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

Optimal. Leaf size=136 \[ a^{5/2} (-B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {15}{8} a^2 A \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )+\frac {1}{8} a \sqrt {a+b x^2} (8 a B+15 A b x)-\frac {\left (a+b x^2\right )^{5/2} (5 A-B x)}{5 x}+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 a B+15 A b x) \]

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Rubi [A]  time = 0.13, antiderivative size = 136, 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, 815, 844, 217, 206, 266, 63, 208} \begin {gather*} \frac {15}{8} a^2 A \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )+a^{5/2} (-B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {1}{8} a \sqrt {a+b x^2} (8 a B+15 A b x)-\frac {\left (a+b x^2\right )^{5/2} (5 A-B x)}{5 x}+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 a B+15 A b x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(8*a*B + 15*A*b*x)*Sqrt[a + b*x^2])/8 + ((4*a*B + 15*A*b*x)*(a + b*x^2)^(3/2))/12 - ((5*A - B*x)*(a + b*x^2
)^(5/2))/(5*x) + (15*a^2*A*Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/8 - a^(5/2)*B*ArcTanh[Sqrt[a + b*x^2]
/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 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 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 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+b x^2\right )^{5/2}}{x^2} \, dx &=-\frac {(5 A-B x) \left (a+b x^2\right )^{5/2}}{5 x}-\frac {1}{2} \int \frac {(-2 a B-10 A b x) \left (a+b x^2\right )^{3/2}}{x} \, dx\\ &=\frac {1}{12} (4 a B+15 A b x) \left (a+b x^2\right )^{3/2}-\frac {(5 A-B x) \left (a+b x^2\right )^{5/2}}{5 x}-\frac {\int \frac {\left (-8 a^2 b B-30 a A b^2 x\right ) \sqrt {a+b x^2}}{x} \, dx}{8 b}\\ &=\frac {1}{8} a (8 a B+15 A b x) \sqrt {a+b x^2}+\frac {1}{12} (4 a B+15 A b x) \left (a+b x^2\right )^{3/2}-\frac {(5 A-B x) \left (a+b x^2\right )^{5/2}}{5 x}-\frac {\int \frac {-16 a^3 b^2 B-30 a^2 A b^3 x}{x \sqrt {a+b x^2}} \, dx}{16 b^2}\\ &=\frac {1}{8} a (8 a B+15 A b x) \sqrt {a+b x^2}+\frac {1}{12} (4 a B+15 A b x) \left (a+b x^2\right )^{3/2}-\frac {(5 A-B x) \left (a+b x^2\right )^{5/2}}{5 x}+\frac {1}{8} \left (15 a^2 A b\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx+\left (a^3 B\right ) \int \frac {1}{x \sqrt {a+b x^2}} \, dx\\ &=\frac {1}{8} a (8 a B+15 A b x) \sqrt {a+b x^2}+\frac {1}{12} (4 a B+15 A b x) \left (a+b x^2\right )^{3/2}-\frac {(5 A-B x) \left (a+b x^2\right )^{5/2}}{5 x}+\frac {1}{8} \left (15 a^2 A b\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )+\frac {1}{2} \left (a^3 B\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=\frac {1}{8} a (8 a B+15 A b x) \sqrt {a+b x^2}+\frac {1}{12} (4 a B+15 A b x) \left (a+b x^2\right )^{3/2}-\frac {(5 A-B x) \left (a+b x^2\right )^{5/2}}{5 x}+\frac {15}{8} a^2 A \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )+\frac {\left (a^3 B\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{b}\\ &=\frac {1}{8} a (8 a B+15 A b x) \sqrt {a+b x^2}+\frac {1}{12} (4 a B+15 A b x) \left (a+b x^2\right )^{3/2}-\frac {(5 A-B x) \left (a+b x^2\right )^{5/2}}{5 x}+\frac {15}{8} a^2 A \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-a^{5/2} B \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.23, size = 117, normalized size = 0.86 \begin {gather*} -a^{5/2} B \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {a^3 A \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (-\frac {5}{2},-\frac {1}{2};\frac {1}{2};-\frac {b x^2}{a}\right )}{x \sqrt {a+b x^2}}+\frac {1}{15} B \sqrt {a+b x^2} \left (23 a^2+11 a b x^2+3 b^2 x^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(B*Sqrt[a + b*x^2]*(23*a^2 + 11*a*b*x^2 + 3*b^2*x^4))/15 - a^(5/2)*B*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]] - (a^3*A
*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[-5/2, -1/2, 1/2, -((b*x^2)/a)])/(x*Sqrt[a + b*x^2])

