3.4.26 \(\int x^4 (A+B x) (a+c x^2)^{3/2} \, dx\)

Optimal. Leaf size=175 \[ \frac {3 a^4 A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{128 c^{5/2}}+\frac {3 a^3 A x \sqrt {a+c x^2}}{128 c^2}+\frac {a^2 A x \left (a+c x^2\right )^{3/2}}{64 c^2}+\frac {a \left (a+c x^2\right )^{5/2} (128 a B-315 A c x)}{5040 c^3}+\frac {A x^3 \left (a+c x^2\right )^{5/2}}{8 c}-\frac {4 a B x^2 \left (a+c x^2\right )^{5/2}}{63 c^2}+\frac {B x^4 \left (a+c x^2\right )^{5/2}}{9 c} \]

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Rubi [A]  time = 0.14, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 8, 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 {3 a^3 A x \sqrt {a+c x^2}}{128 c^2}+\frac {a^2 A x \left (a+c x^2\right )^{3/2}}{64 c^2}+\frac {3 a^4 A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{128 c^{5/2}}+\frac {a \left (a+c x^2\right )^{5/2} (128 a B-315 A c x)}{5040 c^3}+\frac {A x^3 \left (a+c x^2\right )^{5/2}}{8 c}-\frac {4 a B x^2 \left (a+c x^2\right )^{5/2}}{63 c^2}+\frac {B x^4 \left (a+c x^2\right )^{5/2}}{9 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(A + B*x)*(a + c*x^2)^(3/2),x]

[Out]

