3.837 \(\int \frac {(a+b x^2)^2 (c+d x^2)^{3/2}}{(e x)^{5/2}} \, dx\)

Optimal. Leaf size=288 \[ -\frac {2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}-\frac {2 \sqrt {e x} \left (c+d x^2\right )^{3/2} \left (3 b^2 c^2-11 a d (7 a d+6 b c)\right )}{231 c d e^3}-\frac {4 \sqrt {e x} \sqrt {c+d x^2} \left (3 b^2 c^2-11 a d (7 a d+6 b c)\right )}{231 d e^3}-\frac {4 c^{3/4} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (3 b^2 c^2-11 a d (7 a d+6 b c)\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 d^{5/4} e^{5/2} \sqrt {c+d x^2}}+\frac {2 b^2 \sqrt {e x} \left (c+d x^2\right )^{5/2}}{11 d e^3} \]

[Out]

-2/3*a^2*(d*x^2+c)^(5/2)/c/e/(e*x)^(3/2)-2/231*(3*b^2*c^2-11*a*d*(7*a*d+6*b*c))*(d*x^2+c)^(3/2)*(e*x)^(1/2)/c/
d/e^3+2/11*b^2*(d*x^2+c)^(5/2)*(e*x)^(1/2)/d/e^3-4/231*(3*b^2*c^2-11*a*d*(7*a*d+6*b*c))*(e*x)^(1/2)*(d*x^2+c)^
(1/2)/d/e^3-4/231*c^(3/4)*(3*b^2*c^2-11*a*d*(7*a*d+6*b*c))*(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^(5/4)/e^(5/2)/(d*x^2+c)
^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {462, 459, 279, 329, 220} \[ -\frac {2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}-\frac {4 c^{3/4} \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (3 b^2 c^2-11 a d (7 a d+6 b c)\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{231 d^{5/4} e^{5/2} \sqrt {c+d x^2}}-\frac {2 \sqrt {e x} \left (c+d x^2\right )^{3/2} \left (3 b^2 c^2-11 a d (7 a d+6 b c)\right )}{231 c d e^3}-\frac {4 \sqrt {e x} \sqrt {c+d x^2} \left (3 b^2 c^2-11 a d (7 a d+6 b c)\right )}{231 d e^3}+\frac {2 b^2 \sqrt {e x} \left (c+d x^2\right )^{5/2}}{11 d e^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(c + d*x^2)^(3/2))/(e*x)^(5/2),x]

[Out]

(-4*(3*b^2*c^2 - 11*a*d*(6*b*c + 7*a*d))*Sqrt[e*x]*Sqrt[c + d*x^2])/(231*d*e^3) - (2*(3*b^2*c^2 - 11*a*d*(6*b*
c + 7*a*d))*Sqrt[e*x]*(c + d*x^2)^(3/2))/(231*c*d*e^3) - (2*a^2*(c + d*x^2)^(5/2))/(3*c*e*(e*x)^(3/2)) + (2*b^
2*Sqrt[e*x]*(c + d*x^2)^(5/2))/(11*d*e^3) - (4*c^(3/4)*(3*b^2*c^2 - 11*a*d*(6*b*c + 7*a*d))*(Sqrt[c] + Sqrt[d]
*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])
/(231*d^(5/4)*e^(5/2)*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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && 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 462

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{(e x)^{5/2}} \, dx &=-\frac {2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}+\frac {2 \int \frac {\left (\frac {1}{2} a (6 b c+7 a d)+\frac {3}{2} b^2 c x^2\right ) \left (c+d x^2\right )^{3/2}}{\sqrt {e x}} \, dx}{3 c e^2}\\ &=-\frac {2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \left (c+d x^2\right )^{5/2}}{11 d e^3}-\frac {\left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \int \frac {\left (c+d x^2\right )^{3/2}}{\sqrt {e x}} \, dx}{33 c d e^2}\\ &=-\frac {2 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \sqrt {e x} \left (c+d x^2\right )^{3/2}}{231 c d e^3}-\frac {2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \left (c+d x^2\right )^{5/2}}{11 d e^3}-\frac {\left (2 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right )\right ) \int \frac {\sqrt {c+d x^2}}{\sqrt {e x}} \, dx}{77 d e^2}\\ &=-\frac {4 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \sqrt {e x} \sqrt {c+d x^2}}{231 d e^3}-\frac {2 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \sqrt {e x} \left (c+d x^2\right )^{3/2}}{231 c d e^3}-\frac {2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \left (c+d x^2\right )^{5/2}}{11 d e^3}-\frac {\left (4 c \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right )\right ) \int \frac {1}{\sqrt {e x} \sqrt {c+d x^2}} \, dx}{231 d e^2}\\ &=-\frac {4 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \sqrt {e x} \sqrt {c+d x^2}}{231 d e^3}-\frac {2 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \sqrt {e x} \left (c+d x^2\right )^{3/2}}{231 c d e^3}-\frac {2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \left (c+d x^2\right )^{5/2}}{11 d e^3}-\frac {\left (8 c \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{231 d e^3}\\ &=-\frac {4 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \sqrt {e x} \sqrt {c+d x^2}}{231 d e^3}-\frac {2 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \sqrt {e x} \left (c+d x^2\right )^{3/2}}{231 c d e^3}-\frac {2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}+\frac {2 b^2 \sqrt {e x} \left (c+d x^2\right )^{5/2}}{11 d e^3}-\frac {4 c^{3/4} \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \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} e^{5/2} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.29, size = 202, normalized size = 0.70 \[ \frac {x^{5/2} \left (\frac {2 \left (c+d x^2\right ) \left (77 a^2 d \left (d x^2-c\right )+66 a b d x^2 \left (3 c+d x^2\right )+3 b^2 x^2 \left (4 c^2+13 c d x^2+7 d^2 x^4\right )\right )}{d x^{3/2}}+\frac {8 i c x \sqrt {\frac {c}{d x^2}+1} \left (77 a^2 d^2+66 a b c d-3 b^2 c^2\right ) F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}{\sqrt {x}}\right )\right |-1\right )}{d \sqrt {\frac {i \sqrt {c}}{\sqrt {d}}}}\right )}{231 (e x)^{5/2} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(c + d*x^2)^(3/2))/(e*x)^(5/2),x]

