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

Optimal. Leaf size=132 \[ -a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {5 a^3 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}}+\frac {1}{16} a^2 \sqrt {a+b x^2} (16 A+5 B x)+\frac {1}{24} a \left (a+b x^2\right )^{3/2} (8 A+5 B x)+\frac {1}{30} \left (a+b x^2\right )^{5/2} (6 A+5 B x) \]

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Rubi [A]  time = 0.16, antiderivative size = 132, 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 = {815, 844, 217, 206, 266, 63, 208} \begin {gather*} \frac {1}{16} a^2 \sqrt {a+b x^2} (16 A+5 B x)-a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {5 a^3 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}}+\frac {1}{24} a \left (a+b x^2\right )^{3/2} (8 A+5 B x)+\frac {1}{30} \left (a+b x^2\right )^{5/2} (6 A+5 B x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(16*A + 5*B*x)*Sqrt[a + b*x^2])/16 + (a*(8*A + 5*B*x)*(a + b*x^2)^(3/2))/24 + ((6*A + 5*B*x)*(a + b*x^2)^
(5/2))/30 + (5*a^3*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*Sqrt[b]) - a^(5/2)*A*ArcTanh[Sqrt[a + b*x^2]/Sq
rt[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 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} \, dx &=\frac {1}{30} (6 A+5 B x) \left (a+b x^2\right )^{5/2}+\frac {\int \frac {(6 a A b+5 a b B x) \left (a+b x^2\right )^{3/2}}{x} \, dx}{6 b}\\ &=\frac {1}{24} a (8 A+5 B x) \left (a+b x^2\right )^{3/2}+\frac {1}{30} (6 A+5 B x) \left (a+b x^2\right )^{5/2}+\frac {\int \frac {\left (24 a^2 A b^2+15 a^2 b^2 B x\right ) \sqrt {a+b x^2}}{x} \, dx}{24 b^2}\\ &=\frac {1}{16} a^2 (16 A+5 B x) \sqrt {a+b x^2}+\frac {1}{24} a (8 A+5 B x) \left (a+b x^2\right )^{3/2}+\frac {1}{30} (6 A+5 B x) \left (a+b x^2\right )^{5/2}+\frac {\int \frac {48 a^3 A b^3+15 a^3 b^3 B x}{x \sqrt {a+b x^2}} \, dx}{48 b^3}\\ &=\frac {1}{16} a^2 (16 A+5 B x) \sqrt {a+b x^2}+\frac {1}{24} a (8 A+5 B x) \left (a+b x^2\right )^{3/2}+\frac {1}{30} (6 A+5 B x) \left (a+b x^2\right )^{5/2}+\left (a^3 A\right ) \int \frac {1}{x \sqrt {a+b x^2}} \, dx+\frac {1}{16} \left (5 a^3 B\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx\\ &=\frac {1}{16} a^2 (16 A+5 B x) \sqrt {a+b x^2}+\frac {1}{24} a (8 A+5 B x) \left (a+b x^2\right )^{3/2}+\frac {1}{30} (6 A+5 B x) \left (a+b x^2\right )^{5/2}+\frac {1}{2} \left (a^3 A\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )+\frac {1}{16} \left (5 a^3 B\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )\\ &=\frac {1}{16} a^2 (16 A+5 B x) \sqrt {a+b x^2}+\frac {1}{24} a (8 A+5 B x) \left (a+b x^2\right )^{3/2}+\frac {1}{30} (6 A+5 B x) \left (a+b x^2\right )^{5/2}+\frac {5 a^3 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}}+\frac {\left (a^3 A\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}{16} a^2 (16 A+5 B x) \sqrt {a+b x^2}+\frac {1}{24} a (8 A+5 B x) \left (a+b x^2\right )^{3/2}+\frac {1}{30} (6 A+5 B x) \left (a+b x^2\right )^{5/2}+\frac {5 a^3 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}}-a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 139, normalized size = 1.05 \begin {gather*} -a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {5 a^{7/2} B \sqrt {\frac {b x^2}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 \sqrt {b} \sqrt {a+b x^2}}+\frac {1}{240} \sqrt {a+b x^2} \left (a^2 (368 A+165 B x)+2 a b x^2 (88 A+65 B x)+8 b^2 x^4 (6 A+5 B x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(8*b^2*x^4*(6*A + 5*B*x) + 2*a*b*x^2*(88*A + 65*B*x) + a^2*(368*A + 165*B*x)))/240 + (5*a^(7/
2)*B*Sqrt[1 + (b*x^2)/a]*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/(16*Sqrt[b]*Sqrt[a + b*x^2]) - a^(5/2)*A*ArcTanh[Sqrt[a
+ b*x^2]/Sqrt[a]]

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IntegrateAlgebraic [A]  time = 0.46, size = 138, normalized size = 1.05 \begin {gather*} 2 a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}-\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {5 a^3 B \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{16 \sqrt {b}}+\frac {1}{240} \sqrt {a+b x^2} \left (368 a^2 A+165 a^2 B x+176 a A b x^2+130 a b B x^3+48 A b^2 x^4+40 b^2 B x^5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(368*a^2*A + 165*a^2*B*x + 176*a*A*b*x^2 + 130*a*b*B*x^3 + 48*A*b^2*x^4 + 40*b^2*B*x^5))/240
+ 2*a^(5/2)*A*ArcTanh[(Sqrt[b]*x)/Sqrt[a] - Sqrt[a + b*x^2]/Sqrt[a]] - (5*a^3*B*Log[-(Sqrt[b]*x) + Sqrt[a + b*
x^2]])/(16*Sqrt[b])