(3*a^3*A*x*Sqrt[a + c*x^2])/(128*c^2) + (a^2*A*x*(a + c*x^2)^(3/2))/(64*c^2) - (4*a*B*x^2*(a + c*x^2)^(5/2))/(
63*c^2) + (A*x^3*(a + c*x^2)^(5/2))/(8*c) + (B*x^4*(a + c*x^2)^(5/2))/(9*c) + (a*(128*a*B - 315*A*c*x)*(a + c*
x^2)^(5/2))/(5040*c^3) + (3*a^4*A*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(128*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) \left (a+c x^2\right )^{3/2} \, dx &=\frac {B x^4 \left (a+c x^2\right )^{5/2}}{9 c}+\frac {\int x^3 (-4 a B+9 A c x) \left (a+c x^2\right )^{3/2} \, dx}{9 c}\\ &=\frac {A x^3 \left (a+c x^2\right )^{5/2}}{8 c}+\frac {B x^4 \left (a+c x^2\right )^{5/2}}{9 c}+\frac {\int x^2 (-27 a A c-32 a B c x) \left (a+c x^2\right )^{3/2} \, dx}{72 c^2}\\ &=-\frac {4 a B x^2 \left (a+c x^2\right )^{5/2}}{63 c^2}+\frac {A x^3 \left (a+c x^2\right )^{5/2}}{8 c}+\frac {B x^4 \left (a+c x^2\right )^{5/2}}{9 c}+\frac {\int x \left (64 a^2 B c-189 a A c^2 x\right ) \left (a+c x^2\right )^{3/2} \, dx}{504 c^3}\\ &=-\frac {4 a B x^2 \left (a+c x^2\right )^{5/2}}{63 c^2}+\frac {A x^3 \left (a+c x^2\right )^{5/2}}{8 c}+\frac {B x^4 \left (a+c x^2\right )^{5/2}}{9 c}+\frac {a (128 a B-315 A c x) \left (a+c x^2\right )^{5/2}}{5040 c^3}+\frac {\left (a^2 A\right ) \int \left (a+c x^2\right )^{3/2} \, dx}{16 c^2}\\ &=\frac {a^2 A x \left (a+c x^2\right )^{3/2}}{64 c^2}-\frac {4 a B x^2 \left (a+c x^2\right )^{5/2}}{63 c^2}+\frac {A x^3 \left (a+c x^2\right )^{5/2}}{8 c}+\frac {B x^4 \left (a+c x^2\right )^{5/2}}{9 c}+\frac {a (128 a B-315 A c x) \left (a+c x^2\right )^{5/2}}{5040 c^3}+\frac {\left (3 a^3 A\right ) \int \sqrt {a+c x^2} \, dx}{64 c^2}\\ &=\frac {3 a^3 A x \sqrt {a+c x^2}}{128 c^2}+\frac {a^2 A x \left (a+c x^2\right )^{3/2}}{64 c^2}-\frac {4 a B x^2 \left (a+c x^2\right )^{5/2}}{63 c^2}+\frac {A x^3 \left (a+c x^2\right )^{5/2}}{8 c}+\frac {B x^4 \left (a+c x^2\right )^{5/2}}{9 c}+\frac {a (128 a B-315 A c x) \left (a+c x^2\right )^{5/2}}{5040 c^3}+\frac {\left (3 a^4 A\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{128 c^2}\\ &=\frac {3 a^3 A x \sqrt {a+c x^2}}{128 c^2}+\frac {a^2 A x \left (a+c x^2\right )^{3/2}}{64 c^2}-\frac {4 a B x^2 \left (a+c x^2\right )^{5/2}}{63 c^2}+\frac {A x^3 \left (a+c x^2\right )^{5/2}}{8 c}+\frac {B x^4 \left (a+c x^2\right )^{5/2}}{9 c}+\frac {a (128 a B-315 A c x) \left (a+c x^2\right )^{5/2}}{5040 c^3}+\frac {\left (3 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^2}\\ &=\frac {3 a^3 A x \sqrt {a+c x^2}}{128 c^2}+\frac {a^2 A x \left (a+c x^2\right )^{3/2}}{64 c^2}-\frac {4 a B x^2 \left (a+c x^2\right )^{5/2}}{63 c^2}+\frac {A x^3 \left (a+c x^2\right )^{5/2}}{8 c}+\frac {B x^4 \left (a+c x^2\right )^{5/2}}{9 c}+\frac {a (128 a B-315 A c x) \left (a+c x^2\right )^{5/2}}{5040 c^3}+\frac {3 a^4 A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{128 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 132, normalized size = 0.75 \begin {gather*} \frac {\sqrt {a+c x^2} \left (\frac {945 a^{7/2} A \sqrt {c} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {\frac {c x^2}{a}+1}}+1024 a^4 B-a^3 c x (945 A+512 B x)+6 a^2 c^2 x^3 (105 A+64 B x)+40 a c^3 x^5 (189 A+160 B x)+560 c^4 x^7 (9 A+8 B x)\right )}{40320 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(A + B*x)*(a + c*x^2)^(3/2),x]

[Out]

(Sqrt[a + c*x^2]*(1024*a^4*B + 560*c^4*x^7*(9*A + 8*B*x) + 6*a^2*c^2*x^3*(105*A + 64*B*x) + 40*a*c^3*x^5*(189*
A + 160*B*x) - a^3*c*x*(945*A + 512*B*x) + (945*a^(7/2)*A*Sqrt[c]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/Sqrt[1 + (c*x^
2)/a]))/(40320*c^3)

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IntegrateAlgebraic [A]  time = 0.38, size = 140, normalized size = 0.80 \begin {gather*} \frac {\sqrt {a+c x^2} \left (1024 a^4 B-945 a^3 A c x-512 a^3 B c x^2+630 a^2 A c^2 x^3+384 a^2 B c^2 x^4+7560 a A c^3 x^5+6400 a B c^3 x^6+5040 A c^4 x^7+4480 B c^4 x^8\right )}{40320 c^3}-\frac {3 a^4 A \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{128 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^4*(A + B*x)*(a + c*x^2)^(3/2),x]

[Out]

