3.4.47 \(\int \frac {(A+B x) (a+c x^2)^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=137 \[ -\frac {5}{2} a^{3/2} B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )-\frac {5 \left (a+c x^2\right )^{3/2} (a B-A c x)}{6 x^2}-\frac {5 a c \sqrt {a+c x^2} (A-B x)}{2 x}-\frac {\left (a+c x^2\right )^{5/2} (A-B x)}{3 x^3}+\frac {5}{2} a A c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \]

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Rubi [A]  time = 0.11, antiderivative size = 137, 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 = {813, 844, 217, 206, 266, 63, 208} \begin {gather*} -\frac {5}{2} a^{3/2} B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )-\frac {\left (a+c x^2\right )^{5/2} (A-B x)}{3 x^3}-\frac {5 \left (a+c x^2\right )^{3/2} (a B-A c x)}{6 x^2}-\frac {5 a c \sqrt {a+c x^2} (A-B x)}{2 x}+\frac {5}{2} a A 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^4,x]

[Out]

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

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 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^4} \, dx &=-\frac {(A-B x) \left (a+c x^2\right )^{5/2}}{3 x^3}-\frac {5}{18} \int \frac {(-6 a B-6 A c x) \left (a+c x^2\right )^{3/2}}{x^3} \, dx\\ &=-\frac {5 (a B-A c x) \left (a+c x^2\right )^{3/2}}{6 x^2}-\frac {(A-B x) \left (a+c x^2\right )^{5/2}}{3 x^3}+\frac {5}{48} \int \frac {(24 a A c+24 a B c x) \sqrt {a+c x^2}}{x^2} \, dx\\ &=-\frac {5 a c (A-B x) \sqrt {a+c x^2}}{2 x}-\frac {5 (a B-A c x) \left (a+c x^2\right )^{3/2}}{6 x^2}-\frac {(A-B x) \left (a+c x^2\right )^{5/2}}{3 x^3}-\frac {5}{96} \int \frac {-48 a^2 B c-48 a A c^2 x}{x \sqrt {a+c x^2}} \, dx\\ &=-\frac {5 a c (A-B x) \sqrt {a+c x^2}}{2 x}-\frac {5 (a B-A c x) \left (a+c x^2\right )^{3/2}}{6 x^2}-\frac {(A-B x) \left (a+c x^2\right )^{5/2}}{3 x^3}+\frac {1}{2} \left (5 a^2 B c\right ) \int \frac {1}{x \sqrt {a+c x^2}} \, dx+\frac {1}{2} \left (5 a A c^2\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx\\ &=-\frac {5 a c (A-B x) \sqrt {a+c x^2}}{2 x}-\frac {5 (a B-A c x) \left (a+c x^2\right )^{3/2}}{6 x^2}-\frac {(A-B x) \left (a+c x^2\right )^{5/2}}{3 x^3}+\frac {1}{4} \left (5 a^2 B c\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )+\frac {1}{2} \left (5 a A 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 a c (A-B x) \sqrt {a+c x^2}}{2 x}-\frac {5 (a B-A c x) \left (a+c x^2\right )^{3/2}}{6 x^2}-\frac {(A-B x) \left (a+c x^2\right )^{5/2}}{3 x^3}+\frac {5}{2} a A c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )+\frac {1}{2} \left (5 a^2 B\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )\\ &=-\frac {5 a c (A-B x) \sqrt {a+c x^2}}{2 x}-\frac {5 (a B-A c x) \left (a+c x^2\right )^{3/2}}{6 x^2}-\frac {(A-B x) \left (a+c x^2\right )^{5/2}}{3 x^3}+\frac {5}{2} a A c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-\frac {5}{2} a^{3/2} B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 94, normalized size = 0.69 \begin {gather*} \frac {B c \left (a+c x^2\right )^{7/2} \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};\frac {c x^2}{a}+1\right )}{7 a^2}-\frac {a^2 A \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^4,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.59, size = 140, normalized size = 1.02 \begin {gather*} 5 a^{3/2} B c \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )+\frac {\sqrt {a+c x^2} \left (-2 a^2 A-3 a^2 B x-14 a A c x^2+14 a B c x^3+3 A c^2 x^4+2 B c^2 x^5\right )}{6 x^3}-\frac {5}{2} a A 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^4,x]

[Out]