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IntegrateAlgebraic [A]  time = 0.45, size = 141, normalized size = 1.04 \begin {gather*} 2 a^{5/2} B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}-\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {\sqrt {a+b x^2} \left (-120 a^2 A+184 a^2 B x+135 a A b x^2+88 a b B x^3+30 A b^2 x^4+24 b^2 B x^5\right )}{120 x}-\frac {15}{8} a^2 A \sqrt {b} \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(-120*a^2*A + 184*a^2*B*x + 135*a*A*b*x^2 + 88*a*b*B*x^3 + 30*A*b^2*x^4 + 24*b^2*B*x^5))/(120
*x) + 2*a^(5/2)*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a] - Sqrt[a + b*x^2]/Sqrt[a]] - (15*a^2*A*Sqrt[b]*Log[-(Sqrt[b]*x)
+ Sqrt[a + b*x^2]])/8

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fricas [A]  time = 0.95, size = 519, normalized size = 3.82 \begin {gather*} \left [\frac {225 \, A a^{2} \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 120 \, B a^{\frac {5}{2}} x \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (24 \, B b^{2} x^{5} + 30 \, A b^{2} x^{4} + 88 \, B a b x^{3} + 135 \, A a b x^{2} + 184 \, B a^{2} x - 120 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{240 \, x}, -\frac {225 \, A a^{2} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 60 \, B a^{\frac {5}{2}} x \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - {\left (24 \, B b^{2} x^{5} + 30 \, A b^{2} x^{4} + 88 \, B a b x^{3} + 135 \, A a b x^{2} + 184 \, B a^{2} x - 120 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{120 \, x}, \frac {240 \, B \sqrt {-a} a^{2} x \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + 225 \, A a^{2} \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (24 \, B b^{2} x^{5} + 30 \, A b^{2} x^{4} + 88 \, B a b x^{3} + 135 \, A a b x^{2} + 184 \, B a^{2} x - 120 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{240 \, x}, -\frac {225 \, A a^{2} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 120 \, B \sqrt {-a} a^{2} x \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (24 \, B b^{2} x^{5} + 30 \, A b^{2} x^{4} + 88 \, B a b x^{3} + 135 \, A a b x^{2} + 184 \, B a^{2} x - 120 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{120 \, x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/240*(225*A*a^2*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 120*B*a^(5/2)*x*log(-(b*x^2 - 2*
sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(24*B*b^2*x^5 + 30*A*b^2*x^4 + 88*B*a*b*x^3 + 135*A*a*b*x^2 + 184*B*a^
2*x - 120*A*a^2)*sqrt(b*x^2 + a))/x, -1/120*(225*A*a^2*sqrt(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 60*B*a^
(5/2)*x*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - (24*B*b^2*x^5 + 30*A*b^2*x^4 + 88*B*a*b*x^3 + 13
5*A*a*b*x^2 + 184*B*a^2*x - 120*A*a^2)*sqrt(b*x^2 + a))/x, 1/240*(240*B*sqrt(-a)*a^2*x*arctan(sqrt(-a)/sqrt(b*
x^2 + a)) + 225*A*a^2*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(24*B*b^2*x^5 + 30*A*b^2*x
^4 + 88*B*a*b*x^3 + 135*A*a*b*x^2 + 184*B*a^2*x - 120*A*a^2)*sqrt(b*x^2 + a))/x, -1/120*(225*A*a^2*sqrt(-b)*x*
arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 120*B*sqrt(-a)*a^2*x*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (24*B*b^2*x^5 + 3
0*A*b^2*x^4 + 88*B*a*b*x^3 + 135*A*a*b*x^2 + 184*B*a^2*x - 120*A*a^2)*sqrt(b*x^2 + a))/x]

