\(\int \frac {1}{(a+b x^2)^{5/2} (c+d x^2)} \, dx\) [94]

Optimal result

Integrand size = 21, antiderivative size = 122 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )} \, dx=\frac {b x}{3 a (b c-a d) \left (a+b x^2\right )^{3/2}}+\frac {b (2 b c-5 a d) x}{3 a^2 (b c-a d)^2 \sqrt {a+b x^2}}+\frac {d^2 \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{\sqrt {c} (b c-a d)^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {425, 541, 12, 385, 214} \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )} \, dx=\frac {b x (2 b c-5 a d)}{3 a^2 \sqrt {a+b x^2} (b c-a d)^2}+\frac {d^2 \text {arctanh}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{\sqrt {c} (b c-a d)^{5/2}}+\frac {b x}{3 a \left (a+b x^2\right )^{3/2} (b c-a d)} \]

[In]

Int[1/((a + b*x^2)^(5/2)*(c + d*x^2)),x]

[Out]

(b*x)/(3*a*(b*c - a*d)*(a + b*x^2)^(3/2)) + (b*(2*b*c - 5*a*d)*x)/(3*a^2*(b*c - a*d)^2*Sqrt[a + b*x^2]) + (d^2
*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(Sqrt[c]*(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 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 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, 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 {b x}{3 a (b c-a d) \left (a+b x^2\right )^{3/2}}-\frac {\int \frac {-2 b c+3 a d-2 b d x^2}{\left (a+b x^2\right )^{3/2} \left (c+d x^2\right )} \, dx}{3 a (b c-a d)} \\ & = \frac {b x}{3 a (b c-a d) \left (a+b x^2\right )^{3/2}}+\frac {b (2 b c-5 a d) x}{3 a^2 (b c-a d)^2 \sqrt {a+b x^2}}+\frac {\int \frac {3 a^2 d^2}{\sqrt {a+b x^2} \left (c+d x^2\right )} \, dx}{3 a^2 (b c-a d)^2} \\ & = \frac {b x}{3 a (b c-a d) \left (a+b x^2\right )^{3/2}}+\frac {b (2 b c-5 a d) x}{3 a^2 (b c-a d)^2 \sqrt {a+b x^2}}+\frac {d^2 \int \frac {1}{\sqrt {a+b x^2} \left (c+d x^2\right )} \, dx}{(b c-a d)^2} \\ & = \frac {b x}{3 a (b c-a d) \left (a+b x^2\right )^{3/2}}+\frac {b (2 b c-5 a d) x}{3 a^2 (b c-a d)^2 \sqrt {a+b x^2}}+\frac {d^2 \text {Subst}\left (\int \frac {1}{c-(b c-a d) x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{(b c-a d)^2} \\ & = \frac {b x}{3 a (b c-a d) \left (a+b x^2\right )^{3/2}}+\frac {b (2 b c-5 a d) x}{3 a^2 (b c-a d)^2 \sqrt {a+b x^2}}+\frac {d^2 \tanh ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{\sqrt {c} (b c-a d)^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )} \, dx=\frac {b x \left (-6 a^2 d+2 b^2 c x^2+a b \left (3 c-5 d x^2\right )\right )}{3 a^2 (b c-a d)^2 \left (a+b x^2\right )^{3/2}}-\frac {d^2 \arctan \left (\frac {-d x \sqrt {a+b x^2}+\sqrt {b} \left (c+d x^2\right )}{\sqrt {c} \sqrt {-b c+a d}}\right )}{\sqrt {c} (-b c+a d)^{5/2}} \]

[In]

Integrate[1/((a + b*x^2)^(5/2)*(c + d*x^2)),x]

[Out]

(b*x*(-6*a^2*d + 2*b^2*c*x^2 + a*b*(3*c - 5*d*x^2)))/(3*a^2*(b*c - a*d)^2*(a + b*x^2)^(3/2)) - (d^2*ArcTan[(-(
d*x*Sqrt[a + b*x^2]) + Sqrt[b]*(c + d*x^2))/(Sqrt[c]*Sqrt[-(b*c) + a*d])])/(Sqrt[c]*(-(b*c) + a*d)^(5/2))

Maple [A] (verified)

