3.475 \(\int \frac{(e x)^{11/2} (A+B x)}{\left (a+c x^2\right )^{5/2}} \, dx\)

Optimal. Leaf size=398 \[ -\frac{a^{3/4} e^6 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (77 \sqrt{a} B+25 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{20 c^{15/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{77 a^{5/4} B e^6 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{10 c^{15/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{e^3 (e x)^{5/2} (9 A+11 B x)}{6 c^2 \sqrt{a+c x^2}}-\frac{e (e x)^{9/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}+\frac{5 A e^5 \sqrt{e x} \sqrt{a+c x^2}}{2 c^3}-\frac{77 a B e^6 x \sqrt{a+c x^2}}{10 c^{7/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{77 B e^4 (e x)^{3/2} \sqrt{a+c x^2}}{30 c^3} \]

[Out]

-(e*(e*x)^(9/2)*(A + B*x))/(3*c*(a + c*x^2)^(3/2)) - (e^3*(e*x)^(5/2)*(9*A + 11*
B*x))/(6*c^2*Sqrt[a + c*x^2]) + (5*A*e^5*Sqrt[e*x]*Sqrt[a + c*x^2])/(2*c^3) + (7
7*B*e^4*(e*x)^(3/2)*Sqrt[a + c*x^2])/(30*c^3) - (77*a*B*e^6*x*Sqrt[a + c*x^2])/(
10*c^(7/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (77*a^(5/4)*B*e^6*Sqrt[x]*(Sqrt[a]
+ Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1
/4)*Sqrt[x])/a^(1/4)], 1/2])/(10*c^(15/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) - (a^(3/4)*
(77*Sqrt[a]*B + 25*A*Sqrt[c])*e^6*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)
/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(
20*c^(15/4)*Sqrt[e*x]*Sqrt[a + c*x^2])

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Rubi [A]  time = 1.15859, antiderivative size = 398, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292 \[ -\frac{a^{3/4} e^6 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} \left (77 \sqrt{a} B+25 A \sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{20 c^{15/4} \sqrt{e x} \sqrt{a+c x^2}}+\frac{77 a^{5/4} B e^6 \sqrt{x} \left (\sqrt{a}+\sqrt{c} x\right ) \sqrt{\frac{a+c x^2}{\left (\sqrt{a}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{10 c^{15/4} \sqrt{e x} \sqrt{a+c x^2}}-\frac{e^3 (e x)^{5/2} (9 A+11 B x)}{6 c^2 \sqrt{a+c x^2}}-\frac{e (e x)^{9/2} (A+B x)}{3 c \left (a+c x^2\right )^{3/2}}+\frac{5 A e^5 \sqrt{e x} \sqrt{a+c x^2}}{2 c^3}-\frac{77 a B e^6 x \sqrt{a+c x^2}}{10 c^{7/2} \sqrt{e x} \left (\sqrt{a}+\sqrt{c} x\right )}+\frac{77 B e^4 (e x)^{3/2} \sqrt{a+c x^2}}{30 c^3} \]

Antiderivative was successfully verified.

[In]  Int[((e*x)^(11/2)*(A + B*x))/(a + c*x^2)^(5/2),x]

[Out]

-(e*(e*x)^(9/2)*(A + B*x))/(3*c*(a + c*x^2)^(3/2)) - (e^3*(e*x)^(5/2)*(9*A + 11*
B*x))/(6*c^2*Sqrt[a + c*x^2]) + (5*A*e^5*Sqrt[e*x]*Sqrt[a + c*x^2])/(2*c^3) + (7
7*B*e^4*(e*x)^(3/2)*Sqrt[a + c*x^2])/(30*c^3) - (77*a*B*e^6*x*Sqrt[a + c*x^2])/(
10*c^(7/2)*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (77*a^(5/4)*B*e^6*Sqrt[x]*(Sqrt[a]
+ Sqrt[c]*x)*Sqrt[(a + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1
/4)*Sqrt[x])/a^(1/4)], 1/2])/(10*c^(15/4)*Sqrt[e*x]*Sqrt[a + c*x^2]) - (a^(3/4)*
(77*Sqrt[a]*B + 25*A*Sqrt[c])*e^6*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)
/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(
20*c^(15/4)*Sqrt[e*x]*Sqrt[a + c*x^2])

