3.4.37 \(\int \frac {(A+B x) (a+c x^2)^{3/2}}{x^7} \, dx\)

Optimal. Leaf size=124 \[ \frac {A c^3 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{16 a^{3/2}}+\frac {A c^2 \sqrt {a+c x^2}}{16 a x^2}-\frac {A \left (a+c x^2\right )^{5/2}}{6 a x^6}+\frac {A c \left (a+c x^2\right )^{3/2}}{24 a x^4}-\frac {B \left (a+c x^2\right )^{5/2}}{5 a x^5} \]

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Rubi [A]  time = 0.08, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {835, 807, 266, 47, 63, 208} \begin {gather*} \frac {A c^3 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{16 a^{3/2}}+\frac {A c^2 \sqrt {a+c x^2}}{16 a x^2}+\frac {A c \left (a+c x^2\right )^{3/2}}{24 a x^4}-\frac {A \left (a+c x^2\right )^{5/2}}{6 a x^6}-\frac {B \left (a+c x^2\right )^{5/2}}{5 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(3/2))/x^7,x]

[Out]

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

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]

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 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 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])

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 54, normalized size = 0.44 \begin {gather*} \frac {\left (a+c x^2\right )^{5/2} \left (A c^3 x^5 \, _2F_1\left (\frac {5}{2},4;\frac {7}{2};\frac {c x^2}{a}+1\right )-a^3 B\right )}{5 a^4 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(3/2))/x^7,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.78, size = 115, normalized size = 0.93 \begin {gather*} \frac {\sqrt {a+c x^2} \left (-40 a^2 A-48 a^2 B x-70 a A c x^2-96 a B c x^3-15 A c^2 x^4-48 B c^2 x^5\right )}{240 a x^6}-\frac {A c^3 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(a + c*x^2)^(3/2))/x^7,x]

[Out]

(Sqrt[a + c*x^2]*(-40*a^2*A - 48*a^2*B*x - 70*a*A*c*x^2 - 96*a*B*c*x^3 - 15*A*c^2*x^4 - 48*B*c^2*x^5))/(240*a*
x^6) - (A*c^3*ArcTanh[(Sqrt[c]*x)/Sqrt[a] - Sqrt[a + c*x^2]/Sqrt[a]])/(8*a^(3/2))

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fricas [A]  time = 0.47, size = 219, normalized size = 1.77 \begin {gather*} \left [\frac {15 \, A \sqrt {a} c^{3} x^{6} \log \left (-\frac {c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (48 \, B a c^{2} x^{5} + 15 \, A a c^{2} x^{4} + 96 \, B a^{2} c x^{3} + 70 \, A a^{2} c x^{2} + 48 \, B a^{3} x + 40 \, A a^{3}\right )} \sqrt {c x^{2} + a}}{480 \, a^{2} x^{6}}, -\frac {15 \, A \sqrt {-a} c^{3} x^{6} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + {\left (48 \, B a c^{2} x^{5} + 15 \, A a c^{2} x^{4} + 96 \, B a^{2} c x^{3} + 70 \, A a^{2} c x^{2} + 48 \, B a^{3} x + 40 \, A a^{3}\right )} \sqrt {c x^{2} + a}}{240 \, a^{2} x^{6}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^7,x, algorithm="fricas")

[Out]

[1/480*(15*A*sqrt(a)*c^3*x^6*log(-(c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(48*B*a*c^2*x^5 + 15*A*a*
c^2*x^4 + 96*B*a^2*c*x^3 + 70*A*a^2*c*x^2 + 48*B*a^3*x + 40*A*a^3)*sqrt(c*x^2 + a))/(a^2*x^6), -1/240*(15*A*sq
rt(-a)*c^3*x^6*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + (48*B*a*c^2*x^5 + 15*A*a*c^2*x^4 + 96*B*a^2*c*x^3 + 70*A*a^2
*c*x^2 + 48*B*a^3*x + 40*A*a^3)*sqrt(c*x^2 + a))/(a^2*x^6)]

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giac [B]  time = 0.27, size = 379, normalized size = 3.06 \begin {gather*} -\frac {A c^{3} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a} + \frac {15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{11} A c^{3} + 240 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{10} B a c^{\frac {5}{2}} + 235 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{9} A a c^{3} - 240 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{8} B a^{2} c^{\frac {5}{2}} + 390 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{7} A a^{2} c^{3} + 480 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{6} B a^{3} c^{\frac {5}{2}} + 390 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} A a^{3} c^{3} - 480 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} B a^{4} c^{\frac {5}{2}} + 235 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} A a^{4} c^{3} + 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} B a^{5} c^{\frac {5}{2}} + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} A a^{5} c^{3} - 48 \, B a^{6} c^{\frac {5}{2}}}{120 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{6} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^7,x, algorithm="giac")

[Out]

