3.703 \(\int \frac{(d x)^{5/2}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx\)

Optimal. Leaf size=335 \[ \frac{5 d^{5/2} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{256 \sqrt{2} a^{9/4} b^{7/4}}-\frac{5 d^{5/2} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{256 \sqrt{2} a^{9/4} b^{7/4}}-\frac{5 d^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{128 \sqrt{2} a^{9/4} b^{7/4}}+\frac{5 d^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}+1\right )}{128 \sqrt{2} a^{9/4} b^{7/4}}+\frac{5 d (d x)^{3/2}}{64 a^2 b \left (a+b x^2\right )}+\frac{d (d x)^{3/2}}{16 a b \left (a+b x^2\right )^2}-\frac{d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3} \]

[Out]

-(d*(d*x)^(3/2))/(6*b*(a + b*x^2)^3) + (d*(d*x)^(3/2))/(16*a*b*(a + b*x^2)^2) +
(5*d*(d*x)^(3/2))/(64*a^2*b*(a + b*x^2)) - (5*d^(5/2)*ArcTan[1 - (Sqrt[2]*b^(1/4
)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(9/4)*b^(7/4)) + (5*d^(5/2)*ArcT
an[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(9/4)*b^(7
/4)) + (5*d^(5/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - Sqrt[2]*a^(1/4)*b^(1
/4)*Sqrt[d*x]])/(256*Sqrt[2]*a^(9/4)*b^(7/4)) - (5*d^(5/2)*Log[Sqrt[a]*Sqrt[d] +
Sqrt[b]*Sqrt[d]*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(256*Sqrt[2]*a^(9/4)*b^
(7/4))

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Rubi [A]  time = 0.741866, antiderivative size = 335, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357 \[ \frac{5 d^{5/2} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{256 \sqrt{2} a^{9/4} b^{7/4}}-\frac{5 d^{5/2} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d x}+\sqrt{a} \sqrt{d}+\sqrt{b} \sqrt{d} x\right )}{256 \sqrt{2} a^{9/4} b^{7/4}}-\frac{5 d^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}\right )}{128 \sqrt{2} a^{9/4} b^{7/4}}+\frac{5 d^{5/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}}+1\right )}{128 \sqrt{2} a^{9/4} b^{7/4}}+\frac{5 d (d x)^{3/2}}{64 a^2 b \left (a+b x^2\right )}+\frac{d (d x)^{3/2}}{16 a b \left (a+b x^2\right )^2}-\frac{d (d x)^{3/2}}{6 b \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]  Int[(d*x)^(5/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

-(d*(d*x)^(3/2))/(6*b*(a + b*x^2)^3) + (d*(d*x)^(3/2))/(16*a*b*(a + b*x^2)^2) +
(5*d*(d*x)^(3/2))/(64*a^2*b*(a + b*x^2)) - (5*d^(5/2)*ArcTan[1 - (Sqrt[2]*b^(1/4
)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(9/4)*b^(7/4)) + (5*d^(5/2)*ArcT
an[1 + (Sqrt[2]*b^(1/4)*Sqrt[d*x])/(a^(1/4)*Sqrt[d])])/(128*Sqrt[2]*a^(9/4)*b^(7
/4)) + (5*d^(5/2)*Log[Sqrt[a]*Sqrt[d] + Sqrt[b]*Sqrt[d]*x - Sqrt[2]*a^(1/4)*b^(1
/4)*Sqrt[d*x]])/(256*Sqrt[2]*a^(9/4)*b^(7/4)) - (5*d^(5/2)*Log[Sqrt[a]*Sqrt[d] +
Sqrt[b]*Sqrt[d]*x + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[d*x]])/(256*Sqrt[2]*a^(9/4)*b^
(7/4))

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Rubi in Sympy [A]  time = 148.37, size = 309, normalized size = 0.92 \[ - \frac{d \left (d x\right )^{\frac{3}{2}}}{6 b \left (a + b x^{2}\right )^{3}} + \frac{d \left (d x\right )^{\frac{3}{2}}}{16 a b \left (a + b x^{2}\right )^{2}} + \frac{5 d \left (d x\right )^{\frac{3}{2}}}{64 a^{2} b \left (a + b x^{2}\right )} + \frac{5 \sqrt{2} d^{\frac{5}{2}} \log{\left (- \sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d} \sqrt{d x} + \sqrt{a} d + \sqrt{b} d x \right )}}{512 a^{\frac{9}{4}} b^{\frac{7}{4}}} - \frac{5 \sqrt{2} d^{\frac{5}{2}} \log{\left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{d} \sqrt{d x} + \sqrt{a} d + \sqrt{b} d x \right )}}{512 a^{\frac{9}{4}} b^{\frac{7}{4}}} - \frac{5 \sqrt{2} d^{\frac{5}{2}} \operatorname{atan}{\left (1 - \frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}} \right )}}{256 a^{\frac{9}{4}} b^{\frac{7}{4}}} + \frac{5 \sqrt{2} d^{\frac{5}{2}} \operatorname{atan}{\left (1 + \frac{\sqrt{2} \sqrt [4]{b} \sqrt{d x}}{\sqrt [4]{a} \sqrt{d}} \right )}}{256 a^{\frac{9}{4}} b^{\frac{7}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((d*x)**(5/2)/(b**2*x**4+2*a*b*x**2+a**2)**2,x)

