3.8.77 \(\int \frac {(a+b x^2)^{3/2}}{x^3 \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=136 \[ \frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{\sqrt {d}}-\frac {\sqrt {a} (3 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 c^{3/2}}-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 c x^2} \]

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Rubi [A]  time = 0.14, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {446, 98, 157, 63, 217, 206, 93, 208} \begin {gather*} \frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{\sqrt {d}}-\frac {\sqrt {a} (3 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 c^{3/2}}-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/(x^3*Sqrt[c + d*x^2]),x]

[Out]

-(a*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(2*c*x^2) - (Sqrt[a]*(3*b*c - a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqr
t[a]*Sqrt[c + d*x^2])])/(2*c^(3/2)) + (b^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])])/S
qrt[d]

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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, 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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^{3/2}}{x^3 \sqrt {c+d x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^2 \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 c x^2}-\frac {\operatorname {Subst}\left (\int \frac {-\frac {1}{2} a (3 b c-a d)-b^2 c x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{2 c}\\ &=-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 c x^2}+\frac {1}{2} b^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )+\frac {(a (3 b c-a d)) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{4 c}\\ &=-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 c x^2}+b \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x^2}\right )+\frac {(a (3 b c-a d)) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x^2}}{\sqrt {c+d x^2}}\right )}{2 c}\\ &=-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 c x^2}-\frac {\sqrt {a} (3 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 c^{3/2}}+b \operatorname {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x^2}}{\sqrt {c+d x^2}}\right )\\ &=-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 c x^2}-\frac {\sqrt {a} (3 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 c^{3/2}}+\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{\sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 1.01, size = 172, normalized size = 1.26 \begin {gather*} \frac {\sqrt {a} (a d-3 b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 c^{3/2}}-\frac {a \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 c x^2}+\frac {(b c-a d)^{3/2} \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{3/2} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )}{\sqrt {d} \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/(x^3*Sqrt[c + d*x^2]),x]

[Out]

-1/2*(a*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(c*x^2) + ((b*c - a*d)^(3/2)*((b*(c + d*x^2))/(b*c - a*d))^(3/2)*ArcS
inh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]])/(Sqrt[d]*(c + d*x^2)^(3/2)) + (Sqrt[a]*(-3*b*c + a*d)*ArcTanh[
(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*c^(3/2))

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IntegrateAlgebraic [A]  time = 1.66, size = 167, normalized size = 1.23 \begin {gather*} \frac {\left (a^{3/2} d-3 \sqrt {a} b c\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x^2}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{2 c^{3/2}}+\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {a+b x^2}}\right )}{\sqrt {d}}+\frac {a \sqrt {c+d x^2} (a d-b c)}{2 c \sqrt {a+b x^2} \left (c-\frac {a \left (c+d x^2\right )}{a+b x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x^2)^(3/2)/(x^3*Sqrt[c + d*x^2]),x]

[Out]

(a*(-(b*c) + a*d)*Sqrt[c + d*x^2])/(2*c*Sqrt[a + b*x^2]*(c - (a*(c + d*x^2))/(a + b*x^2))) + ((-3*Sqrt[a]*b*c
+ a^(3/2)*d)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x^2])/(Sqrt[c]*Sqrt[a + b*x^2])])/(2*c^(3/2)) + (b^(3/2)*ArcTanh[(Sqr
t[b]*Sqrt[c + d*x^2])/(Sqrt[d]*Sqrt[a + b*x^2])])/Sqrt[d]

