3.1.67 \(\int \frac {(a+b x^2)^{5/2}}{(c+d x^2)^2} \, dx\) [67]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.15, antiderivative size = 175, 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 = {424, 542, 537, 223, 212, 385, 214} \begin {gather*} -\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 )} \end {gather*}

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 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 424

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 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}}{\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 ) \text {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 ) \text {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 = 0.66, size = 187, normalized size = 1.07 \begin {gather*} \frac {\frac {d x \sqrt {a+b x^2} \left (-2 a b c d+a^2 d^2+b^2 c \left (2 c+d x^2\right )\right )}{c \left (c+d x^2\right )}+\frac {\sqrt {-b c+a d} \left (4 b^2 c^2-3 a b c d-a^2 d^2\right ) \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 )}{c^{3/2}}+b^{3/2} (4 b c-5 a d) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{2 d^3} \end {gather*}

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

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(5229\) vs. \(2(147)=294\).
time = 0.13, size = 5230, normalized size = 29.89

method	result	size
risch	\(\text {Expression too large to display}\)	\(3386\)
default	\(\text {Expression too large to display}\)	\(5230\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

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)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.97, size = 1236, normalized size = 7.06 \begin {gather*} \left [-\frac {2 \, {\left (4 \, b^{2} c^{3} - 5 \, a b c^{2} d + {\left (4 \, b^{2} c^{2} d - 5 \, a b c d^{2}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + {\left (4 \, b^{2} c^{3} - 3 \, a b c^{2} d - a^{2} c d^{2} + {\left (4 \, b^{2} c^{2} d - 3 \, a b c d^{2} - a^{2} d^{3}\right )} x^{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 ) - 4 \, {\left (b^{2} c d^{2} x^{3} + {\left (2 \, b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, {\left (c d^{4} x^{2} + c^{2} d^{3}\right )}}, \frac {4 \, {\left (4 \, b^{2} c^{3} - 5 \, a b c^{2} d + {\left (4 \, b^{2} c^{2} d - 5 \, a b c d^{2}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (4 \, b^{2} c^{3} - 3 \, a b c^{2} d - a^{2} c d^{2} + {\left (4 \, b^{2} c^{2} d - 3 \, a b c d^{2} - a^{2} d^{3}\right )} x^{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 ) + 4 \, {\left (b^{2} c d^{2} x^{3} + {\left (2 \, b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, {\left (c d^{4} x^{2} + c^{2} d^{3}\right )}}, -\frac {{\left (4 \, b^{2} c^{3} - 3 \, a b c^{2} d - a^{2} c d^{2} + {\left (4 \, b^{2} c^{2} d - 3 \, a b c d^{2} - a^{2} d^{3}\right )} x^{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 (4 \, b^{2} c^{3} - 5 \, a b c^{2} d + {\left (4 \, b^{2} c^{2} d - 5 \, a b c d^{2}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (b^{2} c d^{2} x^{3} + {\left (2 \, b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{4 \, {\left (c d^{4} x^{2} + c^{2} d^{3}\right )}}, \frac {2 \, {\left (4 \, b^{2} c^{3} - 5 \, a b c^{2} d + {\left (4 \, b^{2} c^{2} d - 5 \, a b c d^{2}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (4 \, b^{2} c^{3} - 3 \, a b c^{2} d - a^{2} c d^{2} + {\left (4 \, b^{2} c^{2} d - 3 \, a b c d^{2} - a^{2} d^{3}\right )} x^{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 ) + 2 \, {\left (b^{2} c d^{2} x^{3} + {\left (2 \, b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{4 \, {\left (c d^{4} x^{2} + c^{2} d^{3}\right )}}\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)^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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x^{2}\right )^{\frac {5}{2}}}{\left (c + d x^{2}\right )^{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)**2,x)

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 405 vs. \(2 (147) = 294\).
time = 0.57, size = 405, normalized size = 2.31 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________