3.4.50 \(\int \frac {(A+B x) (a+c x^2)^{5/2}}{x^7} \, dx\)

Optimal. Leaf size=140 \[ -\frac {c^2 \sqrt {a+c x^2} (5 A+16 B x)}{16 x^2}-\frac {\left (a+c x^2\right )^{5/2} (5 A+6 B x)}{30 x^6}-\frac {c \left (a+c x^2\right )^{3/2} (5 A+8 B x)}{24 x^4}-\frac {5 A c^3 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{16 \sqrt {a}}+B c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \]

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Rubi [A]  time = 0.12, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {811, 844, 217, 206, 266, 63, 208} \begin {gather*} -\frac {c^2 \sqrt {a+c x^2} (5 A+16 B x)}{16 x^2}-\frac {c \left (a+c x^2\right )^{3/2} (5 A+8 B x)}{24 x^4}-\frac {\left (a+c x^2\right )^{5/2} (5 A+6 B x)}{30 x^6}-\frac {5 A c^3 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{16 \sqrt {a}}+B c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(5/2))/x^7,x]

[Out]

-(c^2*(5*A + 16*B*x)*Sqrt[a + c*x^2])/(16*x^2) - (c*(5*A + 8*B*x)*(a + c*x^2)^(3/2))/(24*x^4) - ((5*A + 6*B*x)
*(a + c*x^2)^(5/2))/(30*x^6) + B*c^(5/2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]] - (5*A*c^3*ArcTanh[Sqrt[a + c*x^
2]/Sqrt[a]])/(16*Sqrt[a])

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 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 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 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 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 811

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

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+c x^2\right )^{5/2}}{x^7} \, dx &=-\frac {(5 A+6 B x) \left (a+c x^2\right )^{5/2}}{30 x^6}-\frac {\int \frac {(-10 a A c-12 a B c x) \left (a+c x^2\right )^{3/2}}{x^5} \, dx}{12 a}\\ &=-\frac {c (5 A+8 B x) \left (a+c x^2\right )^{3/2}}{24 x^4}-\frac {(5 A+6 B x) \left (a+c x^2\right )^{5/2}}{30 x^6}+\frac {\int \frac {\left (60 a^2 A c^2+96 a^2 B c^2 x\right ) \sqrt {a+c x^2}}{x^3} \, dx}{96 a^2}\\ &=-\frac {c^2 (5 A+16 B x) \sqrt {a+c x^2}}{16 x^2}-\frac {c (5 A+8 B x) \left (a+c x^2\right )^{3/2}}{24 x^4}-\frac {(5 A+6 B x) \left (a+c x^2\right )^{5/2}}{30 x^6}-\frac {\int \frac {-120 a^3 A c^3-384 a^3 B c^3 x}{x \sqrt {a+c x^2}} \, dx}{384 a^3}\\ &=-\frac {c^2 (5 A+16 B x) \sqrt {a+c x^2}}{16 x^2}-\frac {c (5 A+8 B x) \left (a+c x^2\right )^{3/2}}{24 x^4}-\frac {(5 A+6 B x) \left (a+c x^2\right )^{5/2}}{30 x^6}+\frac {1}{16} \left (5 A c^3\right ) \int \frac {1}{x \sqrt {a+c x^2}} \, dx+\left (B c^3\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx\\ &=-\frac {c^2 (5 A+16 B x) \sqrt {a+c x^2}}{16 x^2}-\frac {c (5 A+8 B x) \left (a+c x^2\right )^{3/2}}{24 x^4}-\frac {(5 A+6 B x) \left (a+c x^2\right )^{5/2}}{30 x^6}+\frac {1}{32} \left (5 A c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )+\left (B c^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )\\ &=-\frac {c^2 (5 A+16 B x) \sqrt {a+c x^2}}{16 x^2}-\frac {c (5 A+8 B x) \left (a+c x^2\right )^{3/2}}{24 x^4}-\frac {(5 A+6 B x) \left (a+c x^2\right )^{5/2}}{30 x^6}+B c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )+\frac {1}{16} \left (5 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 )\\ &=-\frac {c^2 (5 A+16 B x) \sqrt {a+c x^2}}{16 x^2}-\frac {c (5 A+8 B x) \left (a+c x^2\right )^{3/2}}{24 x^4}-\frac {(5 A+6 B x) \left (a+c x^2\right )^{5/2}}{30 x^6}+B c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-\frac {5 A c^3 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{16 \sqrt {a}}\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 125, normalized size = 0.89 \begin {gather*} -\frac {\sqrt {a+c x^2} \left (48 a^3 B x \, _2F_1\left (-\frac {5}{2},-\frac {5}{2};-\frac {3}{2};-\frac {c x^2}{a}\right )+5 a A \sqrt {\frac {c x^2}{a}+1} \left (8 a^2+26 a c x^2+33 c^2 x^4\right )+75 A c^3 x^6 \tanh ^{-1}\left (\sqrt {\frac {c x^2}{a}+1}\right )\right )}{240 a x^6 \sqrt {\frac {c x^2}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(5/2))/x^7,x]