-1/8*A*c^3*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(sqrt(-a)*a) + 1/120*(15*(sqrt(c)*x - sqrt(c*x^2 +
a))^11*A*c^3 + 240*(sqrt(c)*x - sqrt(c*x^2 + a))^10*B*a*c^(5/2) + 235*(sqrt(c)*x - sqrt(c*x^2 + a))^9*A*a*c^3
- 240*(sqrt(c)*x - sqrt(c*x^2 + a))^8*B*a^2*c^(5/2) + 390*(sqrt(c)*x - sqrt(c*x^2 + a))^7*A*a^2*c^3 + 480*(sqr
t(c)*x - sqrt(c*x^2 + a))^6*B*a^3*c^(5/2) + 390*(sqrt(c)*x - sqrt(c*x^2 + a))^5*A*a^3*c^3 - 480*(sqrt(c)*x - s
qrt(c*x^2 + a))^4*B*a^4*c^(5/2) + 235*(sqrt(c)*x - sqrt(c*x^2 + a))^3*A*a^4*c^3 + 48*(sqrt(c)*x - sqrt(c*x^2 +
a))^2*B*a^5*c^(5/2) + 15*(sqrt(c)*x - sqrt(c*x^2 + a))*A*a^5*c^3 - 48*B*a^6*c^(5/2))/(((sqrt(c)*x - sqrt(c*x^
2 + a))^2 - a)^6*a)

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maple [A]  time = 0.07, size = 146, normalized size = 1.18 \begin {gather*} \frac {A \,c^{3} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{16 a^{\frac {3}{2}}}-\frac {\sqrt {c \,x^{2}+a}\, A \,c^{3}}{16 a^{2}}-\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} A \,c^{3}}{48 a^{3}}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} A \,c^{2}}{48 a^{3} x^{2}}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} A c}{24 a^{2} x^{4}}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B}{5 a \,x^{5}}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} A}{6 a \,x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(3/2)/x^7,x)

[Out]

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

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maxima [A]  time = 0.58, size = 134, normalized size = 1.08 \begin {gather*} \frac {A c^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{16 \, a^{\frac {3}{2}}} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} A c^{3}}{48 \, a^{3}} - \frac {\sqrt {c x^{2} + a} A c^{3}}{16 \, a^{2}} + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A c^{2}}{48 \, a^{3} x^{2}} + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A c}{24 \, a^{2} x^{4}} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B}{5 \, a x^{5}} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A}{6 \, a x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^7,x, algorithm="maxima")

[Out]

1/16*A*c^3*arcsinh(a/(sqrt(a*c)*abs(x)))/a^(3/2) - 1/48*(c*x^2 + a)^(3/2)*A*c^3/a^3 - 1/16*sqrt(c*x^2 + a)*A*c
^3/a^2 + 1/48*(c*x^2 + a)^(5/2)*A*c^2/(a^3*x^2) + 1/24*(c*x^2 + a)^(5/2)*A*c/(a^2*x^4) - 1/5*(c*x^2 + a)^(5/2)
*B/(a*x^5) - 1/6*(c*x^2 + a)^(5/2)*A/(a*x^6)

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mupad [B]  time = 3.21, size = 94, normalized size = 0.76 \begin {gather*} \frac {A\,a\,\sqrt {c\,x^2+a}}{16\,x^6}-\frac {A\,{\left (c\,x^2+a\right )}^{3/2}}{6\,x^6}-\frac {A\,{\left (c\,x^2+a\right )}^{5/2}}{16\,a\,x^6}-\frac {B\,{\left (c\,x^2+a\right )}^{5/2}}{5\,a\,x^5}-\frac {A\,c^3\,\mathrm {atan}\left (\frac {\sqrt {c\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{16\,a^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^(3/2)*(A + B*x))/x^7,x)

[Out]

(A*a*(a + c*x^2)^(1/2))/(16*x^6) - (A*c^3*atan(((a + c*x^2)^(1/2)*1i)/a^(1/2))*1i)/(16*a^(3/2)) - (A*(a + c*x^
2)^(3/2))/(6*x^6) - (A*(a + c*x^2)^(5/2))/(16*a*x^6) - (B*(a + c*x^2)^(5/2))/(5*a*x^5)

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sympy [A]  time = 13.39, size = 201, normalized size = 1.62 \begin {gather*} - \frac {A a^{2}}{6 \sqrt {c} x^{7} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {11 A a \sqrt {c}}{24 x^{5} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {17 A c^{\frac {3}{2}}}{48 x^{3} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {A c^{\frac {5}{2}}}{16 a x \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {A c^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{16 a^{\frac {3}{2}}} - \frac {B a \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{5 x^{4}} - \frac {2 B c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{5 x^{2}} - \frac {B c^{\frac {5}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{5 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(3/2)/x**7,x)

[Out]

-A*a**2/(6*sqrt(c)*x**7*sqrt(a/(c*x**2) + 1)) - 11*A*a*sqrt(c)/(24*x**5*sqrt(a/(c*x**2) + 1)) - 17*A*c**(3/2)/
(48*x**3*sqrt(a/(c*x**2) + 1)) - A*c**(5/2)/(16*a*x*sqrt(a/(c*x**2) + 1)) + A*c**3*asinh(sqrt(a)/(sqrt(c)*x))/
(16*a**(3/2)) - B*a*sqrt(c)*sqrt(a/(c*x**2) + 1)/(5*x**4) - 2*B*c**(3/2)*sqrt(a/(c*x**2) + 1)/(5*x**2) - B*c**
(5/2)*sqrt(a/(c*x**2) + 1)/(5*a)

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