Time = 2.40 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02

method	result	size
pseudoelliptic	\(-\frac {a^{2} d^{2} \arctan \left (\frac {c \sqrt {b \,x^{2}+a}}{x \sqrt {\left (a d -b c \right ) c}}\right ) \left (b \,x^{2}+a \right )^{\frac {3}{2}}+2 x \sqrt {\left (a d -b c \right ) c}\, b \left (a^{2} d -\frac {\left (-\frac {5 d \,x^{2}}{3}+c \right ) b a}{2}-\frac {b^{2} c \,x^{2}}{3}\right )}{\left (b \,x^{2}+a \right )^{\frac {3}{2}} \sqrt {\left (a d -b c \right ) c}\, \left (a d -b c \right )^{2} a^{2}}\)	\(124\)
default	\(\text {Expression too large to display}\)	\(1378\)

[In]

int(1/(b*x^2+a)^(5/2)/(d*x^2+c),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 362 vs. \(2 (104) = 208\).

Time = 0.48 (sec) , antiderivative size = 764, normalized size of antiderivative = 6.26 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )} \, dx=\left [\frac {3 \, {\left (a^{2} b^{2} d^{2} x^{4} + 2 \, a^{3} b d^{2} x^{2} + a^{4} d^{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^{4} c^{3} - 7 \, a b^{3} c^{2} d + 5 \, a^{2} b^{2} c d^{2}\right )} x^{3} + 3 \, {\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 2 \, a^{3} b c d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{12 \, {\left (a^{4} b^{3} c^{4} - 3 \, a^{5} b^{2} c^{3} d + 3 \, a^{6} b c^{2} d^{2} - a^{7} c d^{3} + {\left (a^{2} b^{5} c^{4} - 3 \, a^{3} b^{4} c^{3} d + 3 \, a^{4} b^{3} c^{2} d^{2} - a^{5} b^{2} c d^{3}\right )} x^{4} + 2 \, {\left (a^{3} b^{4} c^{4} - 3 \, a^{4} b^{3} c^{3} d + 3 \, a^{5} b^{2} c^{2} d^{2} - a^{6} b c d^{3}\right )} x^{2}\right )}}, -\frac {3 \, {\left (a^{2} b^{2} d^{2} x^{4} + 2 \, a^{3} b d^{2} x^{2} + a^{4} d^{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^{4} c^{3} - 7 \, a b^{3} c^{2} d + 5 \, a^{2} b^{2} c d^{2}\right )} x^{3} + 3 \, {\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 2 \, a^{3} b c d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{4} b^{3} c^{4} - 3 \, a^{5} b^{2} c^{3} d + 3 \, a^{6} b c^{2} d^{2} - a^{7} c d^{3} + {\left (a^{2} b^{5} c^{4} - 3 \, a^{3} b^{4} c^{3} d + 3 \, a^{4} b^{3} c^{2} d^{2} - a^{5} b^{2} c d^{3}\right )} x^{4} + 2 \, {\left (a^{3} b^{4} c^{4} - 3 \, a^{4} b^{3} c^{3} d + 3 \, a^{5} b^{2} c^{2} d^{2} - a^{6} b c d^{3}\right )} x^{2}\right )}}\right ] \]

[In]

integrate(1/(b*x^2+a)^(5/2)/(d*x^2+c),x, algorithm="fricas")

[Out]

