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

Optimal. Leaf size=175 \[ -\frac{b^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 d^3}+\frac{(b c-a d)^{3/2} (a d+4 b c) \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{2 c^{3/2} d^3}+\frac{b x \sqrt{a+b x^2} (2 b c-a d)}{2 c d^2}-\frac{x \left (a+b x^2\right )^{3/2} (b c-a d)}{2 c d \left (c+d x^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.225999, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {413, 528, 523, 217, 206, 377, 208} \[ -\frac{b^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 d^3}+\frac{(b c-a d)^{3/2} (a d+4 b c) \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{2 c^{3/2} d^3}+\frac{b x \sqrt{a+b x^2} (2 b c-a d)}{2 c d^2}-\frac{x \left (a+b x^2\right )^{3/2} (b c-a d)}{2 c d \left (c+d x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 528

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

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

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 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 377

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.160223, size = 144, normalized size = 0.82 \[ \frac{d x \sqrt{a+b x^2} \left (\frac{(b c-a d)^2}{c \left (c+d x^2\right )}+b^2\right )+b^{3/2} (-(4 b c-5 a d)) \log \left (\sqrt{b} \sqrt{a+b x^2}+b x\right )+\frac{(a d-b c)^{3/2} (a d+4 b c) \tan ^{-1}\left (\frac{x \sqrt{a d-b c}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{c^{3/2}}}{2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.02, size = 7345, normalized size = 42. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 6.33458, size = 2584, normalized size = 14.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(2*(4*b^2*c^3 - 5*a*b*c^2*d + (4*b^2*c^2*d - 5*a*b*c*d^2)*x^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*
sqrt(b)*x - a) + (4*b^2*c^3 - 3*a*b*c^2*d - a^2*c*d^2 + (4*b^2*c^2*d - 3*a*b*c*d^2 - a^2*d^3)*x^2)*sqrt((b*c -
a*d)/c)*log(((8*b^2*c^2 - 8*a*b*c*d + a^2*d^2)*x^4 + a^2*c^2 + 2*(4*a*b*c^2 - 3*a^2*c*d)*x^2 - 4*(a*c^2*x + (
2*b*c^2 - a*c*d)*x^3)*sqrt(b*x^2 + a)*sqrt((b*c - a*d)/c))/(d^2*x^4 + 2*c*d*x^2 + c^2)) - 4*(b^2*c*d^2*x^3 + (
2*b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x)*sqrt(b*x^2 + a))/(c*d^4*x^2 + c^2*d^3), 1/8*(4*(4*b^2*c^3 - 5*a*b*c^2*
d + (4*b^2*c^2*d - 5*a*b*c*d^2)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (4*b^2*c^3 - 3*a*b*c^2*d -
a^2*c*d^2 + (4*b^2*c^2*d - 3*a*b*c*d^2 - a^2*d^3)*x^2)*sqrt((b*c - a*d)/c)*log(((8*b^2*c^2 - 8*a*b*c*d + a^2*d
^2)*x^4 + a^2*c^2 + 2*(4*a*b*c^2 - 3*a^2*c*d)*x^2 - 4*(a*c^2*x + (2*b*c^2 - a*c*d)*x^3)*sqrt(b*x^2 + a)*sqrt((
b*c - a*d)/c))/(d^2*x^4 + 2*c*d*x^2 + c^2)) + 4*(b^2*c*d^2*x^3 + (2*b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x)*sqrt
(b*x^2 + a))/(c*d^4*x^2 + c^2*d^3), -1/4*((4*b^2*c^3 - 3*a*b*c^2*d - a^2*c*d^2 + (4*b^2*c^2*d - 3*a*b*c*d^2 -
