3.956 \(\int \frac{x \left (a+b x^2\right )^{5/2}}{\sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=164 \[ -\frac{5 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{16 \sqrt{b} d^{7/2}}+\frac{5 \sqrt{a+b x^2} \sqrt{c+d x^2} (b c-a d)^2}{16 d^3}-\frac{5 \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2} (b c-a d)}{24 d^2}+\frac{\left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{6 d} \]

[Out]

(5*(b*c - a*d)^2*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(16*d^3) - (5*(b*c - a*d)*(a +
b*x^2)^(3/2)*Sqrt[c + d*x^2])/(24*d^2) + ((a + b*x^2)^(5/2)*Sqrt[c + d*x^2])/(6
*d) - (5*(b*c - a*d)^3*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2
])])/(16*Sqrt[b]*d^(7/2))

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Rubi [A]  time = 0.33064, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167 \[ -\frac{5 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b} \sqrt{c+d x^2}}\right )}{16 \sqrt{b} d^{7/2}}+\frac{5 \sqrt{a+b x^2} \sqrt{c+d x^2} (b c-a d)^2}{16 d^3}-\frac{5 \left (a+b x^2\right )^{3/2} \sqrt{c+d x^2} (b c-a d)}{24 d^2}+\frac{\left (a+b x^2\right )^{5/2} \sqrt{c+d x^2}}{6 d} \]

Antiderivative was successfully verified.

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

[Out]

(5*(b*c - a*d)^2*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(16*d^3) - (5*(b*c - a*d)*(a +
b*x^2)^(3/2)*Sqrt[c + d*x^2])/(24*d^2) + ((a + b*x^2)^(5/2)*Sqrt[c + d*x^2])/(6
*d) - (5*(b*c - a*d)^3*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2
])])/(16*Sqrt[b]*d^(7/2))

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Rubi in Sympy [A]  time = 35.8391, size = 146, normalized size = 0.89 \[ \frac{\left (a + b x^{2}\right )^{\frac{5}{2}} \sqrt{c + d x^{2}}}{6 d} + \frac{5 \left (a + b x^{2}\right )^{\frac{3}{2}} \sqrt{c + d x^{2}} \left (a d - b c\right )}{24 d^{2}} + \frac{5 \sqrt{a + b x^{2}} \sqrt{c + d x^{2}} \left (a d - b c\right )^{2}}{16 d^{3}} + \frac{5 \left (a d - b c\right )^{3} \operatorname{atanh}{\left (\frac{\sqrt{b} \sqrt{c + d x^{2}}}{\sqrt{d} \sqrt{a + b x^{2}}} \right )}}{16 \sqrt{b} d^{\frac{7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

(a + b*x**2)**(5/2)*sqrt(c + d*x**2)/(6*d) + 5*(a + b*x**2)**(3/2)*sqrt(c + d*x*
*2)*(a*d - b*c)/(24*d**2) + 5*sqrt(a + b*x**2)*sqrt(c + d*x**2)*(a*d - b*c)**2/(
16*d**3) + 5*(a*d - b*c)**3*atanh(sqrt(b)*sqrt(c + d*x**2)/(sqrt(d)*sqrt(a + b*x
**2)))/(16*sqrt(b)*d**(7/2))

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Mathematica [A]  time = 0.141231, size = 152, normalized size = 0.93 \[ \frac{\sqrt{a+b x^2} \sqrt{c+d x^2} \left (33 a^2 d^2+2 a b d \left (13 d x^2-20 c\right )+b^2 \left (15 c^2-10 c d x^2+8 d^2 x^4\right )\right )}{48 d^3}-\frac{5 (b c-a d)^3 \log \left (2 \sqrt{b} \sqrt{d} \sqrt{a+b x^2} \sqrt{c+d x^2}+a d+b c+2 b d x^2\right )}{32 \sqrt{b} d^{7/2}} \]

Antiderivative was successfully verified.

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

[Out]

(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(33*a^2*d^2 + 2*a*b*d*(-20*c + 13*d*x^2) + b^2*
(15*c^2 - 10*c*d*x^2 + 8*d^2*x^4)))/(48*d^3) - (5*(b*c - a*d)^3*Log[b*c + a*d +
2*b*d*x^2 + 2*Sqrt[b]*Sqrt[d]*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]])/(32*Sqrt[b]*d^(7
/2))

