\(\int \frac {(a+b x^2)^2}{(c+d x^2)^{5/2}} \, dx\) [664]

Optimal result

Integrand size = 21, antiderivative size = 105 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=-\frac {(b c-a d) x \left (a+b x^2\right )}{3 c d \left (c+d x^2\right )^{3/2}}-\frac {(b c-a d) (3 b c+2 a d) x}{3 c^2 d^2 \sqrt {c+d x^2}}+\frac {b^2 \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{d^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 105, 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, 393, 223, 212} \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=-\frac {x (b c-a d) (2 a d+3 b c)}{3 c^2 d^2 \sqrt {c+d x^2}}-\frac {x \left (a+b x^2\right ) (b c-a d)}{3 c d \left (c+d x^2\right )^{3/2}}+\frac {b^2 \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{d^{5/2}} \]

[In]

Int[(a + b*x^2)^2/(c + d*x^2)^(5/2),x]

[Out]

-1/3*((b*c - a*d)*x*(a + b*x^2))/(c*d*(c + d*x^2)^(3/2)) - ((b*c - a*d)*(3*b*c + 2*a*d)*x)/(3*c^2*d^2*Sqrt[c +
d*x^2]) + (b^2*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^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 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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 )}{3 c d \left (c+d x^2\right )^{3/2}}+\frac {\int \frac {a (b c+2 a d)+3 b^2 c x^2}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c d} \\ & = -\frac {(b c-a d) x \left (a+b x^2\right )}{3 c d \left (c+d x^2\right )^{3/2}}-\frac {(b c-a d) (3 b c+2 a d) x}{3 c^2 d^2 \sqrt {c+d x^2}}+\frac {b^2 \int \frac {1}{\sqrt {c+d x^2}} \, dx}{d^2} \\ & = -\frac {(b c-a d) x \left (a+b x^2\right )}{3 c d \left (c+d x^2\right )^{3/2}}-\frac {(b c-a d) (3 b c+2 a d) x}{3 c^2 d^2 \sqrt {c+d x^2}}+\frac {b^2 \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{d^2} \\ & = -\frac {(b c-a d) x \left (a+b x^2\right )}{3 c d \left (c+d x^2\right )^{3/2}}-\frac {(b c-a d) (3 b c+2 a d) x}{3 c^2 d^2 \sqrt {c+d x^2}}+\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{d^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=-\frac {(b c-a d) x \left (3 b c^2+3 a c d+4 b c d x^2+2 a d^2 x^2\right )}{3 c^2 d^2 \left (c+d x^2\right )^{3/2}}-\frac {b^2 \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{d^{5/2}} \]

[In]

Integrate[(a + b*x^2)^2/(c + d*x^2)^(5/2),x]

[Out]

-1/3*((b*c - a*d)*x*(3*b*c^2 + 3*a*c*d + 4*b*c*d*x^2 + 2*a*d^2*x^2))/(c^2*d^2*(c + d*x^2)^(3/2)) - (b^2*Log[-(
Sqrt[d]*x) + Sqrt[c + d*x^2]])/d^(5/2)

Maple [A] (verified)

Time = 2.93 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.99

method	result	size
pseudoelliptic	\(\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right ) b^{2} c^{2}+x \left (a c \left (\frac {2 b \,x^{2}}{3}+a \right ) d^{\frac {5}{2}}-\frac {4 d^{\frac {3}{2}} b^{2} c^{2} x^{2}}{3}-b^{2} c^{3} \sqrt {d}+\frac {2 d^{\frac {7}{2}} a^{2} x^{2}}{3}\right )}{\left (d \,x^{2}+c \right )^{\frac {3}{2}} d^{\frac {5}{2}} c^{2}}\)	\(104\)
default	\(a^{2} \left (\frac {x}{3 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {d \,x^{2}+c}}\right )+b^{2} \left (-\frac {x^{3}}{3 d \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {-\frac {x}{d \sqrt {d \,x^{2}+c}}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}}{d}\right )+2 a b \left (-\frac {x}{2 d \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {c \left (\frac {x}{3 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {d \,x^{2}+c}}\right )}{2 d}\right )\)	\(156\)

