3.1.66 \(\int \frac {(a+b x^2)^{5/2}}{c+d x^2} \, dx\) [66]

Optimal. Leaf size=157 \[ -\frac {b (4 b c-7 a d) x \sqrt {a+b x^2}}{8 d^2}+\frac {b x \left (a+b x^2\right )^{3/2}}{4 d}+\frac {\sqrt {b} \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 d^3}-\frac {(b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{\sqrt {c} d^3} \]

[Out]

1/4*b*x*(b*x^2+a)^(3/2)/d+1/8*(15*a^2*d^2-20*a*b*c*d+8*b^2*c^2)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))*b^(1/2)/d^3
-(-a*d+b*c)^(5/2)*arctanh(x*(-a*d+b*c)^(1/2)/c^(1/2)/(b*x^2+a)^(1/2))/d^3/c^(1/2)-1/8*b*(-7*a*d+4*b*c)*x*(b*x^
2+a)^(1/2)/d^2

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Rubi [A]
time = 0.14, antiderivative size = 157, normalized size of antiderivative = 1.00, 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 = {427, 542, 537, 223, 212, 385, 214} \begin {gather*} \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 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 \sqrt {a+b x^2} (4 b c-7 a d)}{8 d^2}+\frac {b x \left (a+b x^2\right )^{3/2}}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/(c + d*x^2),x]

[Out]

-1/8*(b*(4*b*c - 7*a*d)*x*Sqrt[a + b*x^2])/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 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 214

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

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 385

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 427

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 537

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 542

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]

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 \text {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 ) \text {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*}

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Mathematica [A]
time = 0.39, size = 157, normalized size = 1.00 \begin {gather*} \frac {b d x \sqrt {a+b x^2} \left (-4 b c+9 a d+2 b d x^2\right )-\frac {8 (-b c+a d)^{5/2} \tan ^{-1}\left (\frac {-d x \sqrt {a+b x^2}+\sqrt {b} \left (c+d x^2\right )}{\sqrt {c} \sqrt {-b c+a d}}\right )}{\sqrt {c}}-\sqrt {b} \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 d^3} \end {gather*}

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[(-(d*x*Sqrt[a + b*x^2]) +
Sqrt[b]*(c + d*x^2))/(Sqrt[c]*Sqrt[-(b*c) + a*d])])/Sqrt[c] - Sqrt[b]*(8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*L
og[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(8*d^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2047\) vs. \(2(131)=262\).
time = 0.09, size = 2048, normalized size = 13.04

