\(\int \frac {(a+b x^2)^{3/2}}{(c+d x^2)^3} \, dx\) [59]

Optimal result

Integrand size = 21, antiderivative size = 113 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^3} \, dx=\frac {x \left (a+b x^2\right )^{3/2}}{4 c \left (c+d x^2\right )^2}+\frac {3 a x \sqrt {a+b x^2}}{8 c^2 \left (c+d x^2\right )}+\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{8 c^{5/2} \sqrt {b c-a d}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {386, 385, 214} \[ \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^3} \, dx=\frac {3 a^2 \text {arctanh}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{8 c^{5/2} \sqrt {b c-a d}}+\frac {3 a x \sqrt {a+b x^2}}{8 c^2 \left (c+d x^2\right )}+\frac {x \left (a+b x^2\right )^{3/2}}{4 c \left (c+d x^2\right )^2} \]

[In]

Int[(a + b*x^2)^(3/2)/(c + d*x^2)^3,x]

[Out]

(x*(a + b*x^2)^(3/2))/(4*c*(c + d*x^2)^2) + (3*a*x*Sqrt[a + b*x^2])/(8*c^2*(c + d*x^2)) + (3*a^2*ArcTanh[(Sqrt
[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(8*c^(5/2)*Sqrt[b*c - a*d])

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 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 386

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[(-x)*(a + b*x^n)^(p + 1)*(
(c + d*x^n)^q/(a*n*(p + 1))), x] - Dist[c*(q/(a*(p + 1))), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {x \left (a+b x^2\right )^{3/2}}{4 c \left (c+d x^2\right )^2}+\frac {(3 a) \int \frac {\sqrt {a+b x^2}}{\left (c+d x^2\right )^2} \, dx}{4 c} \\ & = \frac {x \left (a+b x^2\right )^{3/2}}{4 c \left (c+d x^2\right )^2}+\frac {3 a x \sqrt {a+b x^2}}{8 c^2 \left (c+d x^2\right )}+\frac {\left (3 a^2\right ) \int \frac {1}{\sqrt {a+b x^2} \left (c+d x^2\right )} \, dx}{8 c^2} \\ & = \frac {x \left (a+b x^2\right )^{3/2}}{4 c \left (c+d x^2\right )^2}+\frac {3 a x \sqrt {a+b x^2}}{8 c^2 \left (c+d x^2\right )}+\frac {\left (3 a^2\right ) \text {Subst}\left (\int \frac {1}{c-(b c-a d) x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{8 c^2} \\ & = \frac {x \left (a+b x^2\right )^{3/2}}{4 c \left (c+d x^2\right )^2}+\frac {3 a x \sqrt {a+b x^2}}{8 c^2 \left (c+d x^2\right )}+\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{8 c^{5/2} \sqrt {b c-a d}} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 10.51 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^3} \, dx=\frac {x \sqrt {a+b x^2} \left (\frac {\sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \left (5 a c+2 b c x^2+3 a d x^2\right )}{\left (c+d x^2\right ) \sqrt {1+\frac {d x^2}{c}}}+\frac {3 a \arcsin \left (\frac {\sqrt {\left (-\frac {b}{a}+\frac {d}{c}\right ) x^2}}{\sqrt {1+\frac {d x^2}{c}}}\right )}{\sqrt {\frac {(-b c+a d) x^2}{a c}}}\right )}{8 c^3 \sqrt {1+\frac {b x^2}{a}}} \]

[In]

Integrate[(a + b*x^2)^(3/2)/(c + d*x^2)^3,x]

[Out]

(x*Sqrt[a + b*x^2]*((Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*(5*a*c + 2*b*c*x^2 + 3*a*d*x^2))/((c + d*x^2)*Sqrt[
1 + (d*x^2)/c]) + (3*a*ArcSin[Sqrt[(-(b/a) + d/c)*x^2]/Sqrt[1 + (d*x^2)/c]])/Sqrt[((-(b*c) + a*d)*x^2)/(a*c)])
)/(8*c^3*Sqrt[1 + (b*x^2)/a])