[In]

int((b*x^2+a)^2/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/(d*x^2+c)^(3/2)*((d*x^2+c)^(3/2)*arctanh((d*x^2+c)^(1/2)/x/d^(1/2))*b^2*c^2+x*(a*c*(2/3*b*x^2+a)*d^(5/2)-4/3
*d^(3/2)*b^2*c^2*x^2-b^2*c^3*d^(1/2)+2/3*d^(7/2)*a^2*x^2))/d^(5/2)/c^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 321, normalized size of antiderivative = 3.06 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{2} c^{2} d^{2} x^{4} + 2 \, b^{2} c^{3} d x^{2} + b^{2} c^{4}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left (2 \, {\left (2 \, b^{2} c^{2} d^{2} - a b c d^{3} - a^{2} d^{4}\right )} x^{3} + 3 \, {\left (b^{2} c^{3} d - a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{6 \, {\left (c^{2} d^{5} x^{4} + 2 \, c^{3} d^{4} x^{2} + c^{4} d^{3}\right )}}, -\frac {3 \, {\left (b^{2} c^{2} d^{2} x^{4} + 2 \, b^{2} c^{3} d x^{2} + b^{2} c^{4}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (2 \, {\left (2 \, b^{2} c^{2} d^{2} - a b c d^{3} - a^{2} d^{4}\right )} x^{3} + 3 \, {\left (b^{2} c^{3} d - a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{3 \, {\left (c^{2} d^{5} x^{4} + 2 \, c^{3} d^{4} x^{2} + c^{4} d^{3}\right )}}\right ] \]

[In]

integrate((b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(b^2*c^2*d^2*x^4 + 2*b^2*c^3*d*x^2 + b^2*c^4)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c)
- 2*(2*(2*b^2*c^2*d^2 - a*b*c*d^3 - a^2*d^4)*x^3 + 3*(b^2*c^3*d - a^2*c*d^3)*x)*sqrt(d*x^2 + c))/(c^2*d^5*x^4
+ 2*c^3*d^4*x^2 + c^4*d^3), -1/3*(3*(b^2*c^2*d^2*x^4 + 2*b^2*c^3*d*x^2 + b^2*c^4)*sqrt(-d)*arctan(sqrt(-d)*x/s
qrt(d*x^2 + c)) + (2*(2*b^2*c^2*d^2 - a*b*c*d^3 - a^2*d^4)*x^3 + 3*(b^2*c^3*d - a^2*c*d^3)*x)*sqrt(d*x^2 + c))
/(c^2*d^5*x^4 + 2*c^3*d^4*x^2 + c^4*d^3)]

Sympy [F]

\[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\int \frac {\left (a + b x^{2}\right )^{2}}{\left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((b*x**2+a)**2/(d*x**2+c)**(5/2),x)

[Out]

Integral((a + b*x**2)**2/(c + d*x**2)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.40 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=-\frac {1}{3} \, b^{2} x {\left (\frac {3 \, x^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d} + \frac {2 \, c}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}}\right )} + \frac {2 \, a^{2} x}{3 \, \sqrt {d x^{2} + c} c^{2}} + \frac {a^{2} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c} - \frac {b^{2} x}{3 \, \sqrt {d x^{2} + c} d^{2}} - \frac {2 \, a b x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d} + \frac {2 \, a b x}{3 \, \sqrt {d x^{2} + c} c d} + \frac {b^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{d^{\frac {5}{2}}} \]

[In]

integrate((b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=-\frac {x {\left (\frac {2 \, {\left (2 \, b^{2} c^{2} d^{2} - a b c d^{3} - a^{2} d^{4}\right )} x^{2}}{c^{2} d^{3}} + \frac {3 \, {\left (b^{2} c^{3} d - a^{2} c d^{3}\right )}}{c^{2} d^{3}}\right )}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} - \frac {b^{2} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{d^{\frac {5}{2}}} \]

[In]

integrate((b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^2}{{\left (d\,x^2+c\right )}^{5/2}} \,d x \]

[In]

int((a + b*x^2)^2/(c + d*x^2)^(5/2),x)

[Out]

int((a + b*x^2)^2/(c + d*x^2)^(5/2), x)