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

Optimal. Leaf size=106 \[ -\frac {b (3 b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{5/2}}+\frac {b x \sqrt {c+d x^2} (3 b c-2 a d)}{2 c d^2}-\frac {x \left (a+b x^2\right ) (b c-a d)}{c d \sqrt {c+d x^2}} \]

[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

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Rubi [A]  time = 0.05, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {413, 388, 217, 206} \[ \frac {b x \sqrt {c+d x^2} (3 b c-2 a d)}{2 c d^2}-\frac {b (3 b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{5/2}}-\frac {x \left (a+b x^2\right ) (b c-a d)}{c d \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[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 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 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 388

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 413

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*} \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 {\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)) \operatorname {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*}

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Mathematica [A]  time = 0.11, size = 93, normalized size = 0.88 \[ \sqrt {c+d x^2} \left (\frac {x (b c-a d)^2}{c d^2 \left (c+d x^2\right )}+\frac {b^2 x}{2 d^2}\right )-\frac {b (3 b c-4 a d) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{2 d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.95, size = 275, normalized size = 2.59 \[ \left [-\frac {{\left (3 \, b^{2} c^{3} - 4 \, a b c^{2} d + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2}\right )} x^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left (b^{2} c d^{2} x^{3} + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2} + 2 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{4 \, {\left (c d^{4} x^{2} + c^{2} d^{3}\right )}}, \frac {{\left (3 \, b^{2} c^{3} - 4 \, a b c^{2} d + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2}\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (b^{2} c d^{2} x^{3} + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2} + 2 \, a^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{2 \, {\left (c d^{4} x^{2} + c^{2} d^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 0.46, size = 92, normalized size = 0.87 \[ \frac {{\left (\frac {b^{2} x^{2}}{d} + \frac {3 \, b^{2} c^{2} d - 4 \, a b c d^{2} + 2 \, a^{2} d^{3}}{c d^{3}}\right )} x}{2 \, \sqrt {d x^{2} + c}} + \frac {{\left (3 \, b^{2} c - 4 \, a b d\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{2 \, d^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.01, size = 123, normalized size = 1.16 \[ \frac {b^{2} x^{3}}{2 \sqrt {d \,x^{2}+c}\, d}+\frac {a^{2} x}{\sqrt {d \,x^{2}+c}\, c}-\frac {2 a b x}{\sqrt {d \,x^{2}+c}\, d}+\frac {3 b^{2} c x}{2 \sqrt {d \,x^{2}+c}\, d^{2}}+\frac {2 a b \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}-\frac {3 b^{2} c \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.92, size = 108, normalized size = 1.02 \[ \frac {b^{2} x^{3}}{2 \, \sqrt {d x^{2} + c} d} + \frac {a^{2} x}{\sqrt {d x^{2} + c} c} + \frac {3 \, b^{2} c x}{2 \, \sqrt {d x^{2} + c} d^{2}} - \frac {2 \, a b x}{\sqrt {d x^{2} + c} d} - \frac {3 \, b^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, d^{\frac {5}{2}}} + \frac {2 \, a b \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{d^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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