[1/12*(3*(a^2*b^2*d^2*x^4 + 2*a^3*b*d^2*x^2 + a^4*d^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^4*c^3 - 7*a*b^3*c^2*d + 5*a^2*b^2*c*d^2)*x^3 + 3*(a*b^3*c^3 -
3*a^2*b^2*c^2*d + 2*a^3*b*c*d^2)*x)*sqrt(b*x^2 + a))/(a^4*b^3*c^4 - 3*a^5*b^2*c^3*d + 3*a^6*b*c^2*d^2 - a^7*c*
d^3 + (a^2*b^5*c^4 - 3*a^3*b^4*c^3*d + 3*a^4*b^3*c^2*d^2 - a^5*b^2*c*d^3)*x^4 + 2*(a^3*b^4*c^4 - 3*a^4*b^3*c^3
*d + 3*a^5*b^2*c^2*d^2 - a^6*b*c*d^3)*x^2), -1/6*(3*(a^2*b^2*d^2*x^4 + 2*a^3*b*d^2*x^2 + a^4*d^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^4*c^3 - 7*a*b^3*c^2*d + 5*a^2*b^2*c*d^2)*x^3 + 3*(a*b^3*c^3 - 3*a^2*b^2*c^2*
d + 2*a^3*b*c*d^2)*x)*sqrt(b*x^2 + a))/(a^4*b^3*c^4 - 3*a^5*b^2*c^3*d + 3*a^6*b*c^2*d^2 - a^7*c*d^3 + (a^2*b^5
*c^4 - 3*a^3*b^4*c^3*d + 3*a^4*b^3*c^2*d^2 - a^5*b^2*c*d^3)*x^4 + 2*(a^3*b^4*c^4 - 3*a^4*b^3*c^3*d + 3*a^5*b^2
*c^2*d^2 - a^6*b*c*d^3)*x^2)]

Sympy [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )} \, dx=\int \frac {1}{\left (a + b x^{2}\right )^{\frac {5}{2}} \left (c + d x^{2}\right )}\, dx \]

[In]

integrate(1/(b*x**2+a)**(5/2)/(d*x**2+c),x)

[Out]

Integral(1/((a + b*x**2)**(5/2)*(c + d*x**2)), x)

Maxima [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} {\left (d x^{2} + c\right )}} \,d x } \]

[In]

integrate(1/(b*x^2+a)^(5/2)/(d*x^2+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(5/2)*(d*x^2 + c)), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (104) = 208\).

Time = 0.28 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.62 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )} \, dx=-\frac {\sqrt {b} d^{2} \arctan \left (\frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} d + 2 \, b c - a d}{2 \, \sqrt {-b^{2} c^{2} + a b c d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b^{2} c^{2} + a b c d}} + \frac {{\left (\frac {{\left (2 \, b^{6} c^{3} - 9 \, a b^{5} c^{2} d + 12 \, a^{2} b^{4} c d^{2} - 5 \, a^{3} b^{3} d^{3}\right )} x^{2}}{a^{2} b^{5} c^{4} - 4 \, a^{3} b^{4} c^{3} d + 6 \, a^{4} b^{3} c^{2} d^{2} - 4 \, a^{5} b^{2} c d^{3} + a^{6} b d^{4}} + \frac {3 \, {\left (a b^{5} c^{3} - 4 \, a^{2} b^{4} c^{2} d + 5 \, a^{3} b^{3} c d^{2} - 2 \, a^{4} b^{2} d^{3}\right )}}{a^{2} b^{5} c^{4} - 4 \, a^{3} b^{4} c^{3} d + 6 \, a^{4} b^{3} c^{2} d^{2} - 4 \, a^{5} b^{2} c d^{3} + a^{6} b d^{4}}\right )} x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} \]

[In]

integrate(1/(b*x^2+a)^(5/2)/(d*x^2+c),x, algorithm="giac")

[Out]

-sqrt(b)*d^2*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))/((b^2*c^2
- 2*a*b*c*d + a^2*d^2)*sqrt(-b^2*c^2 + a*b*c*d)) + 1/3*((2*b^6*c^3 - 9*a*b^5*c^2*d + 12*a^2*b^4*c*d^2 - 5*a^3*
b^3*d^3)*x^2/(a^2*b^5*c^4 - 4*a^3*b^4*c^3*d + 6*a^4*b^3*c^2*d^2 - 4*a^5*b^2*c*d^3 + a^6*b*d^4) + 3*(a*b^5*c^3
- 4*a^2*b^4*c^2*d + 5*a^3*b^3*c*d^2 - 2*a^4*b^2*d^3)/(a^2*b^5*c^4 - 4*a^3*b^4*c^3*d + 6*a^4*b^3*c^2*d^2 - 4*a^
5*b^2*c*d^3 + a^6*b*d^4))*x/(b*x^2 + a)^(3/2)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^{5/2}\,\left (d\,x^2+c\right )} \,d x \]

[In]

int(1/((a + b*x^2)^(5/2)*(c + d*x^2)),x)

[Out]

int(1/((a + b*x^2)^(5/2)*(c + d*x^2)), x)