
)) + B*c**3*x**9/(8*sqrt(a)*sqrt(1 + c*x**2/a))

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