[Out]

-1/240*(Sqrt[a + c*x^2]*(5*a*A*Sqrt[1 + (c*x^2)/a]*(8*a^2 + 26*a*c*x^2 + 33*c^2*x^4) + 75*A*c^3*x^6*ArcTanh[Sq
rt[1 + (c*x^2)/a]] + 48*a^3*B*x*Hypergeometric2F1[-5/2, -5/2, -3/2, -((c*x^2)/a)]))/(a*x^6*Sqrt[1 + (c*x^2)/a]
)

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IntegrateAlgebraic [A]  time = 0.83, size = 141, normalized size = 1.01 \begin {gather*} \frac {\sqrt {a+c x^2} \left (-40 a^2 A-48 a^2 B x-130 a A c x^2-176 a B c x^3-165 A c^2 x^4-368 B c^2 x^5\right )}{240 x^6}+\frac {5 A c^3 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 \sqrt {a}}-B c^{5/2} \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(a + c*x^2)^(5/2))/x^7,x]

[Out]

(Sqrt[a + c*x^2]*(-40*a^2*A - 48*a^2*B*x - 130*a*A*c*x^2 - 176*a*B*c*x^3 - 165*A*c^2*x^4 - 368*B*c^2*x^5))/(24
0*x^6) + (5*A*c^3*ArcTanh[(Sqrt[c]*x)/Sqrt[a] - Sqrt[a + c*x^2]/Sqrt[a]])/(8*Sqrt[a]) - B*c^(5/2)*Log[-(Sqrt[c
]*x) + Sqrt[a + c*x^2]]

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fricas [A]  time = 0.52, size = 574, normalized size = 4.10 \begin {gather*} \left [\frac {240 \, B a c^{\frac {5}{2}} x^{6} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 75 \, 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 (368 \, B a c^{2} x^{5} + 165 \, A a c^{2} x^{4} + 176 \, B a^{2} c x^{3} + 130 \, A a^{2} c x^{2} + 48 \, B a^{3} x + 40 \, A a^{3}\right )} \sqrt {c x^{2} + a}}{480 \, a x^{6}}, -\frac {480 \, B a \sqrt {-c} c^{2} x^{6} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 75 \, 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 (368 \, B a c^{2} x^{5} + 165 \, A a c^{2} x^{4} + 176 \, B a^{2} c x^{3} + 130 \, A a^{2} c x^{2} + 48 \, B a^{3} x + 40 \, A a^{3}\right )} \sqrt {c x^{2} + a}}{480 \, a x^{6}}, \frac {75 \, A \sqrt {-a} c^{3} x^{6} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + 120 \, B a c^{\frac {5}{2}} x^{6} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) - {\left (368 \, B a c^{2} x^{5} + 165 \, A a c^{2} x^{4} + 176 \, B a^{2} c x^{3} + 130 \, A a^{2} c x^{2} + 48 \, B a^{3} x + 40 \, A a^{3}\right )} \sqrt {c x^{2} + a}}{240 \, a x^{6}}, -\frac {240 \, B a \sqrt {-c} c^{2} x^{6} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 75 \, A \sqrt {-a} c^{3} x^{6} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + {\left (368 \, B a c^{2} x^{5} + 165 \, A a c^{2} x^{4} + 176 \, B a^{2} c x^{3} + 130 \, A a^{2} c x^{2} + 48 \, B a^{3} x + 40 \, A a^{3}\right )} \sqrt {c x^{2} + a}}{240 \, a x^{6}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/x^7,x, algorithm="fricas")