Maple [A] (verified)

Time = 2.50 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.83

method	result	size
pseudoelliptic	\(-\frac {a^{2} \left (-\frac {\sqrt {b \,x^{2}+a}\, \left (3 a d \,x^{2}+2 c b \,x^{2}+5 a c \right ) x}{a^{2} \left (d \,x^{2}+c \right )^{2}}+\frac {3 \arctan \left (\frac {c \sqrt {b \,x^{2}+a}}{x \sqrt {\left (a d -b c \right ) c}}\right )}{\sqrt {\left (a d -b c \right ) c}}\right )}{8 c^{2}}\)	\(94\)
default	\(\text {Expression too large to display}\)	\(6921\)

[In]

int((b*x^2+a)^(3/2)/(d*x^2+c)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (93) = 186\).

Time = 0.33 (sec) , antiderivative size = 526, normalized size of antiderivative = 4.65 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^3} \, dx=\left [\frac {3 \, {\left (a^{2} d^{2} x^{4} + 2 \, a^{2} c d x^{2} + a^{2} c^{2}\right )} \sqrt {b c^{2} - a c d} \log \left (\frac {{\left (8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} + 2 \, {\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2} + 4 \, {\left ({\left (2 \, b c - a d\right )} x^{3} + a c x\right )} \sqrt {b c^{2} - a c d} \sqrt {b x^{2} + a}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{2}}\right ) + 4 \, {\left ({\left (2 \, b^{2} c^{3} + a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3} + 5 \, {\left (a b c^{3} - a^{2} c^{2} d\right )} x\right )} \sqrt {b x^{2} + a}}{32 \, {\left (b c^{6} - a c^{5} d + {\left (b c^{4} d^{2} - a c^{3} d^{3}\right )} x^{4} + 2 \, {\left (b c^{5} d - a c^{4} d^{2}\right )} x^{2}\right )}}, -\frac {3 \, {\left (a^{2} d^{2} x^{4} + 2 \, a^{2} c d x^{2} + a^{2} c^{2}\right )} \sqrt {-b c^{2} + a c d} \arctan \left (\frac {\sqrt {-b c^{2} + a c d} {\left ({\left (2 \, b c - a d\right )} x^{2} + a c\right )} \sqrt {b x^{2} + a}}{2 \, {\left ({\left (b^{2} c^{2} - a b c d\right )} x^{3} + {\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) - 2 \, {\left ({\left (2 \, b^{2} c^{3} + a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3} + 5 \, {\left (a b c^{3} - a^{2} c^{2} d\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, {\left (b c^{6} - a c^{5} d + {\left (b c^{4} d^{2} - a c^{3} d^{3}\right )} x^{4} + 2 \, {\left (b c^{5} d - a c^{4} d^{2}\right )} x^{2}\right )}}\right ] \]

[In]

integrate((b*x^2+a)^(3/2)/(d*x^2+c)^3,x, algorithm="fricas")

[Out]

[1/32*(3*(a^2*d^2*x^4 + 2*a^2*c*d*x^2 + a^2*c^2)*sqrt(b*c^2 - a*c*d)*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*((2*b*c - a*d)*x^3 + a*c*x)*sqrt(b*c^2 - a*c*d)*sqrt(b*x^2 + a
))/(d^2*x^4 + 2*c*d*x^2 + c^2)) + 4*((2*b^2*c^3 + a*b*c^2*d - 3*a^2*c*d^2)*x^3 + 5*(a*b*c^3 - a^2*c^2*d)*x)*sq
rt(b*x^2 + a))/(b*c^6 - a*c^5*d + (b*c^4*d^2 - a*c^3*d^3)*x^4 + 2*(b*c^5*d - a*c^4*d^2)*x^2), -1/16*(3*(a^2*d^
2*x^4 + 2*a^2*c*d*x^2 + a^2*c^2)*sqrt(-b*c^2 + a*c*d)*arctan(1/2*sqrt(-b*c^2 + a*c*d)*((2*b*c - a*d)*x^2 + a*c
)*sqrt(b*x^2 + a)/((b^2*c^2 - a*b*c*d)*x^3 + (a*b*c^2 - a^2*c*d)*x)) - 2*((2*b^2*c^3 + a*b*c^2*d - 3*a^2*c*d^2
)*x^3 + 5*(a*b*c^3 - a^2*c^2*d)*x)*sqrt(b*x^2 + a))/(b*c^6 - a*c^5*d + (b*c^4*d^2 - a*c^3*d^3)*x^4 + 2*(b*c^5*
d - a*c^4*d^2)*x^2)]

