3.1.21 \(\int \frac {(A+B x) (a+b x^2)^{5/2}}{x^3} \, dx\) [21]

Optimal. Leaf size=141 \[ \frac {5}{8} a b (4 A+3 B x) \sqrt {a+b x^2}-\frac {5 (3 a B-2 A b x) \left (a+b x^2\right )^{3/2}}{12 x}-\frac {(2 A-B x) \left (a+b x^2\right )^{5/2}}{4 x^2}+\frac {15}{8} a^2 \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {5}{2} a^{3/2} A b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

[Out]

-5/12*(-2*A*b*x+3*B*a)*(b*x^2+a)^(3/2)/x-1/4*(-B*x+2*A)*(b*x^2+a)^(5/2)/x^2-5/2*a^(3/2)*A*b*arctanh((b*x^2+a)^
(1/2)/a^(1/2))+15/8*a^2*B*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))*b^(1/2)+5/8*a*b*(3*B*x+4*A)*(b*x^2+a)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 141, 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 = {827, 829, 858, 223, 212, 272, 65, 214} \begin {gather*} -\frac {5}{2} a^{3/2} A b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {15}{8} a^2 \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\left (a+b x^2\right )^{5/2} (2 A-B x)}{4 x^2}-\frac {5 \left (a+b x^2\right )^{3/2} (3 a B-2 A b x)}{12 x}+\frac {5}{8} a b \sqrt {a+b x^2} (4 A+3 B x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x^2)^(5/2))/x^3,x]

[Out]

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

Rule 65

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 223

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 272

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 827

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 829

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 858

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^3} \, dx &=-\frac {(2 A-B x) \left (a+b x^2\right )^{5/2}}{4 x^2}-\frac {5}{16} \int \frac {(-4 a B-8 A b x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx\\ &=-\frac {5 (3 a B-2 A b x) \left (a+b x^2\right )^{3/2}}{12 x}-\frac {(2 A-B x) \left (a+b x^2\right )^{5/2}}{4 x^2}+\frac {5}{32} \int \frac {(16 a A b+24 a b B x) \sqrt {a+b x^2}}{x} \, dx\\ &=\frac {5}{8} a b (4 A+3 B x) \sqrt {a+b x^2}-\frac {5 (3 a B-2 A b x) \left (a+b x^2\right )^{3/2}}{12 x}-\frac {(2 A-B x) \left (a+b x^2\right )^{5/2}}{4 x^2}+\frac {5 \int \frac {32 a^2 A b^2+24 a^2 b^2 B x}{x \sqrt {a+b x^2}} \, dx}{64 b}\\ &=\frac {5}{8} a b (4 A+3 B x) \sqrt {a+b x^2}-\frac {5 (3 a B-2 A b x) \left (a+b x^2\right )^{3/2}}{12 x}-\frac {(2 A-B x) \left (a+b x^2\right )^{5/2}}{4 x^2}+\frac {1}{2} \left (5 a^2 A b\right ) \int \frac {1}{x \sqrt {a+b x^2}} \, dx+\frac {1}{8} \left (15 a^2 b B\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx\\ &=\frac {5}{8} a b (4 A+3 B x) \sqrt {a+b x^2}-\frac {5 (3 a B-2 A b x) \left (a+b x^2\right )^{3/2}}{12 x}-\frac {(2 A-B x) \left (a+b x^2\right )^{5/2}}{4 x^2}+\frac {1}{4} \left (5 a^2 A b\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )+\frac {1}{8} \left (15 a^2 b B\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )\\ &=\frac {5}{8} a b (4 A+3 B x) \sqrt {a+b x^2}-\frac {5 (3 a B-2 A b x) \left (a+b x^2\right )^{3/2}}{12 x}-\frac {(2 A-B x) \left (a+b x^2\right )^{5/2}}{4 x^2}+\frac {15}{8} a^2 \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )+\frac {1}{2} \left (5 a^2 A\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )\\ &=\frac {5}{8} a b (4 A+3 B x) \sqrt {a+b x^2}-\frac {5 (3 a B-2 A b x) \left (a+b x^2\right )^{3/2}}{12 x}-\frac {(2 A-B x) \left (a+b x^2\right )^{5/2}}{4 x^2}+\frac {15}{8} a^2 \sqrt {b} B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {5}{2} a^{3/2} A b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.50, size = 132, normalized size = 0.94 \begin {gather*} 5 a^{3/2} A b \tanh ^{-1}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {1}{24} \left (\frac {\sqrt {a+b x^2} \left (-12 a^2 (A+2 B x)+2 b^2 x^4 (4 A+3 B x)+a b x^2 (56 A+27 B x)\right )}{x^2}-45 a^2 \sqrt {b} B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x^2)^(5/2))/x^3,x]

[Out]

5*a^(3/2)*A*b*ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2])/Sqrt[a]] + ((Sqrt[a + b*x^2]*(-12*a^2*(A + 2*B*x) + 2*b^2*
x^4*(4*A + 3*B*x) + a*b*x^2*(56*A + 27*B*x)))/x^2 - 45*a^2*Sqrt[b]*B*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/24

