\(\int \frac {x (a+b x^2)^{3/2}}{\sqrt {c+d x^2}} \, dx\) [946]

Optimal result

Integrand size = 24, antiderivative size = 125 \[ \int \frac {x \left (a+b x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx=-\frac {3 (b c-a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 d}+\frac {3 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 \sqrt {b} d^{5/2}} \]

[Out]

3/8*(-a*d+b*c)^2*arctanh(d^(1/2)*(b*x^2+a)^(1/2)/b^(1/2)/(d*x^2+c)^(1/2))/d^(5/2)/b^(1/2)+1/4*(b*x^2+a)^(3/2)*
(d*x^2+c)^(1/2)/d-3/8*(-a*d+b*c)*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/d^2

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {455, 52, 65, 223, 212} \[ \int \frac {x \left (a+b x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx=\frac {3 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 \sqrt {b} d^{5/2}}-\frac {3 \sqrt {a+b x^2} \sqrt {c+d x^2} (b c-a d)}{8 d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 d} \]

[In]

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

[Out]

(-3*(b*c - a*d)*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(8*d^2) + ((a + b*x^2)^(3/2)*Sqrt[c + d*x^2])/(4*d) + (3*(b*c
- a*d)^2*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])])/(8*Sqrt[b]*d^(5/2))

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 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 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = \frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 d}-\frac {(3 (b c-a d)) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx,x,x^2\right )}{8 d} \\ & = -\frac {3 (b c-a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 d}+\frac {\left (3 (b c-a d)^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{16 d^2} \\ & = -\frac {3 (b c-a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 d}+\frac {\left (3 (b c-a d)^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x^2}\right )}{8 b d^2} \\ & = -\frac {3 (b c-a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 d}+\frac {\left (3 (b c-a d)^2\right ) \text {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 )}{8 b d^2} \\ & = -\frac {3 (b c-a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 d}+\frac {3 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 \sqrt {b} d^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.83 \[ \int \frac {x \left (a+b x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx=\frac {\sqrt {a+b x^2} \sqrt {c+d x^2} \left (-3 b c+5 a d+2 b d x^2\right )}{8 d^2}+\frac {3 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {a+b x^2}}\right )}{8 \sqrt {b} d^{5/2}} \]

[In]

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

[Out]

(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(-3*b*c + 5*a*d + 2*b*d*x^2))/(8*d^2) + (3*(b*c - a*d)^2*ArcTanh[(Sqrt[b]*Sqr
t[c + d*x^2])/(Sqrt[d]*Sqrt[a + b*x^2])])/(8*Sqrt[b]*d^(5/2))

Maple [A] (verified)

Time = 3.12 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.22

method	result	size
risch	\(\frac {\left (2 b d \,x^{2}+5 a d -3 b c \right ) \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}{8 d^{2}}+\frac {3 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}{16 d^{2} \sqrt {b d}\, \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(153\)
default	\(\frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (4 \sqrt {b d}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, b d \,x^{2}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} d^{2}-6 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c d +3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2}+10 \sqrt {b d}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, a d -6 \sqrt {b d}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, b c \right )}{16 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, d^{2} \sqrt {b d}}\)	\(288\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \left (\frac {3 a^{2} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right )}{16 \sqrt {b d}}+\frac {b \,x^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{4 d}+\frac {5 \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, a}{8 d}-\frac {3 b \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, c}{8 d^{2}}-\frac {3 b \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a c}{8 d \sqrt {b d}}+\frac {3 b^{2} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) c^{2}}{16 d^{2} \sqrt {b d}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(306\)

[In]

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

[Out]

1/8*(2*b*d*x^2+5*a*d-3*b*c)*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/d^2+3/16*(a^2*d^2-2*a*b*c*d+b^2*c^2)/d^2*ln((1/2*a
*d+1/2*b*c+b*d*x^2)/(b*d)^(1/2)+(b*d*x^4+(a*d+b*c)*x^2+a*c)^(1/2))/(b*d)^(1/2)*((b*x^2+a)*(d*x^2+c))^(1/2)/(b*
x^2+a)^(1/2)/(d*x^2+c)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.67 \[ \int \frac {x \left (a+b x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx=\left [\frac {3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {b d}\right ) + 4 \, {\left (2 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c d + 5 \, a b d^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{32 \, b d^{3}}, -\frac {3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-b d}}{2 \, {\left (b^{2} d^{2} x^{4} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (2 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c d + 5 \, a b d^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{16 \, b d^{3}}\right ] \]

[In]

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

[Out]

[1/32*(3*(b^2*c^2 - 2*a*b*c*d + a^2*d^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*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b*d)) + 4*(2*b^2*d^2*x^2 -
3*b^2*c*d + 5*a*b*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b*d^3), -1/16*(3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqr
t(-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^2*x^4 + a*b*c*d +
(b^2*c*d + a*b*d^2)*x^2)) - 2*(2*b^2*d^2*x^2 - 3*b^2*c*d + 5*a*b*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b*d^3
)]

Sympy [F]

\[ \int \frac {x \left (a+b x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx=\int \frac {x \left (a + b x^{2}\right )^{\frac {3}{2}}}{\sqrt {c + d x^{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {x \left (a+b x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x*(b*x^2+a)^(3/2)/(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 detail

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.19 \[ \int \frac {x \left (a+b x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx=\frac {{\left (\sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \sqrt {b x^{2} + a} {\left (\frac {2 \, {\left (b x^{2} + a\right )}}{b d} - \frac {3 \, {\left (b c d - a d^{2}\right )}}{b d^{3}}\right )} - \frac {3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {b x^{2} + a} \sqrt {b d} + \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{2}}\right )} b}{8 \, {\left | b \right |}} \]

[In]

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

[Out]

1/8*(sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)*(2*(b*x^2 + a)/(b*d) - 3*(b*c*d - a*d^2)/(b*d^3)) -
3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(abs(-sqrt(b*x^2 + a)*sqrt(b*d) + sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))
)/(sqrt(b*d)*d^2))*b/abs(b)

Mupad [F(-1)]

Timed out. \[ \int \frac {x \left (a+b x^2\right )^{3/2}}{\sqrt {c+d x^2}} \, dx=\int \frac {x\,{\left (b\,x^2+a\right )}^{3/2}}{\sqrt {d\,x^2+c}} \,d x \]

[In]

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

[Out]

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