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Rubi in Sympy [A]  time = 152.163, size = 374, normalized size = 0.94 \[ \frac{5 A e^{5} \sqrt{e x} \sqrt{a + c x^{2}}}{2 c^{3}} + \frac{77 B a^{\frac{5}{4}} e^{6} \sqrt{x} \sqrt{\frac{a + c x^{2}}{\left (\sqrt{a} + \sqrt{c} x\right )^{2}}} \left (\sqrt{a} + \sqrt{c} x\right ) E\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{10 c^{\frac{15}{4}} \sqrt{e x} \sqrt{a + c x^{2}}} - \frac{77 B a e^{6} x \sqrt{a + c x^{2}}}{10 c^{\frac{7}{2}} \sqrt{e x} \left (\sqrt{a} + \sqrt{c} x\right )} + \frac{77 B e^{4} \left (e x\right )^{\frac{3}{2}} \sqrt{a + c x^{2}}}{30 c^{3}} - \frac{a^{\frac{3}{4}} e^{6} \sqrt{x} \sqrt{\frac{a + c x^{2}}{\left (\sqrt{a} + \sqrt{c} x\right )^{2}}} \left (\sqrt{a} + \sqrt{c} x\right ) \left (25 A \sqrt{c} + 77 B \sqrt{a}\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{20 c^{\frac{15}{4}} \sqrt{e x} \sqrt{a + c x^{2}}} - \frac{e \left (e x\right )^{\frac{9}{2}} \left (A + B x\right )}{3 c \left (a + c x^{2}\right )^{\frac{3}{2}}} - \frac{e^{3} \left (e x\right )^{\frac{5}{2}} \left (18 A + 22 B x\right )}{12 c^{2} \sqrt{a + c x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((e*x)**(11/2)*(B*x+A)/(c*x**2+a)**(5/2),x)

[Out]

5*A*e**5*sqrt(e*x)*sqrt(a + c*x**2)/(2*c**3) + 77*B*a**(5/4)*e**6*sqrt(x)*sqrt((
a + c*x**2)/(sqrt(a) + sqrt(c)*x)**2)*(sqrt(a) + sqrt(c)*x)*elliptic_e(2*atan(c*
*(1/4)*sqrt(x)/a**(1/4)), 1/2)/(10*c**(15/4)*sqrt(e*x)*sqrt(a + c*x**2)) - 77*B*
a*e**6*x*sqrt(a + c*x**2)/(10*c**(7/2)*sqrt(e*x)*(sqrt(a) + sqrt(c)*x)) + 77*B*e
**4*(e*x)**(3/2)*sqrt(a + c*x**2)/(30*c**3) - a**(3/4)*e**6*sqrt(x)*sqrt((a + c*
x**2)/(sqrt(a) + sqrt(c)*x)**2)*(sqrt(a) + sqrt(c)*x)*(25*A*sqrt(c) + 77*B*sqrt(
a))*elliptic_f(2*atan(c**(1/4)*sqrt(x)/a**(1/4)), 1/2)/(20*c**(15/4)*sqrt(e*x)*s
qrt(a + c*x**2)) - e*(e*x)**(9/2)*(A + B*x)/(3*c*(a + c*x**2)**(3/2)) - e**3*(e*
x)**(5/2)*(18*A + 22*B*x)/(12*c**2*sqrt(a + c*x**2))

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Mathematica [C]  time = 1.10638, size = 277, normalized size = 0.7 \[ \frac{e^6 \left (231 a^{3/2} B \sqrt{c} x^{3/2} \sqrt{\frac{a}{c x^2}+1} \left (a+c x^2\right ) E\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{c}}}}{\sqrt{x}}\right )\right |-1\right )+\sqrt{\frac{i \sqrt{a}}{\sqrt{c}}} \left (-231 a^3 B+5 a^2 c x (15 A-77 B x)+3 a c^2 x^3 (35 A-44 B x)+4 c^3 x^5 (5 A+3 B x)\right )-3 i a \sqrt{c} x^{3/2} \sqrt{\frac{a}{c x^2}+1} \left (a+c x^2\right ) \left (25 A \sqrt{c}-77 i \sqrt{a} B\right ) F\left (\left .i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{a}}{\sqrt{c}}}}{\sqrt{x}}\right )\right |-1\right )\right )}{30 c^4 \sqrt{\frac{i \sqrt{a}}{\sqrt{c}}} \sqrt{e x} \left (a+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]  Integrate[((e*x)^(11/2)*(A + B*x))/(a + c*x^2)^(5/2),x]

[Out]