[Out]

[1/480*(240*B*a*c^(5/2)*x^6*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 75*A*sqrt(a)*c^3*x^6*log(-(c*x^2
- 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(368*B*a*c^2*x^5 + 165*A*a*c^2*x^4 + 176*B*a^2*c*x^3 + 130*A*a^2*
c*x^2 + 48*B*a^3*x + 40*A*a^3)*sqrt(c*x^2 + a))/(a*x^6), -1/480*(480*B*a*sqrt(-c)*c^2*x^6*arctan(sqrt(-c)*x/sq
rt(c*x^2 + a)) - 75*A*sqrt(a)*c^3*x^6*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(368*B*a*c^2*x^5
+ 165*A*a*c^2*x^4 + 176*B*a^2*c*x^3 + 130*A*a^2*c*x^2 + 48*B*a^3*x + 40*A*a^3)*sqrt(c*x^2 + a))/(a*x^6), 1/24
0*(75*A*sqrt(-a)*c^3*x^6*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + 120*B*a*c^(5/2)*x^6*log(-2*c*x^2 - 2*sqrt(c*x^2 +
a)*sqrt(c)*x - a) - (368*B*a*c^2*x^5 + 165*A*a*c^2*x^4 + 176*B*a^2*c*x^3 + 130*A*a^2*c*x^2 + 48*B*a^3*x + 40*A
*a^3)*sqrt(c*x^2 + a))/(a*x^6), -1/240*(240*B*a*sqrt(-c)*c^2*x^6*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 75*A*sqr
t(-a)*c^3*x^6*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + (368*B*a*c^2*x^5 + 165*A*a*c^2*x^4 + 176*B*a^2*c*x^3 + 130*A*
a^2*c*x^2 + 48*B*a^3*x + 40*A*a^3)*sqrt(c*x^2 + a))/(a*x^6)]

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giac [B]  time = 0.26, size = 397, normalized size = 2.84 \begin {gather*} \frac {5 \, A c^{3} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a}} - B c^{\frac {5}{2}} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right ) + \frac {165 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{11} A c^{3} + 720 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{10} B a c^{\frac {5}{2}} + 25 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{9} A a c^{3} - 2160 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{8} B a^{2} c^{\frac {5}{2}} + 450 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{7} A a^{2} c^{3} + 3680 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{6} B a^{3} c^{\frac {5}{2}} + 450 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} A a^{3} c^{3} - 3360 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} B a^{4} c^{\frac {5}{2}} + 25 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} A a^{4} c^{3} + 1488 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} B a^{5} c^{\frac {5}{2}} + 165 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} A a^{5} c^{3} - 368 \, B a^{6} c^{\frac {5}{2}}}{120 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/x^7,x, algorithm="giac")

[Out]

