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

Optimal. Leaf size=156 \[ \frac{\sqrt{d} \left (8 a^2 d^2-20 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 b^3}+\frac{d x \sqrt{c+d x^2} (7 b c-4 a d)}{8 b^2}+\frac{(b c-a d)^{5/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a} b^3}+\frac{d x \left (c+d x^2\right )^{3/2}}{4 b} \]

[Out]

(d*(7*b*c - 4*a*d)*x*Sqrt[c + d*x^2])/(8*b^2) + (d*x*(c + d*x^2)^(3/2))/(4*b) + ((b*c - a*d)^(5/2)*ArcTan[(Sqr
t[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(Sqrt[a]*b^3) + (Sqrt[d]*(15*b^2*c^2 - 20*a*b*c*d + 8*a^2*d^2)*Arc
Tanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(8*b^3)

________________________________________________________________________________________

Rubi [A]  time = 0.179535, antiderivative size = 156, 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 = {416, 528, 523, 217, 206, 377, 205} \[ \frac{\sqrt{d} \left (8 a^2 d^2-20 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 b^3}+\frac{d x \sqrt{c+d x^2} (7 b c-4 a d)}{8 b^2}+\frac{(b c-a d)^{5/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a} b^3}+\frac{d x \left (c+d x^2\right )^{3/2}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^(5/2)/(a + b*x^2),x]

[Out]

(d*(7*b*c - 4*a*d)*x*Sqrt[c + d*x^2])/(8*b^2) + (d*x*(c + d*x^2)^(3/2))/(4*b) + ((b*c - a*d)^(5/2)*ArcTan[(Sqr
t[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(Sqrt[a]*b^3) + (Sqrt[d]*(15*b^2*c^2 - 20*a*b*c*d + 8*a^2*d^2)*Arc
Tanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(8*b^3)

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[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}}{a+b x^2} \, dx &=\frac{d x \left (c+d x^2\right )^{3/2}}{4 b}+\frac{\int \frac{\sqrt{c+d x^2} \left (c (4 b c-a d)+d (7 b c-4 a d) x^2\right )}{a+b x^2} \, dx}{4 b}\\ &=\frac{d (7 b c-4 a d) x \sqrt{c+d x^2}}{8 b^2}+\frac{d x \left (c+d x^2\right )^{3/2}}{4 b}+\frac{\int \frac{c \left (8 b^2 c^2-9 a b c d+4 a^2 d^2\right )+d \left (15 b^2 c^2-20 a b c d+8 a^2 d^2\right ) x^2}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{8 b^2}\\ &=\frac{d (7 b c-4 a d) x \sqrt{c+d x^2}}{8 b^2}+\frac{d x \left (c+d x^2\right )^{3/2}}{4 b}+\frac{(b c-a d)^3 \int \frac{1}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{b^3}+\frac{\left (d \left (15 b^2 c^2-20 a b c d+8 a^2 d^2\right )\right ) \int \frac{1}{\sqrt{c+d x^2}} \, dx}{8 b^3}\\ &=\frac{d (7 b c-4 a d) x \sqrt{c+d x^2}}{8 b^2}+\frac{d x \left (c+d x^2\right )^{3/2}}{4 b}+\frac{(b c-a d)^3 \operatorname{Subst}\left (\int \frac{1}{a-(-b c+a d) x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{b^3}+\frac{\left (d \left (15 b^2 c^2-20 a b c d+8 a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{8 b^3}\\ &=\frac{d (7 b c-4 a d) x \sqrt{c+d x^2}}{8 b^2}+\frac{d x \left (c+d x^2\right )^{3/2}}{4 b}+\frac{(b c-a d)^{5/2} \tan ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a} b^3}+\frac{\sqrt{d} \left (15 b^2 c^2-20 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 b^3}\\ \end{align*}

Mathematica [A]  time = 0.102895, size = 140, normalized size = 0.9 \[ \frac{\sqrt{d} \left (8 a^2 d^2-20 a b c d+15 b^2 c^2\right ) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )+b d x \sqrt{c+d x^2} \left (-4 a d+9 b c+2 b d x^2\right )+\frac{8 (b c-a d)^{5/2} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a}}}{8 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*x*Sqrt[c + d*x^2]*(9*b*c - 4*a*d + 2*b*d*x^2) + (8*(b*c - a*d)^(5/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*
Sqrt[c + d*x^2])])/Sqrt[a] + Sqrt[d]*(15*b^2*c^2 - 20*a*b*c*d + 8*a^2*d^2)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])
/(8*b^3)

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Maple [B]  time = 0.011, size = 3101, normalized size = 19.9 \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),x)

[Out]