(Sqrt[a + c*x^2]*(1024*a^4*B - 945*a^3*A*c*x - 512*a^3*B*c*x^2 + 630*a^2*A*c^2*x^3 + 384*a^2*B*c^2*x^4 + 7560*
a*A*c^3*x^5 + 6400*a*B*c^3*x^6 + 5040*A*c^4*x^7 + 4480*B*c^4*x^8))/(40320*c^3) - (3*a^4*A*Log[-(Sqrt[c]*x) + S
qrt[a + c*x^2]])/(128*c^(5/2))

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fricas [A]  time = 0.52, size = 272, normalized size = 1.55 \begin {gather*} \left [\frac {945 \, A a^{4} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (4480 \, B c^{4} x^{8} + 5040 \, A c^{4} x^{7} + 6400 \, B a c^{3} x^{6} + 7560 \, A a c^{3} x^{5} + 384 \, B a^{2} c^{2} x^{4} + 630 \, A a^{2} c^{2} x^{3} - 512 \, B a^{3} c x^{2} - 945 \, A a^{3} c x + 1024 \, B a^{4}\right )} \sqrt {c x^{2} + a}}{80640 \, c^{3}}, -\frac {945 \, A a^{4} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (4480 \, B c^{4} x^{8} + 5040 \, A c^{4} x^{7} + 6400 \, B a c^{3} x^{6} + 7560 \, A a c^{3} x^{5} + 384 \, B a^{2} c^{2} x^{4} + 630 \, A a^{2} c^{2} x^{3} - 512 \, B a^{3} c x^{2} - 945 \, A a^{3} c x + 1024 \, B a^{4}\right )} \sqrt {c x^{2} + a}}{40320 \, 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)^(3/2),x, algorithm="fricas")

[Out]

[1/80640*(945*A*a^4*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(4480*B*c^4*x^8 + 5040*A*c^4*x
^7 + 6400*B*a*c^3*x^6 + 7560*A*a*c^3*x^5 + 384*B*a^2*c^2*x^4 + 630*A*a^2*c^2*x^3 - 512*B*a^3*c*x^2 - 945*A*a^3
*c*x + 1024*B*a^4)*sqrt(c*x^2 + a))/c^3, -1/40320*(945*A*a^4*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (44
80*B*c^4*x^8 + 5040*A*c^4*x^7 + 6400*B*a*c^3*x^6 + 7560*A*a*c^3*x^5 + 384*B*a^2*c^2*x^4 + 630*A*a^2*c^2*x^3 -
512*B*a^3*c*x^2 - 945*A*a^3*c*x + 1024*B*a^4)*sqrt(c*x^2 + a))/c^3]

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giac [A]  time = 0.22, size = 130, normalized size = 0.74 \begin {gather*} -\frac {3 \, A a^{4} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{128 \, c^{\frac {5}{2}}} + \frac {1}{40320} \, \sqrt {c x^{2} + a} {\left (\frac {1024 \, B a^{4}}{c^{3}} - {\left (\frac {945 \, A a^{3}}{c^{2}} + 2 \, {\left (\frac {256 \, B a^{3}}{c^{2}} - {\left (\frac {315 \, A a^{2}}{c} + 4 \, {\left (\frac {48 \, B a^{2}}{c} + 5 \, {\left (189 \, A a + 2 \, {\left (80 \, B a + 7 \, {\left (8 \, B c x + 9 \, A c\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)*(c*x^2+a)^(3/2),x, algorithm="giac")

[Out]

-3/128*A*a^4*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(5/2) + 1/40320*sqrt(c*x^2 + a)*(1024*B*a^4/c^3 - (945*A
*a^3/c^2 + 2*(256*B*a^3/c^2 - (315*A*a^2/c + 4*(48*B*a^2/c + 5*(189*A*a + 2*(80*B*a + 7*(8*B*c*x + 9*A*c)*x)*x
)*x)*x)*x)*x)*x)