Sympy [F]

\[ \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^3} \, dx=\int \frac {\left (a + b x^{2}\right )^{\frac {3}{2}}}{\left (c + d x^{2}\right )^{3}}\, dx \]

[In]

integrate((b*x**2+a)**(3/2)/(d*x**2+c)**3,x)

[Out]

Integral((a + b*x**2)**(3/2)/(c + d*x**2)**3, x)

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]

[In]

integrate((b*x^2+a)^(3/2)/(d*x^2+c)^3,x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(3/2)/(d*x^2 + c)^3, x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 451 vs. \(2 (93) = 186\).

Time = 1.67 (sec) , antiderivative size = 451, normalized size of antiderivative = 3.99 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^3} \, dx=-\frac {3 \, a^{2} \sqrt {b} \arctan \left (\frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} d + 2 \, b c - a d}{2 \, \sqrt {-b^{2} c^{2} + a b c d}}\right )}{8 \, \sqrt {-b^{2} c^{2} + a b c d} c^{2}} + \frac {8 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} b^{\frac {5}{2}} c^{2} d - 3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a^{2} \sqrt {b} d^{3} + 16 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} b^{\frac {7}{2}} c^{3} + 8 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a b^{\frac {5}{2}} c^{2} d - 18 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{2} b^{\frac {3}{2}} c d^{2} + 9 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{3} \sqrt {b} d^{3} + 8 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{2} b^{\frac {5}{2}} c^{2} d + 16 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{3} b^{\frac {3}{2}} c d^{2} - 9 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{4} \sqrt {b} d^{3} + 2 \, a^{4} b^{\frac {3}{2}} c d^{2} + 3 \, a^{5} \sqrt {b} d^{3}}{4 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} d + 4 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} b c - 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a d + a^{2} d\right )}^{2} c^{2} d^{2}} \]

[In]

integrate((b*x^2+a)^(3/2)/(d*x^2+c)^3,x, algorithm="giac")

[Out]

-3/8*a^2*sqrt(b)*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^2) + 1/4*(8*(sqrt(b)*x - sqrt(b*x^2 + a))^6*b^(5/2)*c^2*d - 3*(sqrt(b)*x - sqrt(b*x^2 +
a))^6*a^2*sqrt(b)*d^3 + 16*(sqrt(b)*x - sqrt(b*x^2 + a))^4*b^(7/2)*c^3 + 8*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a*b
^(5/2)*c^2*d - 18*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^2*b^(3/2)*c*d^2 + 9*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^3*sq
rt(b)*d^3 + 8*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^2*b^(5/2)*c^2*d + 16*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^3*b^(3/
2)*c*d^2 - 9*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^4*sqrt(b)*d^3 + 2*a^4*b^(3/2)*c*d^2 + 3*a^5*sqrt(b)*d^3)/(((sqr
t(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)^2*c^2*d^2)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^3} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{3/2}}{{\left (d\,x^2+c\right )}^3} \,d x \]

[In]

int((a + b*x^2)^(3/2)/(c + d*x^2)^3,x)

[Out]

int((a + b*x^2)^(3/2)/(c + d*x^2)^3, x)