(Sqrt[a + c*x^2]*(-2*a^2*A - 3*a^2*B*x - 14*a*A*c*x^2 + 14*a*B*c*x^3 + 3*A*c^2*x^4 + 2*B*c^2*x^5))/(6*x^3) + 5
*a^(3/2)*B*c*ArcTanh[(Sqrt[c]*x)/Sqrt[a] - Sqrt[a + c*x^2]/Sqrt[a]] - (5*a*A*c^(3/2)*Log[-(Sqrt[c]*x) + Sqrt[a
+ c*x^2]])/2

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fricas [A]  time = 0.53, size = 529, normalized size = 3.86 \begin {gather*} \left [\frac {15 \, A a c^{\frac {3}{2}} x^{3} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 15 \, B a^{\frac {3}{2}} c x^{3} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, B c^{2} x^{5} + 3 \, A c^{2} x^{4} + 14 \, B a c x^{3} - 14 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{12 \, x^{3}}, -\frac {30 \, A a \sqrt {-c} c x^{3} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 15 \, B a^{\frac {3}{2}} c x^{3} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (2 \, B c^{2} x^{5} + 3 \, A c^{2} x^{4} + 14 \, B a c x^{3} - 14 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{12 \, x^{3}}, \frac {30 \, B \sqrt {-a} a c x^{3} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + 15 \, A a c^{\frac {3}{2}} x^{3} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (2 \, B c^{2} x^{5} + 3 \, A c^{2} x^{4} + 14 \, B a c x^{3} - 14 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{12 \, x^{3}}, -\frac {15 \, A a \sqrt {-c} c x^{3} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 15 \, B \sqrt {-a} a c x^{3} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (2 \, B c^{2} x^{5} + 3 \, A c^{2} x^{4} + 14 \, B a c x^{3} - 14 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{6 \, x^{3}}\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^4,x, algorithm="fricas")

[Out]

[1/12*(15*A*a*c^(3/2)*x^3*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 15*B*a^(3/2)*c*x^3*log(-(c*x^2 - 2
*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(2*B*c^2*x^5 + 3*A*c^2*x^4 + 14*B*a*c*x^3 - 14*A*a*c*x^2 - 3*B*a^2*x
- 2*A*a^2)*sqrt(c*x^2 + a))/x^3, -1/12*(30*A*a*sqrt(-c)*c*x^3*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 15*B*a^(3/2
)*c*x^3*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(2*B*c^2*x^5 + 3*A*c^2*x^4 + 14*B*a*c*x^3 - 14
*A*a*c*x^2 - 3*B*a^2*x - 2*A*a^2)*sqrt(c*x^2 + a))/x^3, 1/12*(30*B*sqrt(-a)*a*c*x^3*arctan(sqrt(-a)/sqrt(c*x^2
+ a)) + 15*A*a*c^(3/2)*x^3*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(2*B*c^2*x^5 + 3*A*c^2*x^4 + 1
4*B*a*c*x^3 - 14*A*a*c*x^2 - 3*B*a^2*x - 2*A*a^2)*sqrt(c*x^2 + a))/x^3, -1/6*(15*A*a*sqrt(-c)*c*x^3*arctan(sqr
t(-c)*x/sqrt(c*x^2 + a)) - 15*B*sqrt(-a)*a*c*x^3*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (2*B*c^2*x^5 + 3*A*c^2*x^4
+ 14*B*a*c*x^3 - 14*A*a*c*x^2 - 3*B*a^2*x - 2*A*a^2)*sqrt(c*x^2 + a))/x^3]

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giac [B]  time = 0.23, size = 239, normalized size = 1.74 \begin {gather*} \frac {5 \, B a^{2} c \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {5}{2} \, A a c^{\frac {3}{2}} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right ) + \frac {1}{6} \, {\left (14 \, B a c + {\left (2 \, B c^{2} x + 3 \, A c^{2}\right )} x\right )} \sqrt {c x^{2} + a} + \frac {3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} B a^{2} c + 18 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} A a^{2} c^{\frac {3}{2}} - 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} A a^{3} c^{\frac {3}{2}} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{4} c + 14 \, A a^{4} c^{\frac {3}{2}}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/x^4,x, algorithm="giac")

[Out]

