3.840 \(\int \frac {(e x)^{3/2} (a+b x^2)^2}{\sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=240 \[ -\frac {c^{3/4} e^{3/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (77 a^2 d^2+5 b c (9 b c-22 a d)\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 d^{13/4} \sqrt {c+d x^2}}+\frac {2 e \sqrt {e x} \sqrt {c+d x^2} \left (77 a^2 d^2+5 b c (9 b c-22 a d)\right )}{231 d^3}-\frac {2 b (e x)^{5/2} \sqrt {c+d x^2} (9 b c-22 a d)}{77 d^2 e}+\frac {2 b^2 (e x)^{9/2} \sqrt {c+d x^2}}{11 d e^3} \]

[Out]

-2/77*b*(-22*a*d+9*b*c)*(e*x)^(5/2)*(d*x^2+c)^(1/2)/d^2/e+2/11*b^2*(e*x)^(9/2)*(d*x^2+c)^(1/2)/d/e^3+2/231*(77
*a^2*d^2+5*b*c*(-22*a*d+9*b*c))*e*(e*x)^(1/2)*(d*x^2+c)^(1/2)/d^3-1/231*c^(3/4)*(77*a^2*d^2+5*b*c*(-22*a*d+9*b
*c))*e^(3/2)*(cos(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(
1/4)/e^(1/2)))*EllipticF(sin(2*arctan(d^(1/4)*(e*x)^(1/2)/c^(1/4)/e^(1/2))),1/2*2^(1/2))*(c^(1/2)+x*d^(1/2))*(
(d*x^2+c)/(c^(1/2)+x*d^(1/2))^2)^(1/2)/d^(13/4)/(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {464, 459, 321, 329, 220} \[ -\frac {c^{3/4} e^{3/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (77 a^2 d^2+5 b c (9 b c-22 a d)\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 d^{13/4} \sqrt {c+d x^2}}+\frac {2 e \sqrt {e x} \sqrt {c+d x^2} \left (77 a^2 d^2+5 b c (9 b c-22 a d)\right )}{231 d^3}-\frac {2 b (e x)^{5/2} \sqrt {c+d x^2} (9 b c-22 a d)}{77 d^2 e}+\frac {2 b^2 (e x)^{9/2} \sqrt {c+d x^2}}{11 d e^3} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(3/2)*(a + b*x^2)^2)/Sqrt[c + d*x^2],x]

[Out]

(2*(77*a^2*d^2 + 5*b*c*(9*b*c - 22*a*d))*e*Sqrt[e*x]*Sqrt[c + d*x^2])/(231*d^3) - (2*b*(9*b*c - 22*a*d)*(e*x)^
(5/2)*Sqrt[c + d*x^2])/(77*d^2*e) + (2*b^2*(e*x)^(9/2)*Sqrt[c + d*x^2])/(11*d*e^3) - (c^(3/4)*(77*a^2*d^2 + 5*
b*c*(9*b*c - 22*a*d))*e^(3/2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcT
an[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(231*d^(13/4)*Sqrt[c + d*x^2])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(d^2*(e*x)^
(m + n + 1)*(a + b*x^n)^(p + 1))/(b*e^(n + 1)*(m + n*(p + 2) + 1)), x] + Dist[1/(b*(m + n*(p + 2) + 1)), Int[(
e*x)^m*(a + b*x^n)^p*Simp[b*c^2*(m + n*(p + 2) + 1) + d*((2*b*c - a*d)*(m + n + 1) + 2*b*c*n*(p + 1))*x^n, x],
x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && NeQ[m + n*(p + 2) + 1, 0]

Rubi steps

\begin {align*} \int \frac {(e x)^{3/2} \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx &=\frac {2 b^2 (e x)^{9/2} \sqrt {c+d x^2}}{11 d e^3}+\frac {2 \int \frac {(e x)^{3/2} \left (\frac {11 a^2 d}{2}-\frac {1}{2} b (9 b c-22 a d) x^2\right )}{\sqrt {c+d x^2}} \, dx}{11 d}\\ &=-\frac {2 b (9 b c-22 a d) (e x)^{5/2} \sqrt {c+d x^2}}{77 d^2 e}+\frac {2 b^2 (e x)^{9/2} \sqrt {c+d x^2}}{11 d e^3}-\frac {1}{77} \left (-77 a^2-\frac {5 b c (9 b c-22 a d)}{d^2}\right ) \int \frac {(e x)^{3/2}}{\sqrt {c+d x^2}} \, dx\\ &=\frac {2 \left (77 a^2+\frac {5 b c (9 b c-22 a d)}{d^2}\right ) e \sqrt {e x} \sqrt {c+d x^2}}{231 d}-\frac {2 b (9 b c-22 a d) (e x)^{5/2} \sqrt {c+d x^2}}{77 d^2 e}+\frac {2 b^2 (e x)^{9/2} \sqrt {c+d x^2}}{11 d e^3}-\frac {\left (c \left (77 a^2+\frac {5 b c (9 b c-22 a d)}{d^2}\right ) e^2\right ) \int \frac {1}{\sqrt {e x} \sqrt {c+d x^2}} \, dx}{231 d}\\ &=\frac {2 \left (77 a^2+\frac {5 b c (9 b c-22 a d)}{d^2}\right ) e \sqrt {e x} \sqrt {c+d x^2}}{231 d}-\frac {2 b (9 b c-22 a d) (e x)^{5/2} \sqrt {c+d x^2}}{77 d^2 e}+\frac {2 b^2 (e x)^{9/2} \sqrt {c+d x^2}}{11 d e^3}-\frac {\left (2 c \left (77 a^2+\frac {5 b c (9 b c-22 a d)}{d^2}\right ) e\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{231 d}\\ &=\frac {2 \left (77 a^2+\frac {5 b c (9 b c-22 a d)}{d^2}\right ) e \sqrt {e x} \sqrt {c+d x^2}}{231 d}-\frac {2 b (9 b c-22 a d) (e x)^{5/2} \sqrt {c+d x^2}}{77 d^2 e}+\frac {2 b^2 (e x)^{9/2} \sqrt {c+d x^2}}{11 d e^3}-\frac {c^{3/4} \left (77 a^2+\frac {5 b c (9 b c-22 a d)}{d^2}\right ) e^{3/2} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 d^{5/4} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.24, size = 190, normalized size = 0.79 \[ \frac {(e x)^{3/2} \left (\frac {2 \sqrt {x} \left (c+d x^2\right ) \left (77 a^2 d^2+22 a b d \left (3 d x^2-5 c\right )+3 b^2 \left (15 c^2-9 c d x^2+7 d^2 x^4\right )\right )}{d^3}-\frac {2 i c x \sqrt {\frac {c}{d x^2}+1} \left (77 a^2 d^2-110 a b c d+45 b^2 c^2\right ) F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}{\sqrt {x}}\right )\right |-1\right )}{d^3 \sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}\right )}{231 x^{3/2} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(3/2)*(a + b*x^2)^2)/Sqrt[c + d*x^2],x]

