3.66 \(\int \frac{(a+b x^2)^{5/2}}{c+d x^2} \, dx\)
Optimal. Leaf size=157 \[ \frac{\sqrt{b} \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 d^3}-\frac{b x \sqrt{a+b x^2} (4 b c-7 a d)}{8 d^2}-\frac{(b c-a d)^{5/2} \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{\sqrt{c} d^3}+\frac{b x \left (a+b x^2\right )^{3/2}}{4 d} \]
[Out]
-(b*(4*b*c - 7*a*d)*x*Sqrt[a + b*x^2])/(8*d^2) + (b*x*(a + b*x^2)^(3/2))/(4*d) + (Sqrt[b]*(8*b^2*c^2 - 20*a*b*
c*d + 15*a^2*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(8*d^3) - ((b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b*c - a*d]*
x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(Sqrt[c]*d^3)
________________________________________________________________________________________
Rubi [A] time = 0.198957, antiderivative size = 157, 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, 208} \[ \frac{\sqrt{b} \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 d^3}-\frac{b x \sqrt{a+b x^2} (4 b c-7 a d)}{8 d^2}-\frac{(b c-a d)^{5/2} \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{\sqrt{c} d^3}+\frac{b x \left (a+b x^2\right )^{3/2}}{4 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*x^2)^(5/2)/(c + d*x^2),x]
[Out]
-(b*(4*b*c - 7*a*d)*x*Sqrt[a + b*x^2])/(8*d^2) + (b*x*(a + b*x^2)^(3/2))/(4*d) + (Sqrt[b]*(8*b^2*c^2 - 20*a*b*
c*d + 15*a^2*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(8*d^3) - ((b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b*c - a*d]*
x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(Sqrt[c]*d^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 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}}{c+d x^2} \, dx &=\frac{b x \left (a+b x^2\right )^{3/2}}{4 d}+\frac{\int \frac{\sqrt{a+b x^2} \left (-a (b c-4 a d)-b (4 b c-7 a d) x^2\right )}{c+d x^2} \, dx}{4 d}\\ &=-\frac{b (4 b c-7 a d) x \sqrt{a+b x^2}}{8 d^2}+\frac{b x \left (a+b x^2\right )^{3/2}}{4 d}+\frac{\int \frac{a \left (4 b^2 c^2-9 a b c d+8 a^2 d^2\right )+b \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right ) x^2}{\sqrt{a+b x^2} \left (c+d x^2\right )} \, dx}{8 d^2}\\ &=-\frac{b (4 b c-7 a d) x \sqrt{a+b x^2}}{8 d^2}+\frac{b x \left (a+b x^2\right )^{3/2}}{4 d}-\frac{(b c-a d)^3 \int \frac{1}{\sqrt{a+b x^2} \left (c+d x^2\right )} \, dx}{d^3}+\frac{\left (b \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right )\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{8 d^3}\\ &=-\frac{b (4 b c-7 a d) x \sqrt{a+b x^2}}{8 d^2}+\frac{b x \left (a+b x^2\right )^{3/2}}{4 d}-\frac{(b c-a d)^3 \operatorname{Subst}\left (\int \frac{1}{c-(b c-a d) x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{d^3}+\frac{\left (b \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{8 d^3}\\ &=-\frac{b (4 b c-7 a d) x \sqrt{a+b x^2}}{8 d^2}+\frac{b x \left (a+b x^2\right )^{3/2}}{4 d}+\frac{\sqrt{b} \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 d^3}-\frac{(b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{c} \sqrt{a+b x^2}}\right )}{\sqrt{c} d^3}\\ \end{align*}
Mathematica [A] time = 0.12079, size = 140, normalized size = 0.89 \[ \frac{\sqrt{b} \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right ) \log \left (\sqrt{b} \sqrt{a+b x^2}+b x\right )+b d x \sqrt{a+b x^2} \left (9 a d-4 b c+2 b d x^2\right )+\frac{8 (a d-b c)^{5/2} \tan ^{-1}\left (\frac{x \sqrt{a d-b c}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{\sqrt{c}}}{8 d^3} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*x^2)^(5/2)/(c + d*x^2),x]
[Out]
(b*d*x*Sqrt[a + b*x^2]*(-4*b*c + 9*a*d + 2*b*d*x^2) + (8*(-(b*c) + a*d)^(5/2)*ArcTan[(Sqrt[-(b*c) + a*d]*x)/(S
