3.988 \(\int \frac{x}{\left (a+b x^2\right )^{7/2} \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=113 \[ -\frac{8 d^2 \sqrt{c+d x^2}}{15 \sqrt{a+b x^2} (b c-a d)^3}+\frac{4 d \sqrt{c+d x^2}}{15 \left (a+b x^2\right )^{3/2} (b c-a d)^2}-\frac{\sqrt{c+d x^2}}{5 \left (a+b x^2\right )^{5/2} (b c-a d)} \]

[Out]

-Sqrt[c + d*x^2]/(5*(b*c - a*d)*(a + b*x^2)^(5/2)) + (4*d*Sqrt[c + d*x^2])/(15*(
b*c - a*d)^2*(a + b*x^2)^(3/2)) - (8*d^2*Sqrt[c + d*x^2])/(15*(b*c - a*d)^3*Sqrt
[a + b*x^2])

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Rubi [A]  time = 0.195863, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125 \[ -\frac{8 d^2 \sqrt{c+d x^2}}{15 \sqrt{a+b x^2} (b c-a d)^3}+\frac{4 d \sqrt{c+d x^2}}{15 \left (a+b x^2\right )^{3/2} (b c-a d)^2}-\frac{\sqrt{c+d x^2}}{5 \left (a+b x^2\right )^{5/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]  Int[x/((a + b*x^2)^(7/2)*Sqrt[c + d*x^2]),x]

[Out]

-Sqrt[c + d*x^2]/(5*(b*c - a*d)*(a + b*x^2)^(5/2)) + (4*d*Sqrt[c + d*x^2])/(15*(
b*c - a*d)^2*(a + b*x^2)^(3/2)) - (8*d^2*Sqrt[c + d*x^2])/(15*(b*c - a*d)^3*Sqrt
[a + b*x^2])

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Rubi in Sympy [A]  time = 23.1907, size = 97, normalized size = 0.86 \[ \frac{8 d^{2} \sqrt{c + d x^{2}}}{15 \sqrt{a + b x^{2}} \left (a d - b c\right )^{3}} + \frac{4 d \sqrt{c + d x^{2}}}{15 \left (a + b x^{2}\right )^{\frac{3}{2}} \left (a d - b c\right )^{2}} + \frac{\sqrt{c + d x^{2}}}{5 \left (a + b x^{2}\right )^{\frac{5}{2}} \left (a d - b c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x/(b*x**2+a)**(7/2)/(d*x**2+c)**(1/2),x)

[Out]

8*d**2*sqrt(c + d*x**2)/(15*sqrt(a + b*x**2)*(a*d - b*c)**3) + 4*d*sqrt(c + d*x*
*2)/(15*(a + b*x**2)**(3/2)*(a*d - b*c)**2) + sqrt(c + d*x**2)/(5*(a + b*x**2)**
(5/2)*(a*d - b*c))

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Mathematica [A]  time = 0.101822, size = 83, normalized size = 0.73 \[ -\frac{\sqrt{c+d x^2} \left (15 a^2 d^2-10 a b d \left (c-2 d x^2\right )+b^2 \left (3 c^2-4 c d x^2+8 d^2 x^4\right )\right )}{15 \left (a+b x^2\right )^{5/2} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]  Integrate[x/((a + b*x^2)^(7/2)*Sqrt[c + d*x^2]),x]

[Out]

-(Sqrt[c + d*x^2]*(15*a^2*d^2 - 10*a*b*d*(c - 2*d*x^2) + b^2*(3*c^2 - 4*c*d*x^2
+ 8*d^2*x^4)))/(15*(b*c - a*d)^3*(a + b*x^2)^(5/2))

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Maple [A]  time = 0.01, size = 113, normalized size = 1. \[{\frac{8\,{b}^{2}{d}^{2}{x}^{4}+20\,ab{d}^{2}{x}^{2}-4\,{b}^{2}cd{x}^{2}+15\,{a}^{2}{d}^{2}-10\,cabd+3\,{b}^{2}{c}^{2}}{15\,{a}^{3}{d}^{3}-45\,{a}^{2}c{d}^{2}b+45\,a{c}^{2}d{b}^{2}-15\,{c}^{3}{b}^{3}}\sqrt{d{x}^{2}+c} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x)

[Out]

1/15*(d*x^2+c)^(1/2)*(8*b^2*d^2*x^4+20*a*b*d^2*x^2-4*b^2*c*d*x^2+15*a^2*d^2-10*a
*b*c*d+3*b^2*c^2)/(b*x^2+a)^(5/2)/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x/((b*x^2 + a)^(7/2)*sqrt(d*x^2 + c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.335065, size = 350, normalized size = 3.1 \[ -\frac{{\left (8 \, b^{2} d^{2} x^{4} + 3 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2} - 4 \,{\left (b^{2} c d - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{15 \,{\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} +{\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{6} + 3 \,{\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{4} + 3 \,{\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x/((b*x^2 + a)^(7/2)*sqrt(d*x^2 + c)),x, algorithm="fricas")

[Out]

-1/15*(8*b^2*d^2*x^4 + 3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2 - 4*(b^2*c*d - 5*a*b*
d^2)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)/(a^3*b^3*c^3 - 3*a^4*b^2*c^2*d + 3*a^5
*b*c*d^2 - a^6*d^3 + (b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*c*d^2 - a^3*b^3*d^3)*x
^6 + 3*(a*b^5*c^3 - 3*a^2*b^4*c^2*d + 3*a^3*b^3*c*d^2 - a^4*b^2*d^3)*x^4 + 3*(a^
2*b^4*c^3 - 3*a^3*b^3*c^2*d + 3*a^4*b^2*c*d^2 - a^5*b*d^3)*x^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{x}{\left (a + b x^{2}\right )^{\frac{7}{2}} \sqrt{c + d x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x/(b*x**2+a)**(7/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(x/((a + b*x**2)**(7/2)*sqrt(c + d*x**2)), x)

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GIAC/XCAS [A]  time = 0.264999, size = 328, normalized size = 2.9 \[ -\frac{16 \,{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 5 \,{\left (\sqrt{b x^{2} + a} \sqrt{b d} - \sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} b^{2} c + 5 \,{\left (\sqrt{b x^{2} + a} \sqrt{b d} - \sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a b d + 10 \,{\left (\sqrt{b x^{2} + a} \sqrt{b d} - \sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d}\right )}^{4}\right )} \sqrt{b d} b^{3} d^{2}}{15 \,{\left (b^{2} c - a b d -{\left (\sqrt{b x^{2} + a} \sqrt{b d} - \sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}\right )}^{5}{\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x/((b*x^2 + a)^(7/2)*sqrt(d*x^2 + c)),x, algorithm="giac")

[Out]

-16/15*(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 5*(sqrt(b*x^2 + a)*sqrt(b*d) - sqr
t(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*b^2*c + 5*(sqrt(b*x^2 + a)*sqrt(b*d) - sqr
t(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a*b*d + 10*(sqrt(b*x^2 + a)*sqrt(b*d) - sq
rt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4)*sqrt(b*d)*b^3*d^2/((b^2*c - a*b*d - (sqr
t(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2)^5*abs(b))