3.338 \(\int \frac{(A+B x) (a+c x^2)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=147 \[ \frac{2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}+\frac{B c^3 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{16 a^{3/2}}-\frac{A \left (a+c x^2\right )^{5/2}}{7 a x^7}+\frac{B c^2 \sqrt{a+c x^2}}{16 a x^2}+\frac{B c \left (a+c x^2\right )^{3/2}}{24 a x^4}-\frac{B \left (a+c x^2\right )^{5/2}}{6 a x^6} \]

[Out]

(B*c^2*Sqrt[a + c*x^2])/(16*a*x^2) + (B*c*(a + c*x^2)^(3/2))/(24*a*x^4) - (A*(a + c*x^2)^(5/2))/(7*a*x^7) - (B
*(a + c*x^2)^(5/2))/(6*a*x^6) + (2*A*c*(a + c*x^2)^(5/2))/(35*a^2*x^5) + (B*c^3*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a
]])/(16*a^(3/2))

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Rubi [A]  time = 0.105246, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {835, 807, 266, 47, 63, 208} \[ \frac{2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}+\frac{B c^3 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{16 a^{3/2}}-\frac{A \left (a+c x^2\right )^{5/2}}{7 a x^7}+\frac{B c^2 \sqrt{a+c x^2}}{16 a x^2}+\frac{B c \left (a+c x^2\right )^{3/2}}{24 a x^4}-\frac{B \left (a+c x^2\right )^{5/2}}{6 a x^6} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(3/2))/x^8,x]

[Out]

(B*c^2*Sqrt[a + c*x^2])/(16*a*x^2) + (B*c*(a + c*x^2)^(3/2))/(24*a*x^4) - (A*(a + c*x^2)^(5/2))/(7*a*x^7) - (B
*(a + c*x^2)^(5/2))/(6*a*x^6) + (2*A*c*(a + c*x^2)^(5/2))/(35*a^2*x^5) + (B*c^3*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a
]])/(16*a^(3/2))

Rule 835

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

Rule 807

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a+c x^2\right )^{3/2}}{x^8} \, dx &=-\frac{A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac{\int \frac{(-7 a B+2 A c x) \left (a+c x^2\right )^{3/2}}{x^7} \, dx}{7 a}\\ &=-\frac{A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac{B \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac{\int \frac{(-12 a A c-7 a B c x) \left (a+c x^2\right )^{3/2}}{x^6} \, dx}{42 a^2}\\ &=-\frac{A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac{B \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac{2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}-\frac{(B c) \int \frac{\left (a+c x^2\right )^{3/2}}{x^5} \, dx}{6 a}\\ &=-\frac{A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac{B \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac{2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}-\frac{(B c) \operatorname{Subst}\left (\int \frac{(a+c x)^{3/2}}{x^3} \, dx,x,x^2\right )}{12 a}\\ &=\frac{B c \left (a+c x^2\right )^{3/2}}{24 a x^4}-\frac{A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac{B \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac{2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}-\frac{\left (B c^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+c x}}{x^2} \, dx,x,x^2\right )}{16 a}\\ &=\frac{B c^2 \sqrt{a+c x^2}}{16 a x^2}+\frac{B c \left (a+c x^2\right )^{3/2}}{24 a x^4}-\frac{A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac{B \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac{2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}-\frac{\left (B c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+c x}} \, dx,x,x^2\right )}{32 a}\\ &=\frac{B c^2 \sqrt{a+c x^2}}{16 a x^2}+\frac{B c \left (a+c x^2\right )^{3/2}}{24 a x^4}-\frac{A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac{B \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac{2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}-\frac{\left (B c^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{c}+\frac{x^2}{c}} \, dx,x,\sqrt{a+c x^2}\right )}{16 a}\\ &=\frac{B c^2 \sqrt{a+c x^2}}{16 a x^2}+\frac{B c \left (a+c x^2\right )^{3/2}}{24 a x^4}-\frac{A \left (a+c x^2\right )^{5/2}}{7 a x^7}-\frac{B \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac{2 A c \left (a+c x^2\right )^{5/2}}{35 a^2 x^5}+\frac{B c^3 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{16 a^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.028797, size = 64, normalized size = 0.44 \[ \frac{\left (a+c x^2\right )^{5/2} \left (a^2 A \left (2 c x^2-5 a\right )+7 B c^3 x^7 \, _2F_1\left (\frac{5}{2},4;\frac{7}{2};\frac{c x^2}{a}+1\right )\right )}{35 a^4 x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(3/2))/x^8,x]