qrt[c]*Sqrt[a + b*x^2])])/Sqrt[c] + Sqrt[b]*(8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*Log[b*x + Sqrt[b]*Sqrt[a + b
*x^2]])/(8*d^3)
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Maple [B] time = 0.014, size = 3053, normalized size = 19.5 \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),x)
[Out]
1/6/(-c*d)^(1/2)*((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(3/2)*a+1/2/(-c*d)
^(1/2)*((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)*a^2-1/6/(-c*d)^(1/2)*(
(x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(3/2)*a-1/2/(-c*d)^(1/2)*((x+(-c*d)^
(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)*a^2-3/2/(-c*d)^(1/2)/d^2/((a*d-b*c)/d)^(
1/2)*ln((2*(a*d-b*c)/d+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+2*((a*d-b*c)/d)^(1/2)*((x-(-c*d)^(1/2)/d)^2*b+2*b
*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2))/(x-(-c*d)^(1/2)/d))*a*b^2*c^2-3/2/(-c*d)^(1/2)/d/((a*d-
b*c)/d)^(1/2)*ln((2*(a*d-b*c)/d-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+2*((a*d-b*c)/d)^(1/2)*((x+(-c*d)^(1/2)/d
)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2))/(x+(-c*d)^(1/2)/d))*a^2*b*c+3/2/(-c*d)^(1/2)/d
^2/((a*d-b*c)/d)^(1/2)*ln((2*(a*d-b*c)/d-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+2*((a*d-b*c)/d)^(1/2)*((x+(-c*d
)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2))/(x+(-c*d)^(1/2)/d))*a*b^2*c^2-1/10/(-
c*d)^(1/2)*((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(5/2)+1/10/(-c*d)^(1/2)*
((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(5/2)-1/4/d^2*b^2*((x+(-c*d)^(1/2)/
d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)*x*c-5/4/d^2*b^(3/2)*ln((-b*(-c*d)^(1/2)/d+b*(x
+(-c*d)^(1/2)/d))/b^(1/2)+((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2))*c*
a-1/2/(-c*d)^(1/2)/d^2*((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)*b^2*c^
2+7/16*b/d*a*((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)*x+7/16*b/d*a*((x
-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)*x-1/6/(-c*d)^(1/2)/d*((x-(-c*d)^
(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(3/2)*b*c-1/4/d^2*b^2*((x-(-c*d)^(1/2)/d)^2*b+
2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)*x*c-5/4/d^2*b^(3/2)*ln((b*(-c*d)^(1/2)/d+b*(x-(-c*d)^
(1/2)/d))/b^(1/2)+((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2))*c*a+1/2/(-
c*d)^(1/2)/d^2*((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)*b^2*c^2-1/2/(-
c*d)^(1/2)/d^3/((a*d-b*c)/d)^(1/2)*ln((2*(a*d-b*c)/d-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+2*((a*d-b*c)/d)^(1/
2)*((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2))/(x+(-c*d)^(1/2)/d))*b^3*c
^3+1/6/(-c*d)^(1/2)/d*((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(3/2)*b*c+3/2
/(-c*d)^(1/2)/d/((a*d-b*c)/d)^(1/2)*ln((2*(a*d-b*c)/d+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+2*((a*d-b*c)/d)^(1
/2)*((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2))/(x-(-c*d)^(1/2)/d))*a^2*
b*c-1/(-c*d)^(1/2)/d*((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)*a*b*c+1/
2/(-c*d)^(1/2)/d^3/((a*d-b*c)/d)^(1/2)*ln((2*(a*d-b*c)/d+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+2*((a*d-b*c)/d)
^(1/2)*((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2))/(x-(-c*d)^(1/2)/d))*b
^3*c^3+1/(-c*d)^(1/2)/d*((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2)*a*b*c
+1/8*b/d*((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(3/2)*x+15/16/d*b^(1/2)*ln
((b*(-c*d)^(1/2)/d+b*(x-(-c*d)^(1/2)/d))/b^(1/2)+((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)
