\(\int \frac {(a+b x^2)^2}{(c+d x^2)^{3/2}} \, dx\) [652]

Optimal result

Integrand size = 21, antiderivative size = 106 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=-\frac {(b c-a d) x \left (a+b x^2\right )}{c d \sqrt {c+d x^2}}+\frac {b (3 b c-2 a d) x \sqrt {c+d x^2}}{2 c d^2}-\frac {b (3 b c-4 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {424, 396, 223, 212} \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=-\frac {b (3 b c-4 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{5/2}}+\frac {b x \sqrt {c+d x^2} (3 b c-2 a d)}{2 c d^2}-\frac {x \left (a+b x^2\right ) (b c-a d)}{c d \sqrt {c+d x^2}} \]

[In]

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

[Out]

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

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 396

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

Rule 424

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a*d - c*b)*x*(a + b*x^n)^(
p + 1)*((c + d*x^n)^(q - 1)/(a*b*n*(p + 1))), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {d} x \left (-4 a b c d+2 a^2 d^2+b^2 c \left (3 c+d x^2\right )\right )}{c \sqrt {c+d x^2}}+b (3 b c-4 a d) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{2 d^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.95 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.87

method	result	size
pseudoelliptic	\(\frac {2 \sqrt {d \,x^{2}+c}\, \left (a d -\frac {3 b c}{4}\right ) b c \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )+x \left (-2 \left (-\frac {b \,x^{2}}{4}+a \right ) b c \,d^{\frac {3}{2}}+\frac {3 b^{2} c^{2} \sqrt {d}}{2}+a^{2} d^{\frac {5}{2}}\right )}{\sqrt {d \,x^{2}+c}\, d^{\frac {5}{2}} c}\)	\(92\)
risch	\(\frac {b^{2} x \sqrt {d \,x^{2}+c}}{2 d^{2}}+\frac {\frac {2 a^{2} d^{2} x}{c \sqrt {d \,x^{2}+c}}-\frac {b^{2} c x}{\sqrt {d \,x^{2}+c}}+\left (4 a b \,d^{2}-3 b^{2} c d \right ) \left (-\frac {x}{d \sqrt {d \,x^{2}+c}}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}\right )}{2 d^{2}}\)	\(115\)
default	\(\frac {a^{2} x}{c \sqrt {d \,x^{2}+c}}+b^{2} \left (\frac {x^{3}}{2 d \sqrt {d \,x^{2}+c}}-\frac {3 c \left (-\frac {x}{d \sqrt {d \,x^{2}+c}}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}\right )}{2 d}\right )+2 a b \left (-\frac {x}{d \sqrt {d \,x^{2}+c}}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}\right )\)	\(123\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.59 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\left [-\frac {{\left (3 \, b^{2} c^{3} - 4 \, a b c^{2} d + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2}\right )} x^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left (b^{2} c d^{2} x^{3} + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2} + 2 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{4 \, {\left (c d^{4} x^{2} + c^{2} d^{3}\right )}}, \frac {{\left (3 \, b^{2} c^{3} - 4 \, a b c^{2} d + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2}\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (b^{2} c d^{2} x^{3} + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2} + 2 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{2 \, {\left (c d^{4} x^{2} + c^{2} d^{3}\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {b^{2} x^{3}}{2 \, \sqrt {d x^{2} + c} d} + \frac {a^{2} x}{\sqrt {d x^{2} + c} c} + \frac {3 \, b^{2} c x}{2 \, \sqrt {d x^{2} + c} d^{2}} - \frac {2 \, a b x}{\sqrt {d x^{2} + c} d} - \frac {3 \, b^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, d^{\frac {5}{2}}} + \frac {2 \, a b \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{d^{\frac {3}{2}}} \]

[In]

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

[Out]

1/2*b^2*x^3/(sqrt(d*x^2 + c)*d) + a^2*x/(sqrt(d*x^2 + c)*c) + 3/2*b^2*c*x/(sqrt(d*x^2 + c)*d^2) - 2*a*b*x/(sqr
t(d*x^2 + c)*d) - 3/2*b^2*c*arcsinh(d*x/sqrt(c*d))/d^(5/2) + 2*a*b*arcsinh(d*x/sqrt(c*d))/d^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {{\left (\frac {b^{2} x^{2}}{d} + \frac {3 \, b^{2} c^{2} d - 4 \, a b c d^{2} + 2 \, a^{2} d^{3}}{c d^{3}}\right )} x}{2 \, \sqrt {d x^{2} + c}} + \frac {{\left (3 \, b^{2} c - 4 \, a b d\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{2 \, d^{\frac {5}{2}}} \]

[In]

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

[Out]

1/2*(b^2*x^2/d + (3*b^2*c^2*d - 4*a*b*c*d^2 + 2*a^2*d^3)/(c*d^3))*x/sqrt(d*x^2 + c) + 1/2*(3*b^2*c - 4*a*b*d)*
log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(5/2)

Mupad [F(-1)]

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

[In]

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

[Out]

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