[Out]

-d*(d*x)**(3/2)/(6*b*(a + b*x**2)**3) + d*(d*x)**(3/2)/(16*a*b*(a + b*x**2)**2)
+ 5*d*(d*x)**(3/2)/(64*a**2*b*(a + b*x**2)) + 5*sqrt(2)*d**(5/2)*log(-sqrt(2)*a*
*(1/4)*b**(1/4)*sqrt(d)*sqrt(d*x) + sqrt(a)*d + sqrt(b)*d*x)/(512*a**(9/4)*b**(7
/4)) - 5*sqrt(2)*d**(5/2)*log(sqrt(2)*a**(1/4)*b**(1/4)*sqrt(d)*sqrt(d*x) + sqrt
(a)*d + sqrt(b)*d*x)/(512*a**(9/4)*b**(7/4)) - 5*sqrt(2)*d**(5/2)*atan(1 - sqrt(
2)*b**(1/4)*sqrt(d*x)/(a**(1/4)*sqrt(d)))/(256*a**(9/4)*b**(7/4)) + 5*sqrt(2)*d*
*(5/2)*atan(1 + sqrt(2)*b**(1/4)*sqrt(d*x)/(a**(1/4)*sqrt(d)))/(256*a**(9/4)*b**
(7/4))

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Mathematica [A]  time = 0.327327, size = 259, normalized size = 0.77 \[ \frac{(d x)^{5/2} \left (\frac{15 \sqrt{2} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{a^{9/4}}-\frac{15 \sqrt{2} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} \sqrt{x}+\sqrt{a}+\sqrt{b} x\right )}{a^{9/4}}-\frac{30 \sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )}{a^{9/4}}+\frac{30 \sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}+1\right )}{a^{9/4}}+\frac{120 b^{3/4} x^{3/2}}{a^2 \left (a+b x^2\right )}+\frac{96 b^{3/4} x^{3/2}}{a \left (a+b x^2\right )^2}-\frac{256 b^{3/4} x^{3/2}}{\left (a+b x^2\right )^3}\right )}{1536 b^{7/4} x^{5/2}} \]

Antiderivative was successfully verified.

[In]  Integrate[(d*x)^(5/2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

((d*x)^(5/2)*((-256*b^(3/4)*x^(3/2))/(a + b*x^2)^3 + (96*b^(3/4)*x^(3/2))/(a*(a
+ b*x^2)^2) + (120*b^(3/4)*x^(3/2))/(a^2*(a + b*x^2)) - (30*Sqrt[2]*ArcTan[1 - (
Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/a^(9/4) + (30*Sqrt[2]*ArcTan[1 + (Sqrt[2]*b^(
1/4)*Sqrt[x])/a^(1/4)])/a^(9/4) + (15*Sqrt[2]*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1
/4)*Sqrt[x] + Sqrt[b]*x])/a^(9/4) - (15*Sqrt[2]*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^
(1/4)*Sqrt[x] + Sqrt[b]*x])/a^(9/4)))/(1536*b^(7/4)*x^(5/2))

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Maple [A]  time = 0.026, size = 277, normalized size = 0.8 \[{\frac{5\,{d}^{3}b}{64\, \left ( b{d}^{2}{x}^{2}+a{d}^{2} \right ) ^{3}{a}^{2}} \left ( dx \right ) ^{{\frac{11}{2}}}}+{\frac{7\,{d}^{5}}{32\, \left ( b{d}^{2}{x}^{2}+a{d}^{2} \right ) ^{3}a} \left ( dx \right ) ^{{\frac{7}{2}}}}-{\frac{5\,{d}^{7}}{192\, \left ( b{d}^{2}{x}^{2}+a{d}^{2} \right ) ^{3}b} \left ( dx \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{d}^{3}\sqrt{2}}{512\,{a}^{2}{b}^{2}}\ln \left ({1 \left ( dx-\sqrt [4]{{\frac{a{d}^{2}}{b}}}\sqrt{dx}\sqrt{2}+\sqrt{{\frac{a{d}^{2}}{b}}} \right ) \left ( dx+\sqrt [4]{{\frac{a{d}^{2}}{b}}}\sqrt{dx}\sqrt{2}+\sqrt{{\frac{a{d}^{2}}{b}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{a{d}^{2}}{b}}}}}}+{\frac{5\,{d}^{3}\sqrt{2}}{256\,{a}^{2}{b}^{2}}\arctan \left ({\sqrt{2}\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{a{d}^{2}}{b}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{a{d}^{2}}{b}}}}}}+{\frac{5\,{d}^{3}\sqrt{2}}{256\,{a}^{2}{b}^{2}}\arctan \left ({\sqrt{2}\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{a{d}^{2}}{b}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{a{d}^{2}}{b}}}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((d*x)^(5/2)/(b^2*x^4+2*a*b*x^2+a^2)^2,x)