5/8*A*c^3*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/sqrt(-a) - B*c^(5/2)*log(abs(-sqrt(c)*x + sqrt(c*x^2
+ a))) + 1/120*(165*(sqrt(c)*x - sqrt(c*x^2 + a))^11*A*c^3 + 720*(sqrt(c)*x - sqrt(c*x^2 + a))^10*B*a*c^(5/2)
+ 25*(sqrt(c)*x - sqrt(c*x^2 + a))^9*A*a*c^3 - 2160*(sqrt(c)*x - sqrt(c*x^2 + a))^8*B*a^2*c^(5/2) + 450*(sqrt
(c)*x - sqrt(c*x^2 + a))^7*A*a^2*c^3 + 3680*(sqrt(c)*x - sqrt(c*x^2 + a))^6*B*a^3*c^(5/2) + 450*(sqrt(c)*x - s
qrt(c*x^2 + a))^5*A*a^3*c^3 - 3360*(sqrt(c)*x - sqrt(c*x^2 + a))^4*B*a^4*c^(5/2) + 25*(sqrt(c)*x - sqrt(c*x^2
+ a))^3*A*a^4*c^3 + 1488*(sqrt(c)*x - sqrt(c*x^2 + a))^2*B*a^5*c^(5/2) + 165*(sqrt(c)*x - sqrt(c*x^2 + a))*A*a
^5*c^3 - 368*B*a^6*c^(5/2))/((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^6

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maple [B]  time = 0.07, size = 281, normalized size = 2.01 \begin {gather*} -\frac {5 A \,c^{3} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{16 \sqrt {a}}+B \,c^{\frac {5}{2}} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )+\frac {\sqrt {c \,x^{2}+a}\, B \,c^{3} x}{a}+\frac {5 \sqrt {c \,x^{2}+a}\, A \,c^{3}}{16 a}+\frac {2 \left (c \,x^{2}+a \right )^{\frac {3}{2}} B \,c^{3} x}{3 a^{2}}+\frac {5 \left (c \,x^{2}+a \right )^{\frac {3}{2}} A \,c^{3}}{48 a^{2}}+\frac {8 \left (c \,x^{2}+a \right )^{\frac {5}{2}} B \,c^{3} x}{15 a^{3}}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} A \,c^{3}}{16 a^{3}}-\frac {8 \left (c \,x^{2}+a \right )^{\frac {7}{2}} B \,c^{2}}{15 a^{3} x}-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} A \,c^{2}}{16 a^{3} x^{2}}-\frac {2 \left (c \,x^{2}+a \right )^{\frac {7}{2}} B c}{15 a^{2} x^{3}}-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} A c}{24 a^{2} x^{4}}-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} B}{5 a \,x^{5}}-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{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)^(5/2)/x^7,x)

[Out]

-1/5*B/a/x^5*(c*x^2+a)^(7/2)-2/15*B*c/a^2/x^3*(c*x^2+a)^(7/2)-8/15*B*c^2/a^3/x*(c*x^2+a)^(7/2)+8/15*B*c^3/a^3*
x*(c*x^2+a)^(5/2)+2/3*B*c^3/a^2*x*(c*x^2+a)^(3/2)+B*c^3/a*x*(c*x^2+a)^(1/2)+B*c^(5/2)*ln(c^(1/2)*x+(c*x^2+a)^(
1/2))-1/6*A/a/x^6*(c*x^2+a)^(7/2)-1/24*A*c/a^2/x^4*(c*x^2+a)^(7/2)-1/16*A*c^2/a^3/x^2*(c*x^2+a)^(7/2)+1/16*A*c
^3/a^3*(c*x^2+a)^(5/2)+5/48*A*c^3/a^2*(c*x^2+a)^(3/2)-5/16*A*c^3/a^(1/2)*ln((2*a+2*(c*x^2+a)^(1/2)*a^(1/2))/x)
+5/16*A*c^3/a*(c*x^2+a)^(1/2)