[Out]

((a + c*x^2)^(5/2)*(a^2*A*(-5*a + 2*c*x^2) + 7*B*c^3*x^7*Hypergeometric2F1[5/2, 4, 7/2, 1 + (c*x^2)/a]))/(35*a
^4*x^7)

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Maple [A]  time = 0.011, size = 165, normalized size = 1.1 \begin{align*} -{\frac{A}{7\,a{x}^{7}} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{2\,Ac}{35\,{a}^{2}{x}^{5}} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{B}{6\,a{x}^{6}} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{Bc}{24\,{a}^{2}{x}^{4}} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{B{c}^{2}}{48\,{a}^{3}{x}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{B{c}^{3}}{48\,{a}^{3}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{B{c}^{3}}{16}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{B{c}^{3}}{16\,{a}^{2}}\sqrt{c{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(3/2)/x^8,x)

[Out]

-1/7*A*(c*x^2+a)^(5/2)/a/x^7+2/35*A*c*(c*x^2+a)^(5/2)/a^2/x^5-1/6*B*(c*x^2+a)^(5/2)/a/x^6+1/24*B/a^2*c/x^4*(c*
x^2+a)^(5/2)+1/48*B/a^3*c^2/x^2*(c*x^2+a)^(5/2)-1/48*B/a^3*c^3*(c*x^2+a)^(3/2)+1/16*B/a^(3/2)*c^3*ln((2*a+2*a^
(1/2)*(c*x^2+a)^(1/2))/x)-1/16*B/a^2*c^3*(c*x^2+a)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^8,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88491, size = 590, normalized size = 4.01 \begin{align*} \left [\frac{105 \, B \sqrt{a} c^{3} x^{7} \log \left (-\frac{c x^{2} + 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (96 \, A c^{3} x^{6} - 105 \, B a c^{2} x^{5} - 48 \, A a c^{2} x^{4} - 490 \, B a^{2} c x^{3} - 384 \, A a^{2} c x^{2} - 280 \, B a^{3} x - 240 \, A a^{3}\right )} \sqrt{c x^{2} + a}}{3360 \, a^{2} x^{7}}, -\frac{105 \, B \sqrt{-a} c^{3} x^{7} \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right ) -{\left (96 \, A c^{3} x^{6} - 105 \, B a c^{2} x^{5} - 48 \, A a c^{2} x^{4} - 490 \, B a^{2} c x^{3} - 384 \, A a^{2} c x^{2} - 280 \, B a^{3} x - 240 \, A a^{3}\right )} \sqrt{c x^{2} + a}}{1680 \, a^{2} x^{7}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^8,x, algorithm="fricas")

[Out]

[1/3360*(105*B*sqrt(a)*c^3*x^7*log(-(c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(96*A*c^3*x^6 - 105*B*a
*c^2*x^5 - 48*A*a*c^2*x^4 - 490*B*a^2*c*x^3 - 384*A*a^2*c*x^2 - 280*B*a^3*x - 240*A*a^3)*sqrt(c*x^2 + a))/(a^2
*x^7), -1/1680*(105*B*sqrt(-a)*c^3*x^7*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (96*A*c^3*x^6 - 105*B*a*c^2*x^5 - 48
*A*a*c^2*x^4 - 490*B*a^2*c*x^3 - 384*A*a^2*c*x^2 - 280*B*a^3*x - 240*A*a^3)*sqrt(c*x^2 + a))/(a^2*x^7)]

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Sympy [B]  time = 16.8209, size = 575, normalized size = 3.91 \begin{align*} - \frac{15 A a^{6} c^{\frac{9}{2}} \sqrt{\frac{a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac{33 A a^{5} c^{\frac{11}{2}} x^{2} \sqrt{\frac{a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac{17 A a^{4} c^{\frac{13}{2}} x^{4} \sqrt{\frac{a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac{3 A a^{3} c^{\frac{15}{2}} x^{6} \sqrt{\frac{a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac{12 A a^{2} c^{\frac{17}{2}} x^{8} \sqrt{\frac{a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac{8 A a c^{\frac{19}{2}} x^{10} \sqrt{\frac{a}{c x^{2}} + 1}}{105 a^{5} c^{4} x^{6} + 210 a^{4} c^{5} x^{8} + 105 a^{3} c^{6} x^{10}} - \frac{A c^{\frac{3}{2}} \sqrt{\frac{a}{c x^{2}} + 1}}{5 x^{4}} - \frac{A c^{\frac{5}{2}} \sqrt{\frac{a}{c x^{2}} + 1}}{15 a x^{2}} + \frac{2 A c^{\frac{7}{2}} \sqrt{\frac{a}{c x^{2}} + 1}}{15 a^{2}} - \frac{B a^{2}}{6 \sqrt{c} x^{7} \sqrt{\frac{a}{c x^{2}} + 1}} - \frac{11 B a \sqrt{c}}{24 x^{5} \sqrt{\frac{a}{c x^{2}} + 1}} - \frac{17 B c^{\frac{3}{2}}}{48 x^{3} \sqrt{\frac{a}{c x^{2}} + 1}} - \frac{B c^{\frac{5}{2}}}{16 a x \sqrt{\frac{a}{c x^{2}} + 1}} + \frac{B c^{3} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{c} x} \right )}}{16 a^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(3/2)/x**8,x)