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fricas [A]  time = 0.93, size = 539, normalized size = 4.08 \begin {gather*} \left [\frac {75 \, B a^{3} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 240 \, A a^{\frac {5}{2}} b \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 130 \, B a b^{2} x^{3} + 176 \, A a b^{2} x^{2} + 165 \, B a^{2} b x + 368 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{480 \, b}, -\frac {75 \, B a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 120 \, A a^{\frac {5}{2}} b \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 130 \, B a b^{2} x^{3} + 176 \, A a b^{2} x^{2} + 165 \, B a^{2} b x + 368 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{240 \, b}, \frac {480 \, A \sqrt {-a} a^{2} b \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + 75 \, B a^{3} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 130 \, B a b^{2} x^{3} + 176 \, A a b^{2} x^{2} + 165 \, B a^{2} b x + 368 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{480 \, b}, -\frac {75 \, B a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 240 \, A \sqrt {-a} a^{2} b \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 130 \, B a b^{2} x^{3} + 176 \, A a b^{2} x^{2} + 165 \, B a^{2} b x + 368 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{240 \, b}\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,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.58, size = 125, normalized size = 0.95 \begin {gather*} \frac {2 \, A a^{3} \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {5 \, B a^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, \sqrt {b}} + \frac {1}{240} \, {\left (368 \, A a^{2} + {\left (165 \, B a^{2} + 2 \, {\left (88 \, A a b + {\left (65 \, B a b + 4 \, {\left (5 \, B b^{2} x + 6 \, A b^{2}\right )} x\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,x, algorithm="giac")

[Out]

2*A*a^3*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) - 5/16*B*a^3*log(abs(-sqrt(b)*x + sqrt(b*x^2
+ a)))/sqrt(b) + 1/240*(368*A*a^2 + (165*B*a^2 + 2*(88*A*a*b + (65*B*a*b + 4*(5*B*b^2*x + 6*A*b^2)*x)*x)*x)*x)
*sqrt(b*x^2 + a)

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maple [A]  time = 0.01, size = 138, normalized size = 1.05 \begin {gather*} -A \,a^{\frac {5}{2}} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )+\frac {5 B \,a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 \sqrt {b}}+\frac {5 \sqrt {b \,x^{2}+a}\, B \,a^{2} x}{16}+\sqrt {b \,x^{2}+a}\, A \,a^{2}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} B a x}{24}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A a}{3}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B x}{6}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*B*x*(b*x^2+a)^(5/2)+5/24*B*a*x*(b*x^2+a)^(3/2)+5/16*B*a^2*x*(b*x^2+a)^(1/2)+5/16*B*a^3/b^(1/2)*ln(b^(1/2)*
x+(b*x^2+a)^(1/2))+1/5*A*(b*x^2+a)^(5/2)+1/3*A*a*(b*x^2+a)^(3/2)-A*a^(5/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/
x)+A*(b*x^2+a)^(1/2)*a^2

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maxima [A]  time = 1.38, size = 119, normalized size = 0.90 \begin {gather*} \frac {1}{6} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B x + \frac {5}{24} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x + \frac {5}{16} \, \sqrt {b x^{2} + a} B a^{2} x + \frac {5 \, B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {b}} - A 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}} A + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a + \sqrt {b x^{2} + a} A a^{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,x, algorithm="maxima")

[Out]

1/6*(b*x^2 + a)^(5/2)*B*x + 5/24*(b*x^2 + a)^(3/2)*B*a*x + 5/16*sqrt(b*x^2 + a)*B*a^2*x + 5/16*B*a^3*arcsinh(b
*x/sqrt(a*b))/sqrt(b) - A*a^(5/2)*arcsinh(a/(sqrt(a*b)*abs(x))) + 1/5*(b*x^2 + a)^(5/2)*A + 1/3*(b*x^2 + a)^(3
/2)*A*a + sqrt(b*x^2 + a)*A*a^2

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mupad [B]  time = 1.25, size = 101, normalized size = 0.77 \begin {gather*} \frac {A\,{\left (b\,x^2+a\right )}^{5/2}}{5}+A\,a^2\,\sqrt {b\,x^2+a}+\frac {A\,a\,{\left (b\,x^2+a\right )}^{3/2}}{3}+\frac {B\,x\,{\left (b\,x^2+a\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{5/2}}+A\,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,x)

[Out]

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

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sympy [A]  time = 40.75, size = 323, normalized size = 2.45 \begin {gather*} - A a^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )} + \frac {A a^{3}}{\sqrt {b} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A a^{2} \sqrt {b} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + 2 A 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 ) + A 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 ) + \frac {B a^{\frac {5}{2}} x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {3 B a^{\frac {5}{2}} x}{16 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {35 B a^{\frac {3}{2}} b x^{3}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {17 B \sqrt {a} b^{2} x^{5}}{24 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 B a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 \sqrt {b}} + \frac {B b^{3} x^{7}}{6 \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,x)

[Out]

-A*a**(5/2)*asinh(sqrt(a)/(sqrt(b)*x)) + A*a**3/(sqrt(b)*x*sqrt(a/(b*x**2) + 1)) + A*a**2*sqrt(b)*x/sqrt(a/(b*
x**2) + 1) + 2*A*a*b*Piecewise((sqrt(a)*x**2/2, Eq(b, 0)), ((a + b*x**2)**(3/2)/(3*b), True)) + A*b**2*Piecewi
se((-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)) + B*a**(5/2)*x*sqrt(1 + b*x**2/a)/2 + 3*B*a**(5/2)*x/(16*sqrt(1 + b*x**2/a)) + 35*B*a*
*(3/2)*b*x**3/(48*sqrt(1 + b*x**2/a)) + 17*B*sqrt(a)*b**2*x**5/(24*sqrt(1 + b*x**2/a)) + 5*B*a**3*asinh(sqrt(b
)*x/sqrt(a))/(16*sqrt(b)) + B*b**3*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a))

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