3.1.91 \(\int \frac {(c+d x^2)^2}{(a+b x^2)^{5/2}} \, dx\) [91]

Optimal. Leaf size=105 \[ \frac {(b c-a d) (2 b c+3 a d) x}{3 a^2 b^2 \sqrt {a+b x^2}}+\frac {(b c-a d) x \left (c+d x^2\right )}{3 a b \left (a+b x^2\right )^{3/2}}+\frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 105, 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 = {424, 393, 223, 212} \begin {gather*} \frac {x (b c-a d) (3 a d+2 b c)}{3 a^2 b^2 \sqrt {a+b x^2}}+\frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{5/2}}+\frac {x \left (c+d x^2\right ) (b c-a d)}{3 a b \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)*(2*b*c + 3*a*d)*x)/(3*a^2*b^2*Sqrt[a + b*x^2]) + ((b*c - a*d)*x*(c + d*x^2))/(3*a*b*(a + b*x^2)^(
3/2)) + (d^2*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/b^(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*} \int \frac {\left (c+d x^2\right )^2}{\left (a+b x^2\right )^{5/2}} \, dx &=\frac {(b c-a d) x \left (c+d x^2\right )}{3 a b \left (a+b x^2\right )^{3/2}}+\frac {\int \frac {c (2 b c+a d)+3 a d^2 x^2}{\left (a+b x^2\right )^{3/2}} \, dx}{3 a b}\\ &=\frac {(b c-a d) (2 b c+3 a d) x}{3 a^2 b^2 \sqrt {a+b x^2}}+\frac {(b c-a d) x \left (c+d x^2\right )}{3 a b \left (a+b x^2\right )^{3/2}}+\frac {d^2 \int \frac {1}{\sqrt {a+b x^2}} \, dx}{b^2}\\ &=\frac {(b c-a d) (2 b c+3 a d) x}{3 a^2 b^2 \sqrt {a+b x^2}}+\frac {(b c-a d) x \left (c+d x^2\right )}{3 a b \left (a+b x^2\right )^{3/2}}+\frac {d^2 \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{b^2}\\ &=\frac {(b c-a d) (2 b c+3 a d) x}{3 a^2 b^2 \sqrt {a+b x^2}}+\frac {(b c-a d) x \left (c+d x^2\right )}{3 a b \left (a+b x^2\right )^{3/2}}+\frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 103, normalized size = 0.98 \begin {gather*} \frac {-3 a^3 d^2 x+2 b^3 c^2 x^3-4 a^2 b d^2 x^3+a b^2 c x \left (3 c+2 d x^2\right )}{3 a^2 b^2 \left (a+b x^2\right )^{3/2}}-\frac {d^2 \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 156, normalized size = 1.49

method	result	size
default	\(d^{2} \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )+2 c d \left (-\frac {x}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {a \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )}{2 b}\right )+c^{2} \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )\)	\(156\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 147, normalized size = 1.40 \begin {gather*} -\frac {1}{3} \, d^{2} x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} + \frac {2 \, c^{2} x}{3 \, \sqrt {b x^{2} + a} a^{2}} + \frac {c^{2} x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} - \frac {2 \, c d x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, c d x}{3 \, \sqrt {b x^{2} + a} a b} - \frac {d^{2} x}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.49, size = 318, normalized size = 3.03 \begin {gather*} \left [\frac {3 \, {\left (a^{2} b^{2} d^{2} x^{4} + 2 \, a^{3} b d^{2} x^{2} + a^{4} d^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, {\left (b^{4} c^{2} + a b^{3} c d - 2 \, a^{2} b^{2} d^{2}\right )} x^{3} + 3 \, {\left (a b^{3} c^{2} - a^{3} b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}, -\frac {3 \, {\left (a^{2} b^{2} d^{2} x^{4} + 2 \, a^{3} b d^{2} x^{2} + a^{4} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, {\left (b^{4} c^{2} + a b^{3} c d - 2 \, a^{2} b^{2} d^{2}\right )} x^{3} + 3 \, {\left (a b^{3} c^{2} - a^{3} b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{3 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c + d x^{2}\right )^{2}}{\left (a + b x^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.58, size = 103, normalized size = 0.98 \begin {gather*} \frac {x {\left (\frac {2 \, {\left (b^{4} c^{2} + a b^{3} c d - 2 \, a^{2} b^{2} d^{2}\right )} x^{2}}{a^{2} b^{3}} + \frac {3 \, {\left (a b^{3} c^{2} - a^{3} b d^{2}\right )}}{a^{2} b^{3}}\right )}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} - \frac {d^{2} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d\,x^2+c\right )}^2}{{\left (b\,x^2+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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