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

Optimal. Leaf size=143 \[ -\frac {5 c \sqrt {a+c x^2} (4 a B-3 A c x)}{8 x}-\frac {\left (a+c x^2\right )^{5/2} (A-2 B x)}{4 x^4}-\frac {5 \left (a+c x^2\right )^{3/2} (4 a B+3 A c x)}{24 x^3}-\frac {15}{8} \sqrt {a} A c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )+\frac {5}{2} a B c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \]

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Rubi [A]  time = 0.12, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {813, 811, 844, 217, 206, 266, 63, 208} \begin {gather*} -\frac {\left (a+c x^2\right )^{5/2} (A-2 B x)}{4 x^4}-\frac {5 \left (a+c x^2\right )^{3/2} (4 a B+3 A c x)}{24 x^3}-\frac {5 c \sqrt {a+c x^2} (4 a B-3 A c x)}{8 x}-\frac {15}{8} \sqrt {a} A c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )+\frac {5}{2} a B c^{3/2} \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^5,x]

[Out]

(-5*c*(4*a*B - 3*A*c*x)*Sqrt[a + c*x^2])/(8*x) - (5*(4*a*B + 3*A*c*x)*(a + c*x^2)^(3/2))/(24*x^3) - ((A - 2*B*
x)*(a + c*x^2)^(5/2))/(4*x^4) + (5*a*B*c^(3/2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/2 - (15*Sqrt[a]*A*c^2*Arc
Tanh[Sqrt[a + c*x^2]/Sqrt[a]])/8

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 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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^5} \, dx &=-\frac {(A-2 B x) \left (a+c x^2\right )^{5/2}}{4 x^4}-\frac {5}{16} \int \frac {(-8 a B-4 A c x) \left (a+c x^2\right )^{3/2}}{x^4} \, dx\\ &=-\frac {5 (4 a B+3 A c x) \left (a+c x^2\right )^{3/2}}{24 x^3}-\frac {(A-2 B x) \left (a+c x^2\right )^{5/2}}{4 x^4}+\frac {5 \int \frac {\left (32 a^2 B c+24 a A c^2 x\right ) \sqrt {a+c x^2}}{x^2} \, dx}{64 a}\\ &=-\frac {5 c (4 a B-3 A c x) \sqrt {a+c x^2}}{8 x}-\frac {5 (4 a B+3 A c x) \left (a+c x^2\right )^{3/2}}{24 x^3}-\frac {(A-2 B x) \left (a+c x^2\right )^{5/2}}{4 x^4}-\frac {5 \int \frac {-48 a^2 A c^2-64 a^2 B c^2 x}{x \sqrt {a+c x^2}} \, dx}{128 a}\\ &=-\frac {5 c (4 a B-3 A c x) \sqrt {a+c x^2}}{8 x}-\frac {5 (4 a B+3 A c x) \left (a+c x^2\right )^{3/2}}{24 x^3}-\frac {(A-2 B x) \left (a+c x^2\right )^{5/2}}{4 x^4}+\frac {1}{8} \left (15 a A c^2\right ) \int \frac {1}{x \sqrt {a+c x^2}} \, dx+\frac {1}{2} \left (5 a B c^2\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx\\ &=-\frac {5 c (4 a B-3 A c x) \sqrt {a+c x^2}}{8 x}-\frac {5 (4 a B+3 A c x) \left (a+c x^2\right )^{3/2}}{24 x^3}-\frac {(A-2 B x) \left (a+c x^2\right )^{5/2}}{4 x^4}+\frac {1}{16} \left (15 a A c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )+\frac {1}{2} \left (5 a B c^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )\\ &=-\frac {5 c (4 a B-3 A c x) \sqrt {a+c x^2}}{8 x}-\frac {5 (4 a B+3 A c x) \left (a+c x^2\right )^{3/2}}{24 x^3}-\frac {(A-2 B x) \left (a+c x^2\right )^{5/2}}{4 x^4}+\frac {5}{2} a B c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )+\frac {1}{8} (15 a A c) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )\\ &=-\frac {5 c (4 a B-3 A c x) \sqrt {a+c x^2}}{8 x}-\frac {5 (4 a B+3 A c x) \left (a+c x^2\right )^{3/2}}{24 x^3}-\frac {(A-2 B x) \left (a+c x^2\right )^{5/2}}{4 x^4}+\frac {5}{2} a B c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-\frac {15}{8} \sqrt {a} A c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 96, normalized size = 0.67 \begin {gather*} -\frac {A c^2 \left (a+c x^2\right )^{7/2} \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};\frac {c x^2}{a}+1\right )}{7 a^3}-\frac {a^2 B \sqrt {a+c x^2} \, _2F_1\left (-\frac {5}{2},-\frac {3}{2};-\frac {1}{2};-\frac {c x^2}{a}\right )}{3 x^3 \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^5,x]

[Out]

-1/3*(a^2*B*Sqrt[a + c*x^2]*Hypergeometric2F1[-5/2, -3/2, -1/2, -((c*x^2)/a)])/(x^3*Sqrt[1 + (c*x^2)/a]) - (A*
c^2*(a + c*x^2)^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, 1 + (c*x^2)/a])/(7*a^3)