5*B*a^2*c*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/sqrt(-a) - 5/2*A*a*c^(3/2)*log(abs(-sqrt(c)*x + sqrt
(c*x^2 + a))) + 1/6*(14*B*a*c + (2*B*c^2*x + 3*A*c^2)*x)*sqrt(c*x^2 + a) + 1/3*(3*(sqrt(c)*x - sqrt(c*x^2 + a)
)^5*B*a^2*c + 18*(sqrt(c)*x - sqrt(c*x^2 + a))^4*A*a^2*c^(3/2) - 24*(sqrt(c)*x - sqrt(c*x^2 + a))^2*A*a^3*c^(3
/2) - 3*(sqrt(c)*x - sqrt(c*x^2 + a))*B*a^4*c + 14*A*a^4*c^(3/2))/((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^3

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maple [A]  time = 0.06, size = 207, normalized size = 1.51 \begin {gather*} \frac {5 A a \,c^{\frac {3}{2}} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2}-\frac {5 B \,a^{\frac {3}{2}} c \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2}+\frac {5 \sqrt {c \,x^{2}+a}\, A \,c^{2} x}{2}+\frac {5 \left (c \,x^{2}+a \right )^{\frac {3}{2}} A \,c^{2} x}{3 a}+\frac {5 \sqrt {c \,x^{2}+a}\, B a c}{2}+\frac {4 \left (c \,x^{2}+a \right )^{\frac {5}{2}} A \,c^{2} x}{3 a^{2}}+\frac {5 \left (c \,x^{2}+a \right )^{\frac {3}{2}} B c}{6}+\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B c}{2 a}-\frac {4 \left (c \,x^{2}+a \right )^{\frac {7}{2}} A c}{3 a^{2} x}-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} B}{2 a \,x^{2}}-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}} A}{3 a \,x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(5/2)/x^4,x)

[Out]

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

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maxima [A]  time = 0.66, size = 169, normalized size = 1.23 \begin {gather*} \frac {5}{2} \, \sqrt {c x^{2} + a} A c^{2} x + \frac {5 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} A c^{2} x}{3 \, a} + \frac {5}{2} \, A a c^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) - \frac {5}{2} \, B a^{\frac {3}{2}} c \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right ) + \frac {5}{6} \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} B c + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B c}{2 \, a} + \frac {5}{2} \, \sqrt {c x^{2} + a} B a c - \frac {4 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} A c}{3 \, a x} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} B}{2 \, a x^{2}} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} A}{3 \, a x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(5/2)/x^4,x, algorithm="maxima")

[Out]

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

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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^4} \,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^4,x)

[Out]

int(((a + c*x^2)^(5/2)*(A + B*x))/x^4, x)

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sympy [A]  time = 9.68, size = 277, normalized size = 2.02 \begin {gather*} - \frac {2 A a^{\frac {3}{2}} c}{x \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {A \sqrt {a} c^{2} x \sqrt {1 + \frac {c x^{2}}{a}}}{2} - \frac {2 A \sqrt {a} c^{2} x}{\sqrt {1 + \frac {c x^{2}}{a}}} - \frac {A a^{2} \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{3 x^{2}} - \frac {A a c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{3} + \frac {5 A a c^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{2} - \frac {5 B a^{\frac {3}{2}} c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{2} - \frac {B a^{2} \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{2 x} + \frac {2 B a^{2} \sqrt {c}}{x \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {2 B a c^{\frac {3}{2}} x}{\sqrt {\frac {a}{c x^{2}} + 1}} + B c^{2} \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: c = 0 \\\frac {\left (a + c x^{2}\right )^{\frac {3}{2}}}{3 c} & \text {otherwise} \end {cases}\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**4,x)

[Out]

-2*A*a**(3/2)*c/(x*sqrt(1 + c*x**2/a)) + A*sqrt(a)*c**2*x*sqrt(1 + c*x**2/a)/2 - 2*A*sqrt(a)*c**2*x/sqrt(1 + c
*x**2/a) - A*a**2*sqrt(c)*sqrt(a/(c*x**2) + 1)/(3*x**2) - A*a*c**(3/2)*sqrt(a/(c*x**2) + 1)/3 + 5*A*a*c**(3/2)
*asinh(sqrt(c)*x/sqrt(a))/2 - 5*B*a**(3/2)*c*asinh(sqrt(a)/(sqrt(c)*x))/2 - B*a**2*sqrt(c)*sqrt(a/(c*x**2) + 1
)/(2*x) + 2*B*a**2*sqrt(c)/(x*sqrt(a/(c*x**2) + 1)) + 2*B*a*c**(3/2)*x/sqrt(a/(c*x**2) + 1) + B*c**2*Piecewise
((sqrt(a)*x**2/2, Eq(c, 0)), ((a + c*x**2)**(3/2)/(3*c), True))

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