[Out]

((e*x)^(3/2)*((2*Sqrt[x]*(c + d*x^2)*(77*a^2*d^2 + 22*a*b*d*(-5*c + 3*d*x^2) + 3*b^2*(15*c^2 - 9*c*d*x^2 + 7*d
^2*x^4)))/d^3 - ((2*I)*c*(45*b^2*c^2 - 110*a*b*c*d + 77*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*EllipticF[I*ArcSinh[Sqr
t[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1])/(Sqrt[(I*Sqrt[c])/Sqrt[d]]*d^3)))/(231*x^(3/2)*Sqrt[c + d*x^2])

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} e x^{5} + 2 \, a b e x^{3} + a^{2} e x\right )} \sqrt {e x}}{\sqrt {d x^{2} + c}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

integral((b^2*e*x^5 + 2*a*b*e*x^3 + a^2*e*x)*sqrt(e*x)/sqrt(d*x^2 + c), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{2} \left (e x\right )^{\frac {3}{2}}}{\sqrt {d x^{2} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2*(e*x)^(3/2)/sqrt(d*x^2 + c), x)

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maple [A]  time = 0.05, size = 405, normalized size = 1.69 \[ -\frac {\sqrt {e x}\, \left (-42 b^{2} d^{4} x^{7}-132 a b \,d^{4} x^{5}+12 b^{2} c \,d^{3} x^{5}-154 a^{2} d^{4} x^{3}+88 a b c \,d^{3} x^{3}-36 b^{2} c^{2} d^{2} x^{3}-154 a^{2} c \,d^{3} x +220 a b \,c^{2} d^{2} x -90 b^{2} c^{3} d x +77 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, a^{2} c \,d^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-110 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, a b \,c^{2} d \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+45 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, b^{2} c^{3} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )\right ) e}{231 \sqrt {d \,x^{2}+c}\, d^{4} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(3/2)*(b*x^2+a)^2/(d*x^2+c)^(1/2),x)

[Out]

-1/231*e/x*(e*x)^(1/2)/(d*x^2+c)^(1/2)*(-42*b^2*d^4*x^7+77*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-
d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^
(1/2),1/2*2^(1/2))*(-c*d)^(1/2)*a^2*c*d^2-110*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1
/2))/(-c*d)^(1/2))^(1/2)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(
1/2))*(-c*d)^(1/2)*a*b*c^2*d+45*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1
/2))^(1/2)*(-1/(-c*d)^(1/2)*d*x)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*(-c*d)^(
1/2)*b^2*c^3-132*a*b*d^4*x^5+12*b^2*c*d^3*x^5-154*a^2*d^4*x^3+88*a*b*c*d^3*x^3-36*b^2*c^2*d^2*x^3-154*a^2*c*d^
3*x+220*a*b*c^2*d^2*x-90*b^2*c^3*d*x)/d^4

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{2} \left (e x\right )^{\frac {3}{2}}}{\sqrt {d x^{2} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2*(e*x)^(3/2)/sqrt(d*x^2 + c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e\,x\right )}^{3/2}\,{\left (b\,x^2+a\right )}^2}{\sqrt {d\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^(3/2)*(a + b*x^2)^2)/(c + d*x^2)^(1/2),x)

[Out]

int(((e*x)^(3/2)*(a + b*x^2)^2)/(c + d*x^2)^(1/2), x)

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sympy [C]  time = 16.25, size = 144, normalized size = 0.60 \[ \frac {a^{2} e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt {c} \Gamma \left (\frac {9}{4}\right )} + \frac {a b e^{\frac {3}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{\sqrt {c} \Gamma \left (\frac {13}{4}\right )} + \frac {b^{2} e^{\frac {3}{2}} x^{\frac {13}{2}} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt {c} \Gamma \left (\frac {17}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(3/2)*(b*x**2+a)**2/(d*x**2+c)**(1/2),x)

[Out]

a**2*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((1/2, 5/4), (9/4,), d*x**2*exp_polar(I*pi)/c)/(2*sqrt(c)*gamma(9/4)) +
a*b*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((1/2, 9/4), (13/4,), d*x**2*exp_polar(I*pi)/c)/(sqrt(c)*gamma(13/4)) +
b**2*e**(3/2)*x**(13/2)*gamma(13/4)*hyper((1/2, 13/4), (17/4,), d*x**2*exp_polar(I*pi)/c)/(2*sqrt(c)*gamma(17
/4))

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