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IntegrateAlgebraic [A]  time = 0.64, size = 144, normalized size = 1.01 \begin {gather*} \frac {\sqrt {a+c x^2} \left (-6 a^2 A-8 a^2 B x-27 a A c x^2-56 a B c x^3+24 A c^2 x^4+12 B c^2 x^5\right )}{24 x^4}+\frac {15}{4} \sqrt {a} A c^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )-\frac {5}{2} a B c^{3/2} \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^5,x]

[Out]

(Sqrt[a + c*x^2]*(-6*a^2*A - 8*a^2*B*x - 27*a*A*c*x^2 - 56*a*B*c*x^3 + 24*A*c^2*x^4 + 12*B*c^2*x^5))/(24*x^4)
+ (15*Sqrt[a]*A*c^2*ArcTanh[(Sqrt[c]*x)/Sqrt[a] - Sqrt[a + c*x^2]/Sqrt[a]])/4 - (5*a*B*c^(3/2)*Log[-(Sqrt[c]*x
) + Sqrt[a + c*x^2]])/2

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fricas [A]  time = 0.54, size = 534, normalized size = 3.73 \begin {gather*} \left [\frac {60 \, B a c^{\frac {3}{2}} x^{4} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 45 \, A \sqrt {a} c^{2} x^{4} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (12 \, B c^{2} x^{5} + 24 \, A c^{2} x^{4} - 56 \, B a c x^{3} - 27 \, A a c x^{2} - 8 \, B a^{2} x - 6 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{48 \, x^{4}}, -\frac {120 \, B a \sqrt {-c} c x^{4} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 45 \, A \sqrt {a} c^{2} x^{4} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (12 \, B c^{2} x^{5} + 24 \, A c^{2} x^{4} - 56 \, B a c x^{3} - 27 \, A a c x^{2} - 8 \, B a^{2} x - 6 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{48 \, x^{4}}, \frac {45 \, A \sqrt {-a} c^{2} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + 30 \, B a c^{\frac {3}{2}} x^{4} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + {\left (12 \, B c^{2} x^{5} + 24 \, A c^{2} x^{4} - 56 \, B a c x^{3} - 27 \, A a c x^{2} - 8 \, B a^{2} x - 6 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{24 \, x^{4}}, -\frac {60 \, B a \sqrt {-c} c x^{4} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 45 \, A \sqrt {-a} c^{2} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (12 \, B c^{2} x^{5} + 24 \, A c^{2} x^{4} - 56 \, B a c x^{3} - 27 \, A a c x^{2} - 8 \, B a^{2} x - 6 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{24 \, x^{4}}\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^5,x, algorithm="fricas")

[Out]

[1/48*(60*B*a*c^(3/2)*x^4*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 45*A*sqrt(a)*c^2*x^4*log(-(c*x^2 -
2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(12*B*c^2*x^5 + 24*A*c^2*x^4 - 56*B*a*c*x^3 - 27*A*a*c*x^2 - 8*B*a^
2*x - 6*A*a^2)*sqrt(c*x^2 + a))/x^4, -1/48*(120*B*a*sqrt(-c)*c*x^4*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 45*A*s
qrt(a)*c^2*x^4*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(12*B*c^2*x^5 + 24*A*c^2*x^4 - 56*B*a*c
*x^3 - 27*A*a*c*x^2 - 8*B*a^2*x - 6*A*a^2)*sqrt(c*x^2 + a))/x^4, 1/24*(45*A*sqrt(-a)*c^2*x^4*arctan(sqrt(-a)/s
qrt(c*x^2 + a)) + 30*B*a*c^(3/2)*x^4*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + (12*B*c^2*x^5 + 24*A*c^
2*x^4 - 56*B*a*c*x^3 - 27*A*a*c*x^2 - 8*B*a^2*x - 6*A*a^2)*sqrt(c*x^2 + a))/x^4, -1/24*(60*B*a*sqrt(-c)*c*x^4*
arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 45*A*sqrt(-a)*c^2*x^4*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (12*B*c^2*x^5 +
24*A*c^2*x^4 - 56*B*a*c*x^3 - 27*A*a*c*x^2 - 8*B*a^2*x - 6*A*a^2)*sqrt(c*x^2 + a))/x^4]