a^2*d^3)*x^2)*sqrt(-(b*c - a*d)/c)*arctan(1/2*((2*b*c - a*d)*x^2 + a*c)*sqrt(b*x^2 + a)*sqrt(-(b*c - a*d)/c)/(
(b^2*c - a*b*d)*x^3 + (a*b*c - a^2*d)*x)) + (4*b^2*c^3 - 5*a*b*c^2*d + (4*b^2*c^2*d - 5*a*b*c*d^2)*x^2)*sqrt(b
)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(b^2*c*d^2*x^3 + (2*b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x
)*sqrt(b*x^2 + a))/(c*d^4*x^2 + c^2*d^3), 1/4*(2*(4*b^2*c^3 - 5*a*b*c^2*d + (4*b^2*c^2*d - 5*a*b*c*d^2)*x^2)*s
qrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (4*b^2*c^3 - 3*a*b*c^2*d - a^2*c*d^2 + (4*b^2*c^2*d - 3*a*b*c*d^2
- a^2*d^3)*x^2)*sqrt(-(b*c - a*d)/c)*arctan(1/2*((2*b*c - a*d)*x^2 + a*c)*sqrt(b*x^2 + a)*sqrt(-(b*c - a*d)/c
)/((b^2*c - a*b*d)*x^3 + (a*b*c - a^2*d)*x)) + 2*(b^2*c*d^2*x^3 + (2*b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x)*sqr
t(b*x^2 + a))/(c*d^4*x^2 + c^2*d^3)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.23954, size = 547, normalized size = 3.13 \begin{align*} \frac{\sqrt{b x^{2} + a} b^{2} x}{2 \, d^{2}} + \frac{{\left (4 \, b^{\frac{5}{2}} c - 5 \, a b^{\frac{3}{2}} d\right )} \log \left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2}\right )}{4 \, d^{3}} - \frac{{\left (4 \, b^{\frac{7}{2}} c^{3} - 7 \, a b^{\frac{5}{2}} c^{2} d + 2 \, a^{2} b^{\frac{3}{2}} c d^{2} + a^{3} \sqrt{b} d^{3}\right )} \arctan \left (\frac{{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} d + 2 \, b c - a d}{2 \, \sqrt{-b^{2} c^{2} + a b c d}}\right )}{2 \, \sqrt{-b^{2} c^{2} + a b c d} c d^{3}} + \frac{2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} b^{\frac{7}{2}} c^{3} - 5 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a b^{\frac{5}{2}} c^{2} d + 4 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a^{2} b^{\frac{3}{2}} c d^{2} -{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a^{3} \sqrt{b} d^{3} + a^{2} b^{\frac{5}{2}} c^{2} d - 2 \, a^{3} b^{\frac{3}{2}} c d^{2} + a^{4} \sqrt{b} d^{3}}{{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} d + 4 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} b c - 2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a d + a^{2} d\right )} c d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(b*x^2 + a)*b^2*x/d^2 + 1/4*(4*b^(5/2)*c - 5*a*b^(3/2)*d)*log((sqrt(b)*x - sqrt(b*x^2 + a))^2)/d^3 - 1
/2*(4*b^(7/2)*c^3 - 7*a*b^(5/2)*c^2*d + 2*a^2*b^(3/2)*c*d^2 + a^3*sqrt(b)*d^3)*arctan(1/2*((sqrt(b)*x - sqrt(b
*x^2 + a))^2*d + 2*b*c - a*d)/sqrt(-b^2*c^2 + a*b*c*d))/(sqrt(-b^2*c^2 + a*b*c*d)*c*d^3) + (2*(sqrt(b)*x - sqr
t(b*x^2 + a))^2*b^(7/2)*c^3 - 5*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a*b^(5/2)*c^2*d + 4*(sqrt(b)*x - sqrt(b*x^2 +
a))^2*a^2*b^(3/2)*c*d^2 - (sqrt(b)*x - sqrt(b*x^2 + a))^2*a^3*sqrt(b)*d^3 + a^2*b^(5/2)*c^2*d - 2*a^3*b^(3/2)*
c*d^2 + a^4*sqrt(b)*d^3)/(((sqrt(b)*x - sqrt(b*x^2 + a))^4*d + 4*(sqrt(b)*x - sqrt(b*x^2 + a))^2*b*c - 2*(sqrt
(b)*x - sqrt(b*x^2 + a))^2*a*d + a^2*d)*c*d^3)