\(\int \frac {x^4}{(a+b x^2)^2 (c+d x^2)^{3/2}} \, dx\) [767]

Optimal result

Integrand size = 24, antiderivative size = 130 \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\frac {(2 b c+a d) x}{2 b (b c-a d)^2 \sqrt {c+d x^2}}+\frac {a x}{2 b (b c-a d) \left (a+b x^2\right ) \sqrt {c+d x^2}}-\frac {3 \sqrt {a} c \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 (b c-a d)^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {481, 541, 12, 385, 211} \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=-\frac {3 \sqrt {a} c \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 (b c-a d)^{5/2}}+\frac {x (a d+2 b c)}{2 b \sqrt {c+d x^2} (b c-a d)^2}+\frac {a x}{2 b \left (a+b x^2\right ) \sqrt {c+d x^2} (b c-a d)} \]

[In]

Int[x^4/((a + b*x^2)^2*(c + d*x^2)^(3/2)),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[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 481

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-a)*e^(
2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Dist[e^
(2*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1)
+ (a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(672\) vs. \(2(130)=260\).

Time = 10.80 (sec) , antiderivative size = 672, normalized size of antiderivative = 5.17 \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\frac {1}{2} \left (-\frac {x \left (3 a c+2 b c x^2+a d x^2\right ) \left (4 c^2+5 c d x^2+d^2 x^4-4 c^{3/2} \sqrt {c+d x^2}-3 \sqrt {c} d x^2 \sqrt {c+d x^2}\right )}{(b c-a d)^2 \left (a+b x^2\right ) \left (c+d x^2\right ) \left (4 c^{3/2}+3 \sqrt {c} d x^2-4 c \sqrt {c+d x^2}-d x^2 \sqrt {c+d x^2}\right )}+\frac {3 \sqrt {a} c \arctan \left (\frac {\sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (\sqrt {c}-\sqrt {c+d x^2}\right )}\right )}{(b c-a d)^2 \sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}}}+\frac {3 \sqrt {a} \sqrt {b} c^{3/2} \arctan \left (\frac {\sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (\sqrt {c}-\sqrt {c+d x^2}\right )}\right )}{(b c-a d)^{5/2} \sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}}}+\frac {3 \sqrt {a} c \arctan \left (\frac {\sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (\sqrt {c}-\sqrt {c+d x^2}\right )}\right )}{(b c-a d)^2 \sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}}}+\frac {3 \sqrt {a} \sqrt {b} c^{3/2} \arctan \left (\frac {\sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (-\sqrt {c}+\sqrt {c+d x^2}\right )}\right )}{(b c-a d)^{5/2} \sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}}}\right ) \]

[In]

Integrate[x^4/((a + b*x^2)^2*(c + d*x^2)^(3/2)),x]

[Out]

(-((x*(3*a*c + 2*b*c*x^2 + a*d*x^2)*(4*c^2 + 5*c*d*x^2 + d^2*x^4 - 4*c^(3/2)*Sqrt[c + d*x^2] - 3*Sqrt[c]*d*x^2
*Sqrt[c + d*x^2]))/((b*c - a*d)^2*(a + b*x^2)*(c + d*x^2)*(4*c^(3/2) + 3*Sqrt[c]*d*x^2 - 4*c*Sqrt[c + d*x^2] -
d*x^2*Sqrt[c + d*x^2]))) + (3*Sqrt[a]*c*ArcTan[(Sqrt[2*b*c - a*d - 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*x)/(Sqr
t[a]*(Sqrt[c] - Sqrt[c + d*x^2]))])/((b*c - a*d)^2*Sqrt[2*b*c - a*d - 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]) + (3
*Sqrt[a]*Sqrt[b]*c^(3/2)*ArcTan[(Sqrt[2*b*c - a*d + 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*x)/(Sqrt[a]*(Sqrt[c] -
Sqrt[c + d*x^2]))])/((b*c - a*d)^(5/2)*Sqrt[2*b*c - a*d + 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]) + (3*Sqrt[a]*c*A
rcTan[(Sqrt[2*b*c - a*d + 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*x)/(Sqrt[a]*(Sqrt[c] - Sqrt[c + d*x^2]))])/((b*c
- a*d)^2*Sqrt[2*b*c - a*d + 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]) + (3*Sqrt[a]*Sqrt[b]*c^(3/2)*ArcTan[(Sqrt[2*b*
c - a*d - 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*x)/(Sqrt[a]*(-Sqrt[c] + Sqrt[c + d*x^2]))])/((b*c - a*d)^(5/2)*Sq
rt[2*b*c - a*d - 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]))/2