(e^6*(Sqrt[(I*Sqrt[a])/Sqrt[c]]*(-231*a^3*B + 5*a^2*c*x*(15*A - 77*B*x) + 3*a*c^
2*x^3*(35*A - 44*B*x) + 4*c^3*x^5*(5*A + 3*B*x)) + 231*a^(3/2)*B*Sqrt[c]*Sqrt[1
+ a/(c*x^2)]*x^(3/2)*(a + c*x^2)*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[a])/Sqrt[c]]/S
qrt[x]], -1] - (3*I)*a*((-77*I)*Sqrt[a]*B + 25*A*Sqrt[c])*Sqrt[c]*Sqrt[1 + a/(c*
x^2)]*x^(3/2)*(a + c*x^2)*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[a])/Sqrt[c]]/Sqrt[x]]
, -1]))/(30*Sqrt[(I*Sqrt[a])/Sqrt[c]]*c^4*Sqrt[e*x]*(a + c*x^2)^(3/2))

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Maple [A]  time = 0.09, size = 615, normalized size = 1.6 \[ -{\frac{{e}^{5}}{60\,x{c}^{4}} \left ( 75\,A\sqrt{-ac}\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ){x}^{2}ac+462\,B\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ){x}^{2}{a}^{2}c-231\,B\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ){x}^{2}{a}^{2}c-24\,B{c}^{3}{x}^{6}+75\,A\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ) \sqrt{-ac}{a}^{2}-40\,A{c}^{3}{x}^{5}+462\,B\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ){a}^{3}-231\,B\sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-ac}}{\sqrt{-ac}}}}\sqrt{-{\frac{cx}{\sqrt{-ac}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-ac}}{\sqrt{-ac}}}},1/2\,\sqrt{2} \right ){a}^{3}-198\,aB{c}^{2}{x}^{4}-210\,aA{c}^{2}{x}^{3}-154\,{a}^{2}Bc{x}^{2}-150\,{a}^{2}Acx \right ) \sqrt{ex} \left ( c{x}^{2}+a \right ) ^{-{\frac{3}{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((e*x)^(11/2)*(B*x+A)/(c*x^2+a)^(5/2),x)

[Out]

-1/60*(75*A*(-a*c)^(1/2)*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+
(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(1/2))^(1/2)*EllipticF(((c*x+(-a*
c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*c+462*B*((c*x+(-a*c)^(1/2))/(-a
*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(
1/2))^(1/2)*EllipticE(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a
^2*c-231*B*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/
(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(1/2))^(1/2)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*
c)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a^2*c-24*B*c^3*x^6+75*A*((c*x+(-a*c)^(1/2))/(-a
*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(
1/2))^(1/2)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*(-a*c
)^(1/2)*a^2-40*A*c^3*x^5+462*B*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*2^(1/2)*(
(-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)^(1/2))^(1/2)*EllipticE(((c*
x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^3-231*B*((c*x+(-a*c)^(1/2))/(
-a*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-x*c/(-a*c)
^(1/2))^(1/2)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*a^3
-198*a*B*c^2*x^4-210*a*A*c^2*x^3-154*a^2*B*c*x^2-150*a^2*A*c*x)*e^5/x*(e*x)^(1/2
)/c^4/(c*x^2+a)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (B x + A\right )} \left (e x\right )^{\frac{11}{2}}}{{\left (c x^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x + A)*(e*x)^(11/2)/(c*x^2 + a)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)*(e*x)^(11/2)/(c*x^2 + a)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (B e^{5} x^{6} + A e^{5} x^{5}\right )} \sqrt{e x}}{{\left (c^{2} x^{4} + 2 \, a c x^{2} + a^{2}\right )} \sqrt{c x^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x + A)*(e*x)^(11/2)/(c*x^2 + a)^(5/2),x, algorithm="fricas")

[Out]

integral((B*e^5*x^6 + A*e^5*x^5)*sqrt(e*x)/((c^2*x^4 + 2*a*c*x^2 + a^2)*sqrt(c*x
^2 + a)), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x)**(11/2)*(B*x+A)/(c*x**2+a)**(5/2),x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (B x + A\right )} \left (e x\right )^{\frac{11}{2}}}{{\left (c x^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x + A)*(e*x)^(11/2)/(c*x^2 + a)^(5/2),x, algorithm="giac")

[Out]

integrate((B*x + A)*(e*x)^(11/2)/(c*x^2 + a)^(5/2), x)