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Maple [A]
time = 0.13, size = 187, normalized size = 1.33

method	result	size
risch	\(-\frac {a^{2} \sqrt {b \,x^{2}+a}\, \left (2 B x +A \right )}{2 x^{2}}+\frac {b^{2} B \,x^{3} \sqrt {b \,x^{2}+a}}{4}+\frac {9 b B a x \sqrt {b \,x^{2}+a}}{8}+\frac {15 \sqrt {b}\, a^{2} B \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{8}+\frac {b^{2} A \,x^{2} \sqrt {b \,x^{2}+a}}{3}+\frac {7 b A a \sqrt {b \,x^{2}+a}}{3}-\frac {5 b \,a^{\frac {3}{2}} A \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2}\)	\(145\)
default	\(A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}}+\frac {5 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )}{2 a}\right )+B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{a x}+\frac {6 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{a}\right )\)	\(187\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b*x^2+a)^(5/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 143, normalized size = 1.01 \begin {gather*} \frac {5}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b x + \frac {15}{8} \, \sqrt {b x^{2} + a} B a b x + \frac {15}{8} \, B a^{2} \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {5}{2} \, A a^{\frac {3}{2}} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {5}{6} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A b}{2 \, a} + \frac {5}{2} \, \sqrt {b x^{2} + a} A a b - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B}{x} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A}{2 \, a x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x^2+a)^(5/2)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 4.25, size = 535, normalized size = 3.79 \begin {gather*} \left [\frac {45 \, B a^{2} \sqrt {b} x^{2} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 60 \, A a^{\frac {3}{2}} b x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (6 \, B b^{2} x^{5} + 8 \, A b^{2} x^{4} + 27 \, B a b x^{3} + 56 \, A a b x^{2} - 24 \, B a^{2} x - 12 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{48 \, x^{2}}, -\frac {45 \, B a^{2} \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 30 \, A a^{\frac {3}{2}} b x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - {\left (6 \, B b^{2} x^{5} + 8 \, A b^{2} x^{4} + 27 \, B a b x^{3} + 56 \, A a b x^{2} - 24 \, B a^{2} x - 12 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{24 \, x^{2}}, \frac {120 \, A \sqrt {-a} a b x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + 45 \, B a^{2} \sqrt {b} x^{2} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (6 \, B b^{2} x^{5} + 8 \, A b^{2} x^{4} + 27 \, B a b x^{3} + 56 \, A a b x^{2} - 24 \, B a^{2} x - 12 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{48 \, x^{2}}, -\frac {45 \, B a^{2} \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 60 \, A \sqrt {-a} a b x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (6 \, B b^{2} x^{5} + 8 \, A b^{2} x^{4} + 27 \, B a b x^{3} + 56 \, A a b x^{2} - 24 \, B a^{2} x - 12 \, A a^{2}\right )} \sqrt {b x^{2} + a}}{24 \, x^{2}}\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^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 6.21, size = 279, normalized size = 1.98 \begin {gather*} - \frac {5 A a^{\frac {3}{2}} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2} - \frac {A a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} + \frac {2 A a^{2} \sqrt {b}}{x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {2 A a b^{\frac {3}{2}} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + A b^{2} \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 ) - \frac {B a^{\frac {5}{2}}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + B a^{\frac {3}{2}} b x \sqrt {1 + \frac {b x^{2}}{a}} - \frac {7 B a^{\frac {3}{2}} b x}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 B \sqrt {a} b^{2} x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {15 B a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8} + \frac {B b^{3} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {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**3,x)

[Out]

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

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Giac [A]
time = 0.76, size = 219, normalized size = 1.55 \begin {gather*} \frac {5 \, A a^{2} b \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {15}{8} \, B a^{2} \sqrt {b} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right ) + \frac {1}{24} \, {\left (56 \, A a b + {\left (27 \, B a b + 2 \, {\left (3 \, B b^{2} x + 4 \, A b^{2}\right )} x\right )} x\right )} \sqrt {b x^{2} + a} + \frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} A a^{2} b + 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{3} \sqrt {b} + {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} A a^{3} b - 2 \, B a^{4} \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x^2+a)^(5/2)/x^3,x, algorithm="giac")

[Out]

5*A*a^2*b*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) - 15/8*B*a^2*sqrt(b)*log(abs(-sqrt(b)*x + s
qrt(b*x^2 + a))) + 1/24*(56*A*a*b + (27*B*a*b + 2*(3*B*b^2*x + 4*A*b^2)*x)*x)*sqrt(b*x^2 + a) + ((sqrt(b)*x -
sqrt(b*x^2 + a))^3*A*a^2*b + 2*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a^3*sqrt(b) + (sqrt(b)*x - sqrt(b*x^2 + a))*A
*a^3*b - 2*B*a^4*sqrt(b))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^2

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Mupad [B]
time = 2.59, size = 111, normalized size = 0.79 \begin {gather*} \frac {A\,b\,{\left (b\,x^2+a\right )}^{3/2}}{3}+2\,A\,a\,b\,\sqrt {b\,x^2+a}-\frac {A\,a^2\,\sqrt {b\,x^2+a}}{2\,x^2}-\frac {B\,{\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}}+\frac {A\,a^{3/2}\,b\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^(5/2)*(A + B*x))/x^3,x)

[Out]

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