method	result	size
risch	\(\frac {b x \left (2 b d \,x^{2}+9 a d -4 b c \right ) \sqrt {b \,x^{2}+a}}{8 d^{2}}+\frac {15 \sqrt {b}\, \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) a^{2}}{8 d}-\frac {5 b^{\frac {3}{2}} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) a c}{2 d^{2}}+\frac {b^{\frac {5}{2}} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) c^{2}}{d^{3}}-\frac {\ln \left (\frac {\frac {2 a d -2 b c}{d}+\frac {2 b \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}{d}+2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {\left (x -\frac {\sqrt {-c d}}{d}\right )^{2} b +\frac {2 b \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}{d}+\frac {a d -b c}{d}}}{x -\frac {\sqrt {-c d}}{d}}\right ) a^{3}}{2 \sqrt {-c d}\, \sqrt {\frac {a d -b c}{d}}}+\frac {3 \ln \left (\frac {\frac {2 a d -2 b c}{d}+\frac {2 b \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}{d}+2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {\left (x -\frac {\sqrt {-c d}}{d}\right )^{2} b +\frac {2 b \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}{d}+\frac {a d -b c}{d}}}{x -\frac {\sqrt {-c d}}{d}}\right ) a^{2} b c}{2 d \sqrt {-c d}\, \sqrt {\frac {a d -b c}{d}}}-\frac {3 \ln \left (\frac {\frac {2 a d -2 b c}{d}+\frac {2 b \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}{d}+2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {\left (x -\frac {\sqrt {-c d}}{d}\right )^{2} b +\frac {2 b \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}{d}+\frac {a d -b c}{d}}}{x -\frac {\sqrt {-c d}}{d}}\right ) a \,b^{2} c^{2}}{2 d^{2} \sqrt {-c d}\, \sqrt {\frac {a d -b c}{d}}}+\frac {\ln \left (\frac {\frac {2 a d -2 b c}{d}+\frac {2 b \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}{d}+2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {\left (x -\frac {\sqrt {-c d}}{d}\right )^{2} b +\frac {2 b \sqrt {-c d}\, \left (x -\frac {\sqrt {-c d}}{d}\right )}{d}+\frac {a d -b c}{d}}}{x -\frac {\sqrt {-c d}}{d}}\right ) b^{3} c^{3}}{2 d^{3} \sqrt {-c d}\, \sqrt {\frac {a d -b c}{d}}}+\frac {\ln \left (\frac {\frac {2 a d -2 b c}{d}-\frac {2 b \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}{d}+2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {\left (x +\frac {\sqrt {-c d}}{d}\right )^{2} b -\frac {2 b \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}{d}+\frac {a d -b c}{d}}}{x +\frac {\sqrt {-c d}}{d}}\right ) a^{3}}{2 \sqrt {-c d}\, \sqrt {\frac {a d -b c}{d}}}-\frac {3 \ln \left (\frac {\frac {2 a d -2 b c}{d}-\frac {2 b \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}{d}+2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {\left (x +\frac {\sqrt {-c d}}{d}\right )^{2} b -\frac {2 b \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}{d}+\frac {a d -b c}{d}}}{x +\frac {\sqrt {-c d}}{d}}\right ) a^{2} b c}{2 d \sqrt {-c d}\, \sqrt {\frac {a d -b c}{d}}}+\frac {3 \ln \left (\frac {\frac {2 a d -2 b c}{d}-\frac {2 b \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}{d}+2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {\left (x +\frac {\sqrt {-c d}}{d}\right )^{2} b -\frac {2 b \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}{d}+\frac {a d -b c}{d}}}{x +\frac {\sqrt {-c d}}{d}}\right ) a \,b^{2} c^{2}}{2 d^{2} \sqrt {-c d}\, \sqrt {\frac {a d -b c}{d}}}-\frac {\ln \left (\frac {\frac {2 a d -2 b c}{d}-\frac {2 b \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}{d}+2 \sqrt {\frac {a d -b c}{d}}\, \sqrt {\left (x +\frac {\sqrt {-c d}}{d}\right )^{2} b -\frac {2 b \sqrt {-c d}\, \left (x +\frac {\sqrt {-c d}}{d}\right )}{d}+\frac {a d -b c}{d}}}{x +\frac {\sqrt {-c d}}{d}}\right ) b^{3} c^{3}}{2 d^{3} \sqrt {-c d}\, \sqrt {\frac {a d -b c}{d}}}\)	\(1365\)
default	\(\text {Expression too large to display}\)	\(2048\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(-c*d)^(1/2)*(1/5*((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)+b*(-c*d
)^(1/2)/d*(1/8*(2*b*(x-(-c*d)^(1/2)/d)+2*b*(-c*d)^(1/2)/d)/b*((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)+3/16*(4*b*(a*d-b*c)/d+4*b^2*c/d)/b*(1/4*(2*b*(x-(-c*d)^(1/2)/d)+2*b*(-c*d)^(1/
2)/d)/b*((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)+1/8*(4*b*(a*d-b*c)/d+
4*b^2*c/d)/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))))+(a*d-b*c)/d*(1/3*((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*d)^(1/2)/d*(1/4*(2*b*(x-(-c*d)^(1/2)/d)+2*b*(-c*d)^(1/2)/d)/b*((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)+1/8*(4*b*(a*d-b*c)/d+4*b^2*c/d)/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)))+(a*d-b*c)/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)+b^(1/2)*(-c*d)^(1/2)/d*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*d-b*c)/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)))))-1/2/(-c*d)^(1/2)*(1/5*((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)-b*(-c*d)^(1/2)/d*(1/8*(2*b*(x+(-c*d)^(1/2)/d)-2*b*(-c*d)^(1/2)/d)/b
*((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)+3/16*(4*b*(a*d-b*c)/d+4*b^2*
c/d)/b*(1/4*(2*b*(x+(-c*d)^(1/2)/d)-2*b*(-c*d)^(1/2)/d)/b*((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)+1/8*(4*b*(a*d-b*c)/d+4*b^2*c/d)/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))))+(a*d-b*c)/d*(1/
3*((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*d)^(1/2)/d*(1/4*(2*b*
(x+(-c*d)^(1/2)/d)-2*b*(-c*d)^(1/2)/d)/b*((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)+1/8*(4*b*(a*d-b*c)/d+4*b^2*c/d)/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)))+(a*d-b*c)/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)-b^(1/2)*(-c*d)^(1/2)/d*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
*d-b*c)/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)))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [A]
time = 1.17, size = 935, normalized size = 5.96 \begin {gather*} \left [\frac {{\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 4 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{c}} \log \left (\frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} + 2 \, {\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2} - 4 \, {\left (a c^{2} x + {\left (2 \, b c^{2} - a c d\right )} x^{3}\right )} \sqrt {b x^{2} + a} \sqrt {\frac {b c - a d}{c}}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{2}}\right ) + 2 \, {\left (2 \, b^{2} d^{2} x^{3} - {\left (4 \, b^{2} c d - 9 \, a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, d^{3}}, -\frac {{\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{c}} \log \left (\frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} + 2 \, {\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2} - 4 \, {\left (a c^{2} x + {\left (2 \, b c^{2} - a c d\right )} x^{3}\right )} \sqrt {b x^{2} + a} \sqrt {\frac {b c - a d}{c}}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{2}}\right ) - {\left (2 \, b^{2} d^{2} x^{3} - {\left (4 \, b^{2} c d - 9 \, a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, d^{3}}, \frac {8 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{c}} \arctan \left (\frac {{\left ({\left (2 \, b c - a d\right )} x^{2} + a c\right )} \sqrt {b x^{2} + a} \sqrt {-\frac {b c - a d}{c}}}{2 \, {\left ({\left (b^{2} c - a b d\right )} x^{3} + {\left (a b c - a^{2} d\right )} x\right )}}\right ) + {\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} 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 \, b^{2} d^{2} x^{3} - {\left (4 \, b^{2} c d - 9 \, a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, d^{3}}, -\frac {{\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 4 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{c}} \arctan \left (\frac {{\left ({\left (2 \, b c - a d\right )} x^{2} + a c\right )} \sqrt {b x^{2} + a} \sqrt {-\frac {b c - a d}{c}}}{2 \, {\left ({\left (b^{2} c - a b d\right )} x^{3} + {\left (a b c - a^{2} d\right )} x\right )}}\right ) - {\left (2 \, b^{2} d^{2} x^{3} - {\left (4 \, b^{2} c d - 9 \, a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, d^{3}}\right ] \end {gather*}

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

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 [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/2)/(c + d*x^2),x)

[Out]

int((a + b*x^2)^(5/2)/(c + d*x^2), x)

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