[Out]

-15*A*a**6*c**(9/2)*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10) - 33*
A*a**5*c**(11/2)*x**2*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10) - 1
7*A*a**4*c**(13/2)*x**4*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10) -
3*A*a**3*c**(15/2)*x**6*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10)
- 12*A*a**2*c**(17/2)*x**8*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10
) - 8*A*a*c**(19/2)*x**10*sqrt(a/(c*x**2) + 1)/(105*a**5*c**4*x**6 + 210*a**4*c**5*x**8 + 105*a**3*c**6*x**10)
- A*c**(3/2)*sqrt(a/(c*x**2) + 1)/(5*x**4) - A*c**(5/2)*sqrt(a/(c*x**2) + 1)/(15*a*x**2) + 2*A*c**(7/2)*sqrt(
a/(c*x**2) + 1)/(15*a**2) - B*a**2/(6*sqrt(c)*x**7*sqrt(a/(c*x**2) + 1)) - 11*B*a*sqrt(c)/(24*x**5*sqrt(a/(c*x
**2) + 1)) - 17*B*c**(3/2)/(48*x**3*sqrt(a/(c*x**2) + 1)) - B*c**(5/2)/(16*a*x*sqrt(a/(c*x**2) + 1)) + B*c**3*
asinh(sqrt(a)/(sqrt(c)*x))/(16*a**(3/2))

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Giac [B]  time = 1.20186, size = 512, normalized size = 3.48 \begin{align*} -\frac{B c^{3} \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + a}}{\sqrt{-a}}\right )}{8 \, \sqrt{-a} a} + \frac{105 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{13} B c^{3} + 1540 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{11} B a c^{3} + 3360 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{10} A a c^{\frac{7}{2}} + 1085 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{9} B a^{2} c^{3} + 3360 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{8} A a^{2} c^{\frac{7}{2}} + 6720 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{6} A a^{3} c^{\frac{7}{2}} - 1085 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{5} B a^{4} c^{3} + 1344 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{4} A a^{4} c^{\frac{7}{2}} - 1540 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} B a^{5} c^{3} + 672 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} A a^{5} c^{\frac{7}{2}} - 105 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} B a^{6} c^{3} - 96 \, A a^{6} c^{\frac{7}{2}}}{840 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} - a\right )}^{7} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^8,x, algorithm="giac")

[Out]

-1/8*B*c^3*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(sqrt(-a)*a) + 1/840*(105*(sqrt(c)*x - sqrt(c*x^2 +
a))^13*B*c^3 + 1540*(sqrt(c)*x - sqrt(c*x^2 + a))^11*B*a*c^3 + 3360*(sqrt(c)*x - sqrt(c*x^2 + a))^10*A*a*c^(7
/2) + 1085*(sqrt(c)*x - sqrt(c*x^2 + a))^9*B*a^2*c^3 + 3360*(sqrt(c)*x - sqrt(c*x^2 + a))^8*A*a^2*c^(7/2) + 67
20*(sqrt(c)*x - sqrt(c*x^2 + a))^6*A*a^3*c^(7/2) - 1085*(sqrt(c)*x - sqrt(c*x^2 + a))^5*B*a^4*c^3 + 1344*(sqrt
(c)*x - sqrt(c*x^2 + a))^4*A*a^4*c^(7/2) - 1540*(sqrt(c)*x - sqrt(c*x^2 + a))^3*B*a^5*c^3 + 672*(sqrt(c)*x - s
qrt(c*x^2 + a))^2*A*a^5*c^(7/2) - 105*(sqrt(c)*x - sqrt(c*x^2 + a))*B*a^6*c^3 - 96*A*a^6*c^(7/2))/(((sqrt(c)*x
- sqrt(c*x^2 + a))^2 - a)^7*a)