3.448 \(\int \frac {(A+B x) (a+c x^2)^{3/2}}{(e x)^{7/2}} \, dx\)

Optimal. Leaf size=339 \[ \frac {4 \sqrt [4]{a} c^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (5 \sqrt {a} B+9 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 e^3 \sqrt {e x} \sqrt {a+c x^2}}-\frac {4 c \sqrt {a+c x^2} (9 A-5 B x)}{15 e^3 \sqrt {e x}}-\frac {2 \left (a+c x^2\right )^{3/2} (3 A+5 B x)}{15 e (e x)^{5/2}}+\frac {24 A c^{3/2} x \sqrt {a+c x^2}}{5 e^3 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {24 \sqrt [4]{a} A c^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 e^3 \sqrt {e x} \sqrt {a+c x^2}} \]

[Out]

-2/15*(5*B*x+3*A)*(c*x^2+a)^(3/2)/e/(e*x)^(5/2)-4/15*c*(-5*B*x+9*A)*(c*x^2+a)^(1/2)/e^3/(e*x)^(1/2)+24/5*A*c^(
3/2)*x*(c*x^2+a)^(1/2)/e^3/(a^(1/2)+x*c^(1/2))/(e*x)^(1/2)-24/5*a^(1/4)*A*c^(5/4)*(cos(2*arctan(c^(1/4)*x^(1/2
)/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/a^(1/4))),1
/2*2^(1/2))*(a^(1/2)+x*c^(1/2))*x^(1/2)*((c*x^2+a)/(a^(1/2)+x*c^(1/2))^2)^(1/2)/e^3/(e*x)^(1/2)/(c*x^2+a)^(1/2
)+4/15*a^(1/4)*c^(3/4)*(cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/a^(1/4)))
*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(5*B*a^(1/2)+9*A*c^(1/2))*(a^(1/2)+x*c^(1/2))*x
^(1/2)*((c*x^2+a)/(a^(1/2)+x*c^(1/2))^2)^(1/2)/e^3/(e*x)^(1/2)/(c*x^2+a)^(1/2)

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Rubi [A]  time = 0.34, antiderivative size = 339, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {811, 813, 842, 840, 1198, 220, 1196} \[ \frac {4 \sqrt [4]{a} c^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (5 \sqrt {a} B+9 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 e^3 \sqrt {e x} \sqrt {a+c x^2}}-\frac {4 c \sqrt {a+c x^2} (9 A-5 B x)}{15 e^3 \sqrt {e x}}-\frac {2 \left (a+c x^2\right )^{3/2} (3 A+5 B x)}{15 e (e x)^{5/2}}+\frac {24 A c^{3/2} x \sqrt {a+c x^2}}{5 e^3 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {24 \sqrt [4]{a} A c^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 e^3 \sqrt {e x} \sqrt {a+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(3/2))/(e*x)^(7/2),x]

[Out]

(-4*c*(9*A - 5*B*x)*Sqrt[a + c*x^2])/(15*e^3*Sqrt[e*x]) + (24*A*c^(3/2)*x*Sqrt[a + c*x^2])/(5*e^3*Sqrt[e*x]*(S
qrt[a] + Sqrt[c]*x)) - (2*(3*A + 5*B*x)*(a + c*x^2)^(3/2))/(15*e*(e*x)^(5/2)) - (24*a^(1/4)*A*c^(5/4)*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])/(5*e^3*Sqrt[e*x]*Sqrt[a + c*x^2]) + (4*a^(1/4)*(5*Sqrt[a]*B + 9*A*Sqrt[c])*c^(3/4)*Sqrt[x]*(Sqrt[a] + S
qrt[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])/(15*e
^3*Sqrt[e*x]*Sqrt[a + c*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 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 840

Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2, Subst[Int[(f + g*x^2)/Sqrt[
a + c*x^4], x], x, Sqrt[x]], x] /; FreeQ[{a, c, f, g}, x]