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maxima [B]  time = 0.76, size = 243, normalized size = 1.74 \begin {gather*} \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} B c^{3} x}{3 \, a^{2}} + \frac {\sqrt {c x^{2} + a} B c^{3} x}{a} + B c^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) - \frac {5 \, A c^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{16 \, \sqrt {a}} + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A c^{3}}{16 \, a^{3}} + \frac {5 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} A c^{3}}{48 \, a^{2}} + \frac {5 \, \sqrt {c x^{2} + a} A c^{3}}{16 \, a} - \frac {8 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} B c^{2}}{15 \, a^{2} x} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} A c^{2}}{16 \, a^{3} x^{2}} - \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {7}{2}} B c}{15 \, a^{2} x^{3}} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} A c}{24 \, a^{2} x^{4}} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} B}{5 \, a x^{5}} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{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)^(5/2)/x^7,x, algorithm="maxima")

[Out]

2/3*(c*x^2 + a)^(3/2)*B*c^3*x/a^2 + sqrt(c*x^2 + a)*B*c^3*x/a + B*c^(5/2)*arcsinh(c*x/sqrt(a*c)) - 5/16*A*c^3*
arcsinh(a/(sqrt(a*c)*abs(x)))/sqrt(a) + 1/16*(c*x^2 + a)^(5/2)*A*c^3/a^3 + 5/48*(c*x^2 + a)^(3/2)*A*c^3/a^2 +
5/16*sqrt(c*x^2 + a)*A*c^3/a - 8/15*(c*x^2 + a)^(5/2)*B*c^2/(a^2*x) - 1/16*(c*x^2 + a)^(7/2)*A*c^2/(a^3*x^2) -
2/15*(c*x^2 + a)^(7/2)*B*c/(a^2*x^3) - 1/24*(c*x^2 + a)^(7/2)*A*c/(a^2*x^4) - 1/5*(c*x^2 + a)^(7/2)*B/(a*x^5)
- 1/6*(c*x^2 + a)^(7/2)*A/(a*x^6)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+a\right )}^{5/2}\,\left (A+B\,x\right )}{x^7} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^(5/2)*(A + B*x))/x^7,x)

[Out]

int(((a + c*x^2)^(5/2)*(A + B*x))/x^7, x)

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sympy [B]  time = 17.16, size = 299, normalized size = 2.14 \begin {gather*} - \frac {A a^{3}}{6 \sqrt {c} x^{7} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {17 A a^{2} \sqrt {c}}{24 x^{5} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {35 A a c^{\frac {3}{2}}}{48 x^{3} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {A c^{\frac {5}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{2 x} - \frac {3 A c^{\frac {5}{2}}}{16 x \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {5 A c^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{16 \sqrt {a}} - \frac {B \sqrt {a} c^{2}}{x \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {B a^{2} \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{5 x^{4}} - \frac {11 B a c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{15 x^{2}} - \frac {8 B c^{\frac {5}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{15} + B c^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )} - \frac {B c^{3} x}{\sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(5/2)/x**7,x)

[Out]

-A*a**3/(6*sqrt(c)*x**7*sqrt(a/(c*x**2) + 1)) - 17*A*a**2*sqrt(c)/(24*x**5*sqrt(a/(c*x**2) + 1)) - 35*A*a*c**(
3/2)/(48*x**3*sqrt(a/(c*x**2) + 1)) - A*c**(5/2)*sqrt(a/(c*x**2) + 1)/(2*x) - 3*A*c**(5/2)/(16*x*sqrt(a/(c*x**
2) + 1)) - 5*A*c**3*asinh(sqrt(a)/(sqrt(c)*x))/(16*sqrt(a)) - B*sqrt(a)*c**2/(x*sqrt(1 + c*x**2/a)) - B*a**2*s
qrt(c)*sqrt(a/(c*x**2) + 1)/(5*x**4) - 11*B*a*c**(3/2)*sqrt(a/(c*x**2) + 1)/(15*x**2) - 8*B*c**(5/2)*sqrt(a/(c
*x**2) + 1)/15 + B*c**(5/2)*asinh(sqrt(c)*x/sqrt(a)) - B*c**3*x/(sqrt(a)*sqrt(1 + c*x**2/a))

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