[Out]

5/64*d^3/(b*d^2*x^2+a*d^2)^3/a^2*b*(d*x)^(11/2)+7/32*d^5/(b*d^2*x^2+a*d^2)^3/a*(
d*x)^(7/2)-5/192*d^7/(b*d^2*x^2+a*d^2)^3/b*(d*x)^(3/2)+5/512*d^3/a^2/b^2/(a*d^2/
b)^(1/4)*2^(1/2)*ln((d*x-(a*d^2/b)^(1/4)*(d*x)^(1/2)*2^(1/2)+(a*d^2/b)^(1/2))/(d
*x+(a*d^2/b)^(1/4)*(d*x)^(1/2)*2^(1/2)+(a*d^2/b)^(1/2)))+5/256*d^3/a^2/b^2/(a*d^
2/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a*d^2/b)^(1/4)*(d*x)^(1/2)+1)+5/256*d^3/a^2/b
^2/(a*d^2/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a*d^2/b)^(1/4)*(d*x)^(1/2)-1)

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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((d*x)^(5/2)/(b^2*x^4 + 2*a*b*x^2 + a^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.288391, size = 505, normalized size = 1.51 \[ \frac{60 \,{\left (a^{2} b^{4} x^{6} + 3 \, a^{3} b^{3} x^{4} + 3 \, a^{4} b^{2} x^{2} + a^{5} b\right )} \left (-\frac{d^{10}}{a^{9} b^{7}}\right )^{\frac{1}{4}} \arctan \left (\frac{125 \, a^{7} b^{5} \left (-\frac{d^{10}}{a^{9} b^{7}}\right )^{\frac{3}{4}}}{125 \, \sqrt{d x} d^{7} + \sqrt{-15625 \, a^{5} b^{3} d^{10} \sqrt{-\frac{d^{10}}{a^{9} b^{7}}} + 15625 \, d^{15} x}}\right ) + 15 \,{\left (a^{2} b^{4} x^{6} + 3 \, a^{3} b^{3} x^{4} + 3 \, a^{4} b^{2} x^{2} + a^{5} b\right )} \left (-\frac{d^{10}}{a^{9} b^{7}}\right )^{\frac{1}{4}} \log \left (125 \, a^{7} b^{5} \left (-\frac{d^{10}}{a^{9} b^{7}}\right )^{\frac{3}{4}} + 125 \, \sqrt{d x} d^{7}\right ) - 15 \,{\left (a^{2} b^{4} x^{6} + 3 \, a^{3} b^{3} x^{4} + 3 \, a^{4} b^{2} x^{2} + a^{5} b\right )} \left (-\frac{d^{10}}{a^{9} b^{7}}\right )^{\frac{1}{4}} \log \left (-125 \, a^{7} b^{5} \left (-\frac{d^{10}}{a^{9} b^{7}}\right )^{\frac{3}{4}} + 125 \, \sqrt{d x} d^{7}\right ) + 4 \,{\left (15 \, b^{2} d^{2} x^{5} + 42 \, a b d^{2} x^{3} - 5 \, a^{2} d^{2} x\right )} \sqrt{d x}}{768 \,{\left (a^{2} b^{4} x^{6} + 3 \, a^{3} b^{3} x^{4} + 3 \, a^{4} b^{2} x^{2} + a^{5} b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((d*x)^(5/2)/(b^2*x^4 + 2*a*b*x^2 + a^2)^2,x, algorithm="fricas")

[Out]