+(a*d-b*c)/d)^(1/2))*a^2+1/2/d^3*b^(5/2)*ln((b*(-c*d)^(1/2)/d+b*(x-(-c*d)^(1/2)/d))/b^(1/2)+((x-(-c*d)^(1/2)/d
)^2*b+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2))*c^2-1/2/(-c*d)^(1/2)/((a*d-b*c)/d)^(1/2)*ln((2
*(a*d-b*c)/d+2*b*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+2*((a*d-b*c)/d)^(1/2)*((x-(-c*d)^(1/2)/d)^2*b+2*b*(-c*d)^(1
/2)/d*(x-(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2))/(x-(-c*d)^(1/2)/d))*a^3+1/8*b/d*((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d
)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(3/2)*x+15/16/d*b^(1/2)*ln((-b*(-c*d)^(1/2)/d+b*(x+(-c*d)^(1/2)/d))/
b^(1/2)+((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1/2))*a^2+1/2/d^3*b^(5/2)*
ln((-b*(-c*d)^(1/2)/d+b*(x+(-c*d)^(1/2)/d))/b^(1/2)+((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)
/d)+(a*d-b*c)/d)^(1/2))*c^2+1/2/(-c*d)^(1/2)/((a*d-b*c)/d)^(1/2)*ln((2*(a*d-b*c)/d-2*b*(-c*d)^(1/2)/d*(x+(-c*d
)^(1/2)/d)+2*((a*d-b*c)/d)^(1/2)*((x+(-c*d)^(1/2)/d)^2*b-2*b*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+(a*d-b*c)/d)^(1
/2))/(x+(-c*d)^(1/2)/d))*a^3
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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((b*x^2+a)^(5/2)/(d*x^2+c),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 8.08999, size = 2022, normalized size = 12.88 \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),x, algorithm="fricas")
[Out]
[1/16*((8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 4*(b^2*
c^2 - 2*a*b*c*d + a^2*d^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)) + 2*(2*b^2*d^2*x^3 - (4*b^2*c*d - 9*a*b*d^2)*x)*sqrt(b*x^2 + a))/d^3, -1/8*((8*b^2*c^2 - 20*a*
b*c*d + 15*a^2*d^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^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)) - (2*b^2*d^2*x^3 - (4
*b^2*c*d - 9*a*b*d^2)*x)*sqrt(b*x^2 + a))/d^3, 1/16*(8*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-(b*c - a*d)/c)*ar
ctan(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)) + (8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*b^
2*d^2*x^3 - (4*b^2*c*d - 9*a*b*d^2)*x)*sqrt(b*x^2 + a))/d^3, -1/8*((8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*sqrt(
-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 4*(b^2*c^2 - 2*a*b*c*d + a^2*d^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*d^2*x^3 - (4*b^2*c*d - 9*a*b*d^2)*x)*sqrt(b*x^2 + a))/d^3]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right )^{\frac{5}{2}}}{c + d x^{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),x)
[Out]
Integral((a + b*x**2)**(5/2)/(c + d*x**2), x)
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Giac [A] time = 1.1661, size = 290, normalized size = 1.85 \begin{align*} \frac{1}{8} \, \sqrt{b x^{2} + a}{\left (\frac{2 \, b^{2} x^{2}}{d} - \frac{4 \, b^{4} c d^{4} - 9 \, a b^{3} d^{5}}{b^{2} d^{6}}\right )} x - \frac{{\left (8 \, b^{\frac{5}{2}} c^{2} - 20 \, a b^{\frac{3}{2}} c d + 15 \, a^{2} \sqrt{b} d^{2}\right )} \log \left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2}\right )}{16 \, d^{3}} + \frac{{\left (b^{\frac{7}{2}} c^{3} - 3 \, a b^{\frac{5}{2}} c^{2} d + 3 \, 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 )}{\sqrt{-b^{2} c^{2} + a b c d} 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),x, algorithm="giac")
[Out]
1/8*sqrt(b*x^2 + a)*(2*b^2*x^2/d - (4*b^4*c*d^4 - 9*a*b^3*d^5)/(b^2*d^6))*x - 1/16*(8*b^(5/2)*c^2 - 20*a*b^(3/
2)*c*d + 15*a^2*sqrt(b)*d^2)*log((sqrt(b)*x - sqrt(b*x^2 + a))^2)/d^3 + (b^(7/2)*c^3 - 3*a*b^(5/2)*c^2*d + 3*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)*d^3)