1/2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)
/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^
(1/2)))*c^3+1/10/(-a*b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(
5/2)-1/10/(-a*b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(5/2)-3/
2/(-a*b)^(1/2)/b/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)
/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^
(1/2)))*a*d*c^2-3/2/(-a*b)^(1/2)/b^2/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^
(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(
1/2))/(x-1/b*(-a*b)^(1/2)))*a^2*d^2*c+3/2/(-a*b)^(1/2)/b/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1
/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(
1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))*a*d*c^2+3/2/(-a*b)^(1/2)/b^2/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-
b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/
2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^(1/2)))*a^2*d^2*c-5/4/b^2*d^(3/2)*ln((-d*(-a*b)^(1
/2)/b+(x+1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b
*c)/b)^(1/2))*a*c-1/4/b^2*d^2*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(
1/2)*x*a+1/2/(-a*b)^(1/2)/b^3/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+
2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(
x-1/b*(-a*b)^(1/2)))*a^3*d^3-1/(-a*b)^(1/2)/b*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2)
)-(a*d-b*c)/b)^(1/2)*a*d*c+1/(-a*b)^(1/2)/b*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-
(a*d-b*c)/b)^(1/2)*a*d*c-1/2/(-a*b)^(1/2)/b^3/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/
b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-
b*c)/b)^(1/2))/(x+1/b*(-a*b)^(1/2)))*a^3*d^3-1/2/(-a*b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+
1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*c^2-1/6/(-a*b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*
(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)*c+1/2/(-a*b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)
^(1/2))-(a*d-b*c)/b)^(1/2)*c^2+1/6/(-a*b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/
2))-(a*d-b*c)/b)^(3/2)*c+7/16*d/b*c*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c
)/b)^(1/2)*x-1/6/(-a*b)^(1/2)/b*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)
^(3/2)*a*d-1/4/b^2*d^2*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x*
a-5/4/b^2*d^(3/2)*ln((d*(-a*b)^(1/2)/b+(x-1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1
/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))*a*c+1/2/(-a*b)^(1/2)/b^2*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(
1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*a^2*d^2-1/2/(-a*b)^(1/2)/b^2*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*
b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*a^2*d^2+7/16*d/b*c*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/
2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x+1/6/(-a*b)^(1/2)/b*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b
*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)*a*d+1/8*d/b*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b
)^(1/2))-(a*d-b*c)/b)^(3/2)*x+15/16/b*d^(1/2)*ln((d*(-a*b)^(1/2)/b+(x-1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x-1/b*(-a
*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))*c^2+1/2/b^3*d^(5/2)*ln((d*(-a*b)^(1
/2)/b+(x-1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b
*c)/b)^(1/2))*a^2-1/2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1
/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/
2))/(x-1/b*(-a*b)^(1/2)))*c^3+1/8*d/b*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b
*c)/b)^(3/2)*x+15/16/b*d^(1/2)*ln((-d*(-a*b)^(1/2)/b+(x+1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x+1/b*(-a*b)^(1/2))^2*d
-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))*c^2+1/2/b^3*d^(5/2)*ln((-d*(-a*b)^(1/2)/b+(x+1/b*
(-a*b)^(1/2))*d)/d^(1/2)+((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))
*a^2

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 8.13208, size = 2022, normalized size = 12.96 \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),x, algorithm="fricas")

[Out]

[1/16*((15*b^2*c^2 - 20*a*b*c*d + 8*a^2*d^2)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 4*(b^2*
c^2 - 2*a*b*c*d + a^2*d^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)) + 2*(2*b^2*d^2*x^3 + (9*b^2*c*d - 4*a*b*d^2)*x)*sqrt(d*x^2 + c))/b^3, -1/8*((15*b^2*c^2 - 20
*a*b*c*d + 8*a^2*d^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^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)) - (2*b^2*d^2*x^3
+ (9*b^2*c*d - 4*a*b*d^2)*x)*sqrt(d*x^2 + c))/b^3, 1/16*(8*(b^2*c^2 - 2*a*b*c*d + a^2*d^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)) + (15*b^2*c^2 - 20*a*b*c*d + 8*a^2*d^2)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(2*
b^2*d^2*x^3 + (9*b^2*c*d - 4*a*b*d^2)*x)*sqrt(d*x^2 + c))/b^3, -1/8*((15*b^2*c^2 - 20*a*b*c*d + 8*a^2*d^2)*sqr
t(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - 4*(b^2*c^2 - 2*a*b*c*d + a^2*d^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*b^
2*d^2*x^3 + (9*b^2*c*d - 4*a*b*d^2)*x)*sqrt(d*x^2 + c))/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}}}{a + b x^{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),x)

[Out]

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

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Giac [A]  time = 1.14431, size = 290, normalized size = 1.86 \begin{align*} \frac{1}{8} \, \sqrt{d x^{2} + c}{\left (\frac{2 \, d^{2} x^{2}}{b} + \frac{9 \, b^{5} c d^{3} - 4 \, a b^{4} d^{4}}{b^{6} d^{2}}\right )} x - \frac{{\left (15 \, b^{2} c^{2} \sqrt{d} - 20 \, a b c d^{\frac{3}{2}} + 8 \, a^{2} d^{\frac{5}{2}}\right )} \log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right )}{16 \, b^{3}} - \frac{{\left (b^{3} c^{3} \sqrt{d} - 3 \, a b^{2} c^{2} d^{\frac{3}{2}} + 3 \, a^{2} b c d^{\frac{5}{2}} - 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 )}{\sqrt{a b c d - a^{2} d^{2}} 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),x, algorithm="giac")

[Out]

1/8*sqrt(d*x^2 + c)*(2*d^2*x^2/b + (9*b^5*c*d^3 - 4*a*b^4*d^4)/(b^6*d^2))*x - 1/16*(15*b^2*c^2*sqrt(d) - 20*a*
b*c*d^(3/2) + 8*a^2*d^(5/2))*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/b^3 - (b^3*c^3*sqrt(d) - 3*a*b^2*c^2*d^(3/2)
+ 3*a^2*b*c*d^(5/2) - 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)*b^3)