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fricas [B]  time = 3.08, size = 958, normalized size = 7.04 \begin {gather*} \left [\frac {2 \, b c x^{2} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b d^{2} x^{2} + b c d + a d^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {\frac {b}{d}}\right ) - {\left (3 \, b c - a d\right )} x^{2} \sqrt {\frac {a}{c}} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2} + 4 \, {\left (2 \, a c^{2} + {\left (b c^{2} + a c d\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {\frac {a}{c}}}{x^{4}}\right ) - 4 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} a}{8 \, c x^{2}}, -\frac {4 \, b c x^{2} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{4} + a b c + {\left (b^{2} c + a b d\right )} x^{2}\right )}}\right ) + {\left (3 \, b c - a d\right )} x^{2} \sqrt {\frac {a}{c}} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2} + 4 \, {\left (2 \, a c^{2} + {\left (b c^{2} + a c d\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {\frac {a}{c}}}{x^{4}}\right ) + 4 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} a}{8 \, c x^{2}}, \frac {b c x^{2} \sqrt {\frac {b}{d}} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b d^{2} x^{2} + b c d + a d^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {\frac {b}{d}}\right ) + {\left (3 \, b c - a d\right )} x^{2} \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b d x^{4} + a^{2} c + {\left (a b c + a^{2} d\right )} x^{2}\right )}}\right ) - 2 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} a}{4 \, c x^{2}}, -\frac {2 \, b c x^{2} \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} d x^{4} + a b c + {\left (b^{2} c + a b d\right )} x^{2}\right )}}\right ) - {\left (3 \, b c - a d\right )} x^{2} \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b d x^{4} + a^{2} c + {\left (a b c + a^{2} d\right )} x^{2}\right )}}\right ) + 2 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} a}{4 \, c x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/x^3/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(2*b*c*x^2*sqrt(b/d)*log(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 + 4*(2
*b*d^2*x^2 + b*c*d + a*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b/d)) - (3*b*c - a*d)*x^2*sqrt(a/c)*log(((b^2
*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2 + 8*(a*b*c^2 + a^2*c*d)*x^2 + 4*(2*a*c^2 + (b*c^2 + a*c*d)*x^2)*sq
rt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(a/c))/x^4) - 4*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*a)/(c*x^2), -1/8*(4*b*c*x^2*
sqrt(-b/d)*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b/d)/(b^2*d*x^4 + a*b*c +
(b^2*c + a*b*d)*x^2)) + (3*b*c - a*d)*x^2*sqrt(a/c)*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2 + 8*(
a*b*c^2 + a^2*c*d)*x^2 + 4*(2*a*c^2 + (b*c^2 + a*c*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(a/c))/x^4) + 4
*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*a)/(c*x^2), 1/4*(b*c*x^2*sqrt(b/d)*log(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d +
a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 + 4*(2*b*d^2*x^2 + b*c*d + a*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b/d
)) + (3*b*c - a*d)*x^2*sqrt(-a/c)*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-a
/c)/(a*b*d*x^4 + a^2*c + (a*b*c + a^2*d)*x^2)) - 2*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*a)/(c*x^2), -1/4*(2*b*c*x^2
*sqrt(-b/d)*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b/d)/(b^2*d*x^4 + a*b*c +
(b^2*c + a*b*d)*x^2)) - (3*b*c - a*d)*x^2*sqrt(-a/c)*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqr
t(d*x^2 + c)*sqrt(-a/c)/(a*b*d*x^4 + a^2*c + (a*b*c + a^2*d)*x^2)) + 2*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*a)/(c*x
^2)]

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giac [B]  time = 0.75, size = 498, normalized size = 3.66 \begin {gather*} -\frac {{\left (\frac {\sqrt {b d} b \log \left ({\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}\right )}{d} + \frac {{\left (3 \, \sqrt {b d} a b^{2} c - \sqrt {b d} a^{2} b d\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} b c} + \frac {2 \, {\left (\sqrt {b d} a b^{4} c^{2} - 2 \, \sqrt {b d} a^{2} b^{3} c d + \sqrt {b d} a^{3} b^{2} d^{2} - \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a b^{2} c - \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a^{2} b d\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4}\right )} c}\right )} b}{2 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/x^3/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

-1/2*(sqrt(b*d)*b*log((sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2)/d + (3*sqrt(b*d)*
a*b^2*c - sqrt(b*d)*a^2*b*d)*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a
)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*b*c) + 2*(sqrt(b*d)*a*b^4*c^2 - 2*sqrt(b*d)*a^2*b^3*c*d
+ sqrt(b*d)*a^3*b^2*d^2 - sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a*b
^2*c - sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a^2*b*d)/((b^4*c^2 - 2*
a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*b^2*c - 2*(s
qrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(
b^2*c + (b*x^2 + a)*b*d - a*b*d))^4)*c))*b/abs(b)

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maple [B]  time = 0.02, size = 298, normalized size = 2.19 \begin {gather*} \frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (\sqrt {b d}\, a^{2} d \,x^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )-3 \sqrt {b d}\, a b c \,x^{2} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+2 \sqrt {a c}\, b^{2} c \,x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, \sqrt {a c}\, a \right )}{4 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\, \sqrt {a c}\, c \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)/x^3/(d*x^2+c)^(1/2),x)

[Out]

1/4*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/c*(2*ln(1/2*(2*b*d*x^2+a*d+b*c+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)
^(1/2))/(b*d)^(1/2))*x^2*b^2*c*(a*c)^(1/2)+ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*
c)^(1/2))/x^2)*x^2*a^2*d*(b*d)^(1/2)-3*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(
1/2))/x^2)*x^2*a*b*c*(b*d)^(1/2)-2*a*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2))/(b*d*x^4+a*d
*x^2+b*c*x^2+a*c)^(1/2)/x^2/(b*d)^(1/2)/(a*c)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/x^3/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^{3/2}}{x^3\,\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/2)/(x^3*(c + d*x^2)^(1/2)),x)

[Out]

int((a + b*x^2)^(3/2)/(x^3*(c + d*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x^{2}\right )^{\frac {3}{2}}}{x^{3} \sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)/x**3/(d*x**2+c)**(1/2),x)

[Out]

Integral((a + b*x**2)**(3/2)/(x**3*sqrt(c + d*x**2)), x)

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