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giac [A]  time = 0.50, size = 150, normalized size = 1.10 \begin {gather*} \frac {2 \, B a^{3} \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {15}{8} \, A a^{2} \sqrt {b} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right ) + \frac {2 \, A a^{3} \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} + \frac {1}{120} \, {\left (184 \, B a^{2} + {\left (135 \, A a b + 2 \, {\left (44 \, B a b + 3 \, {\left (4 \, B b^{2} x + 5 \, A b^{2}\right )} x\right )} x\right )} x\right )} \sqrt {b x^{2} + a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B*a^3*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) - 15/8*A*a^2*sqrt(b)*log(abs(-sqrt(b)*x + sqr
t(b*x^2 + a))) + 2*A*a^3*sqrt(b)/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a) + 1/120*(184*B*a^2 + (135*A*a*b + 2*(44
*B*a*b + 3*(4*B*b^2*x + 5*A*b^2)*x)*x)*x)*sqrt(b*x^2 + a)

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maple [A]  time = 0.01, size = 158, normalized size = 1.16 \begin {gather*} \frac {15 A \,a^{2} \sqrt {b}\, \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8}-B \,a^{\frac {5}{2}} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )+\frac {15 \sqrt {b \,x^{2}+a}\, A a b x}{8}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} A b x}{4}+\sqrt {b \,x^{2}+a}\, B \,a^{2}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A b x}{a}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B a}{3}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B}{5}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} A}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.40, size = 120, normalized size = 0.88 \begin {gather*} \frac {5}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b x + \frac {15}{8} \, \sqrt {b x^{2} + a} A a b x + \frac {15}{8} \, A a^{2} \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - B a^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{5} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a + \sqrt {b x^{2} + a} B a^{2} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.16, size = 104, normalized size = 0.76 \begin {gather*} \frac {B\,{\left (b\,x^2+a\right )}^{5/2}}{5}+B\,a^2\,\sqrt {b\,x^2+a}+\frac {B\,a\,{\left (b\,x^2+a\right )}^{3/2}}{3}-\frac {A\,{\left (b\,x^2+a\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b\,x^2}{a}\right )}{x\,{\left (\frac {b\,x^2}{a}+1\right )}^{5/2}}+B\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*(a + b*x^2)^(5/2))/5 + B*a^2*(a + b*x^2)^(1/2) + B*a^(5/2)*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*1i + (B*a*(
a + b*x^2)^(3/2))/3 - (A*(a + b*x^2)^(5/2)*hypergeom([-5/2, -1/2], 1/2, -(b*x^2)/a))/(x*((b*x^2)/a + 1)^(5/2))

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sympy [A]  time = 18.91, size = 318, normalized size = 2.34 \begin {gather*} - \frac {A a^{\frac {5}{2}}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + A a^{\frac {3}{2}} b x \sqrt {1 + \frac {b x^{2}}{a}} - \frac {7 A a^{\frac {3}{2}} b x}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 A \sqrt {a} b^{2} x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {15 A a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8} + \frac {A b^{3} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} - B a^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )} + \frac {B a^{3}}{\sqrt {b} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B a^{2} \sqrt {b} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + 2 B a b \left (\begin {cases} \frac {\sqrt {a} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\left (a + b x^{2}\right )^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases}\right ) + B b^{2} \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {a x^{2} \sqrt {a + b x^{2}}}{15 b} + \frac {x^{4} \sqrt {a + b x^{2}}}{5} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A*a**(5/2)/(x*sqrt(1 + b*x**2/a)) + A*a**(3/2)*b*x*sqrt(1 + b*x**2/a) - 7*A*a**(3/2)*b*x/(8*sqrt(1 + b*x**2/a
)) + 3*A*sqrt(a)*b**2*x**3/(8*sqrt(1 + b*x**2/a)) + 15*A*a**2*sqrt(b)*asinh(sqrt(b)*x/sqrt(a))/8 + A*b**3*x**5
/(4*sqrt(a)*sqrt(1 + b*x**2/a)) - B*a**(5/2)*asinh(sqrt(a)/(sqrt(b)*x)) + B*a**3/(sqrt(b)*x*sqrt(a/(b*x**2) +
1)) + B*a**2*sqrt(b)*x/sqrt(a/(b*x**2) + 1) + 2*B*a*b*Piecewise((sqrt(a)*x**2/2, Eq(b, 0)), ((a + b*x**2)**(3/
2)/(3*b), True)) + B*b**2*Piecewise((-2*a**2*sqrt(a + b*x**2)/(15*b**2) + a*x**2*sqrt(a + b*x**2)/(15*b) + x**
4*sqrt(a + b*x**2)/5, Ne(b, 0)), (sqrt(a)*x**4/4, True))

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