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Maple [B]  time = 0.022, size = 529, normalized size = 3.2 \[{\frac{1}{96\,{d}^{3}}\sqrt{b{x}^{2}+a}\sqrt{d{x}^{2}+c} \left ( 16\,{x}^{4}{b}^{2}{d}^{2}\sqrt{bd}\sqrt{bd{x}^{4}+ad{x}^{2}+c{x}^{2}b+ac}+52\,\sqrt{bd{x}^{4}+ad{x}^{2}+c{x}^{2}b+ac}{x}^{2}ab{d}^{2}\sqrt{bd}-20\,\sqrt{bd{x}^{4}+ad{x}^{2}+c{x}^{2}b+ac}{x}^{2}c{b}^{2}d\sqrt{bd}+15\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+c{x}^{2}b+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{3}{d}^{3}-45\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+c{x}^{2}b+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}cb{d}^{2}+45\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+c{x}^{2}b+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){c}^{2}a{b}^{2}d-15\,{b}^{3}\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+c{x}^{2}b+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){c}^{3}+66\,\sqrt{bd{x}^{4}+ad{x}^{2}+c{x}^{2}b+ac}{a}^{2}{d}^{2}\sqrt{bd}-80\,\sqrt{bd{x}^{4}+ad{x}^{2}+c{x}^{2}b+ac}acbd\sqrt{bd}+30\,\sqrt{bd{x}^{4}+ad{x}^{2}+c{x}^{2}b+ac}{c}^{2}{b}^{2}\sqrt{bd} \right ){\frac{1}{\sqrt{bd{x}^{4}+ad{x}^{2}+c{x}^{2}b+ac}}}{\frac{1}{\sqrt{bd}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/96*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(16*x^4*b^2*d^2*(b*d)^(1/2)*(b*d*x^4+a*d*x^
2+b*c*x^2+a*c)^(1/2)+52*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^2*a*b*d^2*(b*d)^(1
/2)-20*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^2*c*b^2*d*(b*d)^(1/2)+15*ln(1/2*(2*
b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*
a^3*d^3-45*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a
*d+b*c)/(b*d)^(1/2))*a^2*c*b*d^2+45*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2
+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*c^2*a*b^2*d-15*b^3*ln(1/2*(2*b*d*x
^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*c^3+6
6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*a^2*d^2*(b*d)^(1/2)-80*(b*d*x^4+a*d*x^2+b*
c*x^2+a*c)^(1/2)*a*c*b*d*(b*d)^(1/2)+30*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*c^2*
b^2*(b*d)^(1/2))/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)/d^3/(b*d)^(1/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.271252, size = 1, normalized size = 0.01 \[ \left [\frac{4 \,{\left (8 \, b^{2} d^{2} x^{4} + 15 \, b^{2} c^{2} - 40 \, a b c d + 33 \, a^{2} d^{2} - 2 \,{\left (5 \, b^{2} c d - 13 \, a b d^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} \sqrt{b d} - 15 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (4 \,{\left (2 \, b^{2} d^{2} x^{2} + b^{2} c d + a b d^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} +{\left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )} \sqrt{b d}\right )}{192 \, \sqrt{b d} d^{3}}, \frac{2 \,{\left (8 \, b^{2} d^{2} x^{4} + 15 \, b^{2} c^{2} - 40 \, a b c d + 33 \, a^{2} d^{2} - 2 \,{\left (5 \, b^{2} c d - 13 \, a b d^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} \sqrt{-b d} - 15 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac{{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt{-b d}}{2 \, \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} b d}\right )}{96 \, \sqrt{-b d} d^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^2 + a)^(5/2)*x/sqrt(d*x^2 + c),x, algorithm="fricas")

[Out]

[1/192*(4*(8*b^2*d^2*x^4 + 15*b^2*c^2 - 40*a*b*c*d + 33*a^2*d^2 - 2*(5*b^2*c*d -
13*a*b*d^2)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b*d) - 15*(b^3*c^3 - 3*a*
b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(4*(2*b^2*d^2*x^2 + b^2*c*d + a*b*d^2)*s
qrt(b*x^2 + a)*sqrt(d*x^2 + c) + (8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2
+ 8*(b^2*c*d + a*b*d^2)*x^2)*sqrt(b*d)))/(sqrt(b*d)*d^3), 1/96*(2*(8*b^2*d^2*x^4
+ 15*b^2*c^2 - 40*a*b*c*d + 33*a^2*d^2 - 2*(5*b^2*c*d - 13*a*b*d^2)*x^2)*sqrt(b
*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b*d) - 15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d
^2 - a^3*d^3)*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(-b*d)/(sqrt(b*x^2 + a)*sqr
t(d*x^2 + c)*b*d)))/(sqrt(-b*d)*d^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{x \left (a + b x^{2}\right )^{\frac{5}{2}}}{\sqrt{c + d x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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GIAC/XCAS [A]  time = 0.249203, size = 284, normalized size = 1.73 \[ \frac{{\left (\sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d} \sqrt{b x^{2} + a}{\left (2 \,{\left (b x^{2} + a\right )}{\left (\frac{4 \,{\left (b x^{2} + a\right )}}{b d} - \frac{5 \,{\left (b c d^{3} - a d^{4}\right )}}{b d^{5}}\right )} + \frac{15 \,{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )}}{b d^{5}}\right )} + \frac{15 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}{\rm ln}\left ({\left | -\sqrt{b x^{2} + a} \sqrt{b d} + \sqrt{b^{2} c +{\left (b x^{2} + a\right )} b d - a b d} \right |}\right )}{\sqrt{b d} d^{3}}\right )} b}{48 \,{\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^2 + a)^(5/2)*x/sqrt(d*x^2 + c),x, algorithm="giac")

[Out]

1/48*(sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)*(2*(b*x^2 + a)*(4*(b
*x^2 + a)/(b*d) - 5*(b*c*d^3 - a*d^4)/(b*d^5)) + 15*(b^2*c^2*d^2 - 2*a*b*c*d^3 +
a^2*d^4)/(b*d^5)) + 15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*ln(a
bs(-sqrt(b*x^2 + a)*sqrt(b*d) + sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)))/(sqrt(b*
d)*d^3))*b/abs(b)