1/768*(60*(a^2*b^4*x^6 + 3*a^3*b^3*x^4 + 3*a^4*b^2*x^2 + a^5*b)*(-d^10/(a^9*b^7)
)^(1/4)*arctan(125*a^7*b^5*(-d^10/(a^9*b^7))^(3/4)/(125*sqrt(d*x)*d^7 + sqrt(-15
625*a^5*b^3*d^10*sqrt(-d^10/(a^9*b^7)) + 15625*d^15*x))) + 15*(a^2*b^4*x^6 + 3*a
^3*b^3*x^4 + 3*a^4*b^2*x^2 + a^5*b)*(-d^10/(a^9*b^7))^(1/4)*log(125*a^7*b^5*(-d^
10/(a^9*b^7))^(3/4) + 125*sqrt(d*x)*d^7) - 15*(a^2*b^4*x^6 + 3*a^3*b^3*x^4 + 3*a
^4*b^2*x^2 + a^5*b)*(-d^10/(a^9*b^7))^(1/4)*log(-125*a^7*b^5*(-d^10/(a^9*b^7))^(
3/4) + 125*sqrt(d*x)*d^7) + 4*(15*b^2*d^2*x^5 + 42*a*b*d^2*x^3 - 5*a^2*d^2*x)*sq
rt(d*x))/(a^2*b^4*x^6 + 3*a^3*b^3*x^4 + 3*a^4*b^2*x^2 + a^5*b)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\left (d x\right )^{\frac{5}{2}}}{\left (a + b x^{2}\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((d*x)**(5/2)/(b**2*x**4+2*a*b*x**2+a**2)**2,x)

[Out]

Integral((d*x)**(5/2)/(a + b*x**2)**4, x)

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GIAC/XCAS [A]  time = 0.277815, size = 409, normalized size = 1.22 \[ \frac{1}{1536} \, d{\left (\frac{30 \, \sqrt{2} \left (a b^{3} d^{2}\right )^{\frac{3}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a d^{2}}{b}\right )^{\frac{1}{4}} + 2 \, \sqrt{d x}\right )}}{2 \, \left (\frac{a d^{2}}{b}\right )^{\frac{1}{4}}}\right )}{a^{3} b^{4}} + \frac{30 \, \sqrt{2} \left (a b^{3} d^{2}\right )^{\frac{3}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{a d^{2}}{b}\right )^{\frac{1}{4}} - 2 \, \sqrt{d x}\right )}}{2 \, \left (\frac{a d^{2}}{b}\right )^{\frac{1}{4}}}\right )}{a^{3} b^{4}} - \frac{15 \, \sqrt{2} \left (a b^{3} d^{2}\right )^{\frac{3}{4}}{\rm ln}\left (d x + \sqrt{2} \left (\frac{a d^{2}}{b}\right )^{\frac{1}{4}} \sqrt{d x} + \sqrt{\frac{a d^{2}}{b}}\right )}{a^{3} b^{4}} + \frac{15 \, \sqrt{2} \left (a b^{3} d^{2}\right )^{\frac{3}{4}}{\rm ln}\left (d x - \sqrt{2} \left (\frac{a d^{2}}{b}\right )^{\frac{1}{4}} \sqrt{d x} + \sqrt{\frac{a d^{2}}{b}}\right )}{a^{3} b^{4}} + \frac{8 \,{\left (15 \, \sqrt{d x} b^{2} d^{7} x^{5} + 42 \, \sqrt{d x} a b d^{7} x^{3} - 5 \, \sqrt{d x} a^{2} d^{7} x\right )}}{{\left (b d^{2} x^{2} + a d^{2}\right )}^{3} a^{2} b}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((d*x)^(5/2)/(b^2*x^4 + 2*a*b*x^2 + a^2)^2,x, algorithm="giac")

[Out]

1/1536*d*(30*sqrt(2)*(a*b^3*d^2)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(a*d^2/b)^(1/
4) + 2*sqrt(d*x))/(a*d^2/b)^(1/4))/(a^3*b^4) + 30*sqrt(2)*(a*b^3*d^2)^(3/4)*arct
an(-1/2*sqrt(2)*(sqrt(2)*(a*d^2/b)^(1/4) - 2*sqrt(d*x))/(a*d^2/b)^(1/4))/(a^3*b^
4) - 15*sqrt(2)*(a*b^3*d^2)^(3/4)*ln(d*x + sqrt(2)*(a*d^2/b)^(1/4)*sqrt(d*x) + s
qrt(a*d^2/b))/(a^3*b^4) + 15*sqrt(2)*(a*b^3*d^2)^(3/4)*ln(d*x - sqrt(2)*(a*d^2/b
)^(1/4)*sqrt(d*x) + sqrt(a*d^2/b))/(a^3*b^4) + 8*(15*sqrt(d*x)*b^2*d^7*x^5 + 42*
sqrt(d*x)*a*b*d^7*x^3 - 5*sqrt(d*x)*a^2*d^7*x)/((b*d^2*x^2 + a*d^2)^3*a^2*b))