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maple [A]  time = 0.06, size = 155, normalized size = 0.89 \begin {gather*} \frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B \,x^{4}}{9 c}+\frac {3 A \,a^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{128 c^{\frac {5}{2}}}+\frac {3 \sqrt {c \,x^{2}+a}\, A \,a^{3} x}{128 c^{2}}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} A \,x^{3}}{8 c}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} A \,a^{2} x}{64 c^{2}}-\frac {4 \left (c \,x^{2}+a \right )^{\frac {5}{2}} B a \,x^{2}}{63 c^{2}}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} A a x}{16 c^{2}}+\frac {8 \left (c \,x^{2}+a \right )^{\frac {5}{2}} B \,a^{2}}{315 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)^(3/2),x)

[Out]

1/9*B*x^4*(c*x^2+a)^(5/2)/c-4/63*a*B*x^2*(c*x^2+a)^(5/2)/c^2+8/315*B*a^2/c^3*(c*x^2+a)^(5/2)+1/8*A*x^3*(c*x^2+
a)^(5/2)/c-1/16*A*a/c^2*x*(c*x^2+a)^(5/2)+1/64*a^2*A*x*(c*x^2+a)^(3/2)/c^2+3/128*a^3*A*x*(c*x^2+a)^(1/2)/c^2+3
/128*A*a^4/c^(5/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))

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maxima [A]  time = 0.67, size = 147, normalized size = 0.84 \begin {gather*} \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B x^{4}}{9 \, c} + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A x^{3}}{8 \, c} - \frac {4 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} B a x^{2}}{63 \, c^{2}} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A a x}{16 \, c^{2}} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} A a^{2} x}{64 \, c^{2}} + \frac {3 \, \sqrt {c x^{2} + a} A a^{3} x}{128 \, c^{2}} + \frac {3 \, A a^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{128 \, c^{\frac {5}{2}}} + \frac {8 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} B a^{2}}{315 \, 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)^(3/2),x, algorithm="maxima")

[Out]

1/9*(c*x^2 + a)^(5/2)*B*x^4/c + 1/8*(c*x^2 + a)^(5/2)*A*x^3/c - 4/63*(c*x^2 + a)^(5/2)*B*a*x^2/c^2 - 1/16*(c*x
^2 + a)^(5/2)*A*a*x/c^2 + 1/64*(c*x^2 + a)^(3/2)*A*a^2*x/c^2 + 3/128*sqrt(c*x^2 + a)*A*a^3*x/c^2 + 3/128*A*a^4
*arcsinh(c*x/sqrt(a*c))/c^(5/2) + 8/315*(c*x^2 + a)^(5/2)*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\,{\left (c\,x^2+a\right )}^{3/2}\,\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)^(3/2)*(A + B*x),x)

[Out]

int(x^4*(a + c*x^2)^(3/2)*(A + B*x), x)

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sympy [A]  time = 19.24, size = 366, normalized size = 2.09 \begin {gather*} - \frac {3 A a^{\frac {7}{2}} x}{128 c^{2} \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {A a^{\frac {5}{2}} x^{3}}{128 c \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {13 A a^{\frac {3}{2}} x^{5}}{64 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {5 A \sqrt {a} c x^{7}}{16 \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {3 A a^{4} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{128 c^{\frac {5}{2}}} + \frac {A c^{2} x^{9}}{8 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} + B a \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 \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**4*(B*x+A)*(c*x**2+a)**(3/2),x)

[Out]

-3*A*a**(7/2)*x/(128*c**2*sqrt(1 + c*x**2/a)) - A*a**(5/2)*x**3/(128*c*sqrt(1 + c*x**2/a)) + 13*A*a**(3/2)*x**
5/(64*sqrt(1 + c*x**2/a)) + 5*A*sqrt(a)*c*x**7/(16*sqrt(1 + c*x**2/a)) + 3*A*a**4*asinh(sqrt(c)*x/sqrt(a))/(12
8*c**(5/2)) + A*c**2*x**9/(8*sqrt(a)*sqrt(1 + c*x**2/a)) + B*a*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*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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