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

Optimal. Leaf size=174 \[ \frac{(b c-a d)^{3/2} (4 a d+b c) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{3/2} b^3}+\frac{d^{3/2} (5 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 b^3}-\frac{d x \sqrt{c+d x^2} (b c-2 a d)}{2 a b^2}+\frac{x \left (c+d x^2\right )^{3/2} (b c-a d)}{2 a b \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.21934, antiderivative size = 174, 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, 205} \[ \frac{(b c-a d)^{3/2} (4 a d+b c) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{3/2} b^3}+\frac{d^{3/2} (5 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 b^3}-\frac{d x \sqrt{c+d x^2} (b c-2 a d)}{2 a b^2}+\frac{x \left (c+d x^2\right )^{3/2} (b c-a d)}{2 a b \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(b*c - 2*a*d)*x*Sqrt[c + d*x^2])/(2*a*b^2) + ((b*c - a*d)*x*(c + d*x^2)^(3/2))/(2*a*b*(a + b*x^2)) + ((b*c
- a*d)^(3/2)*(b*c + 4*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(3/2)*b^3) + (d^(3/2)*
(5*b*c - 4*a*d)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(2*b^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 205

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

Rubi steps

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

Mathematica [A]  time = 0.187233, size = 143, normalized size = 0.82 \[ \frac{\frac{(b c-a d)^{3/2} (4 a d+b c) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{3/2}}+b x \sqrt{c+d x^2} \left (\frac{(b c-a d)^2}{a \left (a+b x^2\right )}+d^2\right )+d^{3/2} (5 b c-4 a d) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.012, size = 7451, normalized size = 42.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 6.08956, size = 2585, normalized size = 14.86 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.1818, size = 549, normalized size = 3.16 \begin{align*} \frac{\sqrt{d x^{2} + c} d^{2} x}{2 \, b^{2}} - \frac{{\left (5 \, b c d^{\frac{3}{2}} - 4 \, a d^{\frac{5}{2}}\right )} \log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right )}{4 \, b^{3}} - \frac{{\left (b^{3} c^{3} \sqrt{d} + 2 \, a b^{2} c^{2} d^{\frac{3}{2}} - 7 \, a^{2} b c d^{\frac{5}{2}} + 4 \, a^{3} d^{\frac{7}{2}}\right )} \arctan \left (\frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt{a b c d - a^{2} d^{2}}}\right )}{2 \, \sqrt{a b c d - a^{2} d^{2}} a b^{3}} - \frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b^{3} c^{3} \sqrt{d} - 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b^{2} c^{2} d^{\frac{3}{2}} + 5 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{2} b c d^{\frac{5}{2}} - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{3} d^{\frac{7}{2}} - b^{3} c^{4} \sqrt{d} + 2 \, a b^{2} c^{3} d^{\frac{3}{2}} - a^{2} b c^{2} d^{\frac{5}{2}}}{{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c + 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} a b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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