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giac [B]  time = 0.24, size = 316, normalized size = 2.21 \begin {gather*} \frac {15 \, A a c^{2} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a}} - \frac {5}{2} \, B a c^{\frac {3}{2}} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right ) + \frac {1}{2} \, {\left (B c^{2} x + 2 \, A c^{2}\right )} \sqrt {c x^{2} + a} + \frac {27 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{7} A a c^{2} + 72 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{6} B a^{2} c^{\frac {3}{2}} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} A a^{2} c^{2} - 168 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} B a^{3} c^{\frac {3}{2}} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} A a^{3} c^{2} + 152 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} B a^{4} c^{\frac {3}{2}} + 27 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} A a^{4} c^{2} - 56 \, B a^{5} c^{\frac {3}{2}}}{12 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/4*A*a*c^2*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/sqrt(-a) - 5/2*B*a*c^(3/2)*log(abs(-sqrt(c)*x + s
qrt(c*x^2 + a))) + 1/2*(B*c^2*x + 2*A*c^2)*sqrt(c*x^2 + a) + 1/12*(27*(sqrt(c)*x - sqrt(c*x^2 + a))^7*A*a*c^2
+ 72*(sqrt(c)*x - sqrt(c*x^2 + a))^6*B*a^2*c^(3/2) - 3*(sqrt(c)*x - sqrt(c*x^2 + a))^5*A*a^2*c^2 - 168*(sqrt(c
)*x - sqrt(c*x^2 + a))^4*B*a^3*c^(3/2) - 3*(sqrt(c)*x - sqrt(c*x^2 + a))^3*A*a^3*c^2 + 152*(sqrt(c)*x - sqrt(c
*x^2 + a))^2*B*a^4*c^(3/2) + 27*(sqrt(c)*x - sqrt(c*x^2 + a))*A*a^4*c^2 - 56*B*a^5*c^(3/2))/((sqrt(c)*x - sqrt
(c*x^2 + a))^2 - a)^4

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maple [B]  time = 0.06, size = 236, normalized size = 1.65 \begin {gather*} -\frac {15 A \sqrt {a}\, c^{2} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8}+\frac {5 B a \,c^{\frac {3}{2}} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2}+\frac {5 \sqrt {c \,x^{2}+a}\, B \,c^{2} x}{2}+\frac {15 \sqrt {c \,x^{2}+a}\, A \,c^{2}}{8}+\frac {5 \left (c \,x^{2}+a \right )^{\frac {3}{2}} B \,c^{2} x}{3 a}+\frac {5 \left (c \,x^{2}+a \right )^{\frac {3}{2}} A \,c^{2}}{8 a}+\frac {4 \left (c \,x^{2}+a \right )^{\frac {5}{2}} B \,c^{2} x}{3 a^{2}}+\frac {3 \left (c \,x^{2}+a \right )^{\frac {5}{2}} A \,c^{2}}{8 a^{2}}-\frac {4 \left (c \,x^{2}+a \right )^{\frac {7}{2}} B c}{3 a^{2} x}-\frac {3 \left (c \,x^{2}+a \right )^{\frac {7}{2}} A c}{8 a^{2} x^{2}}-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} B}{3 a \,x^{3}}-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} A}{4 a \,x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.65, size = 198, normalized size = 1.38 \begin {gather*} \frac {5}{2} \, \sqrt {c x^{2} + a} B c^{2} x + \frac {5 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} B c^{2} x}{3 \, a} + \frac {5}{2} \, B a c^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) - \frac {15}{8} \, A \sqrt {a} c^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right ) + \frac {15}{8} \, \sqrt {c x^{2} + a} A c^{2} + \frac {3 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} A c^{2}}{8 \, a^{2}} + \frac {5 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} A c^{2}}{8 \, a} - \frac {4 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} B c}{3 \, a x} - \frac {3 \, {\left (c x^{2} + a\right )}^{\frac {7}{2}} A c}{8 \, a^{2} x^{2}} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} B}{3 \, a x^{3}} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} A}{4 \, a x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^5} \,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^5,x)

[Out]

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

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sympy [B]  time = 12.86, size = 299, normalized size = 2.09 \begin {gather*} - \frac {15 A \sqrt {a} c^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{8} - \frac {A a^{3}}{4 \sqrt {c} x^{5} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {3 A a^{2} \sqrt {c}}{8 x^{3} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {A a c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{x} + \frac {7 A a c^{\frac {3}{2}}}{8 x \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {A c^{\frac {5}{2}} x}{\sqrt {\frac {a}{c x^{2}} + 1}} - \frac {2 B a^{\frac {3}{2}} c}{x \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {B \sqrt {a} c^{2} x \sqrt {1 + \frac {c x^{2}}{a}}}{2} - \frac {2 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}}{3 x^{2}} - \frac {B a c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{3} + \frac {5 B a c^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15*A*sqrt(a)*c**2*asinh(sqrt(a)/(sqrt(c)*x))/8 - A*a**3/(4*sqrt(c)*x**5*sqrt(a/(c*x**2) + 1)) - 3*A*a**2*sqrt
(c)/(8*x**3*sqrt(a/(c*x**2) + 1)) - A*a*c**(3/2)*sqrt(a/(c*x**2) + 1)/x + 7*A*a*c**(3/2)/(8*x*sqrt(a/(c*x**2)
+ 1)) + A*c**(5/2)*x/sqrt(a/(c*x**2) + 1) - 2*B*a**(3/2)*c/(x*sqrt(1 + c*x**2/a)) + B*sqrt(a)*c**2*x*sqrt(1 +
c*x**2/a)/2 - 2*B*sqrt(a)*c**2*x/sqrt(1 + c*x**2/a) - B*a**2*sqrt(c)*sqrt(a/(c*x**2) + 1)/(3*x**2) - B*a*c**(3
/2)*sqrt(a/(c*x**2) + 1)/3 + 5*B*a*c**(3/2)*asinh(sqrt(c)*x/sqrt(a))/2

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