[Out]

(x^(5/2)*((2*(c + d*x^2)*(77*a^2*d*(-c + d*x^2) + 66*a*b*d*x^2*(3*c + d*x^2) + 3*b^2*x^2*(4*c^2 + 13*c*d*x^2 +
7*d^2*x^4)))/(d*x^(3/2)) + ((8*I)*c*(-3*b^2*c^2 + 66*a*b*c*d + 77*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*EllipticF[I*
ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1])/(Sqrt[(I*Sqrt[c])/Sqrt[d]]*d)))/(231*(e*x)^(5/2)*Sqrt[c + d*x
^2])

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} d x^{6} + {\left (b^{2} c + 2 \, a b d\right )} x^{4} + a^{2} c + {\left (2 \, a b c + a^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {e x}}{e^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/(e*x)^(5/2),x, algorithm="fricas")

[Out]

integral((b^2*d*x^6 + (b^2*c + 2*a*b*d)*x^4 + a^2*c + (2*a*b*c + a^2*d)*x^2)*sqrt(d*x^2 + c)*sqrt(e*x)/(e^3*x^
3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/(e*x)^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 415, normalized size = 1.44 \[ \frac {\frac {2 b^{2} d^{4} x^{8}}{11}+\frac {4 a b \,d^{4} x^{6}}{7}+\frac {40 b^{2} c \,d^{3} x^{6}}{77}+\frac {2 a^{2} d^{4} x^{4}}{3}+\frac {16 a b c \,d^{3} x^{4}}{7}+\frac {34 b^{2} c^{2} d^{2} x^{4}}{77}+\frac {12 a b \,c^{2} d^{2} x^{2}}{7}+\frac {8 b^{2} c^{3} d \,x^{2}}{77}+\frac {4 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, a^{2} c \,d^{2} x \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{3}+\frac {8 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, a b \,c^{2} d x \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{7}-\frac {4 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, \sqrt {-c d}\, b^{2} c^{3} x \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{77}-\frac {2 a^{2} c^{2} d^{2}}{3}}{\sqrt {d \,x^{2}+c}\, \sqrt {e x}\, d^{2} e^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/231/(d*x^2+c)^(1/2)/x*(21*b^2*d^4*x^8+154*((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)*x*a^2*c*d^2+132*((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)*x*a*b*c^2*d-6*((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)*x*b^2*c
^3+66*a*b*d^4*x^6+60*b^2*c*d^3*x^6+77*a^2*d^4*x^4+264*a*b*c*d^3*x^4+51*b^2*c^2*d^2*x^4+198*a*b*c^2*d^2*x^2+12*
b^2*c^3*d*x^2-77*a^2*c^2*d^2)/d^2/e^2/(e*x)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/(e*x)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{3/2}}{{\left (e\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 24.59, size = 309, normalized size = 1.07 \[ \frac {a^{2} c^{\frac {3}{2}} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {a^{2} \sqrt {c} d \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {a b c^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{e^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {a b \sqrt {c} d x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{e^{\frac {5}{2}} \Gamma \left (\frac {9}{4}\right )} + \frac {b^{2} c^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac {5}{2}} \Gamma \left (\frac {9}{4}\right )} + \frac {b^{2} \sqrt {c} d x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac {5}{2}} \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(3/2)/(e*x)**(5/2),x)

[Out]

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

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