Maple [A] (verified)

Time = 3.00 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.73

method	result	size
pseudoelliptic	\(-\frac {c \left (-a \left (\frac {\sqrt {d \,x^{2}+c}\, x}{c \left (b \,x^{2}+a \right )}-\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}\right )-\frac {2 x}{\sqrt {d \,x^{2}+c}}\right )}{2 \left (a d -b c \right )^{2}}\)	\(95\)
default	\(\text {Expression too large to display}\)	\(1943\)

[In]

int(x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (110) = 220\).

Time = 0.43 (sec) , antiderivative size = 552, normalized size of antiderivative = 4.25 \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (b c d x^{4} + a c^{2} + {\left (b c^{2} + a c d\right )} x^{2}\right )} \sqrt {-\frac {a}{b c - a d}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{3} - {\left (a b c^{2} - a^{2} c d\right )} x\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {a}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left ({\left (2 \, b c + a d\right )} x^{3} + 3 \, a c x\right )} \sqrt {d x^{2} + c}}{8 \, {\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{4} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{2}\right )}}, \frac {3 \, {\left (b c d x^{4} + a c^{2} + {\left (b c^{2} + a c d\right )} x^{2}\right )} \sqrt {\frac {a}{b c - a d}} \arctan \left (-\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {a}{b c - a d}}}{2 \, {\left (a d x^{3} + a c x\right )}}\right ) + 2 \, {\left ({\left (2 \, b c + a d\right )} x^{3} + 3 \, a c x\right )} \sqrt {d x^{2} + c}}{4 \, {\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{4} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{2}\right )}}\right ] \]

[In]

integrate(x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(3*(b*c*d*x^4 + a*c^2 + (b*c^2 + a*c*d)*x^2)*sqrt(-a/(b*c - a*d))*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*((b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^3 - (a*b*c^2 - a^2*c*d)
*x)*sqrt(d*x^2 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*((2*b*c + a*d)*x^3 + 3*a*c*x)*sqrt(
d*x^2 + c))/(a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^4 + (b^3*c^3 -
a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x^2), 1/4*(3*(b*c*d*x^4 + a*c^2 + (b*c^2 + a*c*d)*x^2)*sqrt(a/(b*c - a*d)
)*arctan(-1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sqrt(a/(b*c - a*d))/(a*d*x^3 + a*c*x)) + 2*((2*b*c + a
*d)*x^3 + 3*a*c*x)*sqrt(d*x^2 + c))/(a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*
b*d^3)*x^4 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x^2)]

Sympy [F]

\[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\int \frac {x^{4}}{\left (a + b x^{2}\right )^{2} \left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x**4/(b*x**2+a)**2/(d*x**2+c)**(3/2),x)

[Out]

Integral(x**4/((a + b*x**2)**2*(c + d*x**2)**(3/2)), x)

Maxima [F]

\[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\int { \frac {x^{4}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^4/((b*x^2 + a)^2*(d*x^2 + c)^(3/2)), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (110) = 220\).

Time = 0.90 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.29 \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\frac {3 \, a c \sqrt {d} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {a b c d - a^{2} d^{2}}} + \frac {c x}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {d x^{2} + c}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} d^{\frac {3}{2}} - a b c^{2} \sqrt {d}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )}} \]

[In]

integrate(x^4/(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

3/2*a*c*sqrt(d)*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))/((b^2*c^
2 - 2*a*b*c*d + a^2*d^2)*sqrt(a*b*c*d - a^2*d^2)) + c*x/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(d*x^2 + c)) - ((
sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c*sqrt(d) - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*d^(3/2) - a*b*c^2*sqrt(d)
)/(((sqrt(d)*x - sqrt(d*x^2 + c))^4*b - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c + 4*(sqrt(d)*x - sqrt(d*x^2 + c)
)^2*a*d + b*c^2)*(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx=\int \frac {x^4}{{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{3/2}} \,d x \]

[In]

int(x^4/((a + b*x^2)^2*(c + d*x^2)^(3/2)),x)

[Out]

int(x^4/((a + b*x^2)^2*(c + d*x^2)^(3/2)), x)