Rule 842

Int[((f_) + (g_.)*(x_))/(Sqrt[(e_)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[x]/Sqrt[e*x], Int[
(f + g*x)/(Sqrt[x]*Sqrt[a + c*x^2]), x], x] /; FreeQ[{a, c, e, f, g}, x]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1198

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/q, Int
[1/Sqrt[a + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+c x^2\right )^{3/2}}{(e x)^{7/2}} \, dx &=-\frac {2 (3 A+5 B x) \left (a+c x^2\right )^{3/2}}{15 e (e x)^{5/2}}-\frac {2 \int \frac {\left (-3 a A c e^2-5 a B c e^2 x\right ) \sqrt {a+c x^2}}{(e x)^{3/2}} \, dx}{5 a e^4}\\ &=-\frac {4 c (9 A-5 B x) \sqrt {a+c x^2}}{15 e^3 \sqrt {e x}}-\frac {2 (3 A+5 B x) \left (a+c x^2\right )^{3/2}}{15 e (e x)^{5/2}}+\frac {4 \int \frac {5 a^2 B c e^3+9 a A c^2 e^3 x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{15 a e^6}\\ &=-\frac {4 c (9 A-5 B x) \sqrt {a+c x^2}}{15 e^3 \sqrt {e x}}-\frac {2 (3 A+5 B x) \left (a+c x^2\right )^{3/2}}{15 e (e x)^{5/2}}+\frac {\left (4 \sqrt {x}\right ) \int \frac {5 a^2 B c e^3+9 a A c^2 e^3 x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{15 a e^6 \sqrt {e x}}\\ &=-\frac {4 c (9 A-5 B x) \sqrt {a+c x^2}}{15 e^3 \sqrt {e x}}-\frac {2 (3 A+5 B x) \left (a+c x^2\right )^{3/2}}{15 e (e x)^{5/2}}+\frac {\left (8 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {5 a^2 B c e^3+9 a A c^2 e^3 x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{15 a e^6 \sqrt {e x}}\\ &=-\frac {4 c (9 A-5 B x) \sqrt {a+c x^2}}{15 e^3 \sqrt {e x}}-\frac {2 (3 A+5 B x) \left (a+c x^2\right )^{3/2}}{15 e (e x)^{5/2}}+\frac {\left (8 \sqrt {a} \left (5 \sqrt {a} B+9 A \sqrt {c}\right ) c \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{15 e^3 \sqrt {e x}}-\frac {\left (24 \sqrt {a} A c^{3/2} \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{5 e^3 \sqrt {e x}}\\ &=-\frac {4 c (9 A-5 B x) \sqrt {a+c x^2}}{15 e^3 \sqrt {e x}}+\frac {24 A c^{3/2} x \sqrt {a+c x^2}}{5 e^3 \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {2 (3 A+5 B x) \left (a+c x^2\right )^{3/2}}{15 e (e x)^{5/2}}-\frac {24 \sqrt [4]{a} A c^{5/4} \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 )}{5 e^3 \sqrt {e x} \sqrt {a+c x^2}}+\frac {4 \sqrt [4]{a} \left (5 \sqrt {a} B+9 A \sqrt {c}\right ) c^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 e^3 \sqrt {e x} \sqrt {a+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 84, normalized size = 0.25 \[ -\frac {2 a x \sqrt {a+c x^2} \left (3 A \, _2F_1\left (-\frac {3}{2},-\frac {5}{4};-\frac {1}{4};-\frac {c x^2}{a}\right )+5 B x \, _2F_1\left (-\frac {3}{2},-\frac {3}{4};\frac {1}{4};-\frac {c x^2}{a}\right )\right )}{15 (e x)^{7/2} \sqrt {\frac {c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(3/2))/(e*x)^(7/2),x]

[Out]

(-2*a*x*Sqrt[a + c*x^2]*(3*A*Hypergeometric2F1[-3/2, -5/4, -1/4, -((c*x^2)/a)] + 5*B*x*Hypergeometric2F1[-3/2,
-3/4, 1/4, -((c*x^2)/a)]))/(15*(e*x)^(7/2)*Sqrt[1 + (c*x^2)/a])

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B c x^{3} + A c x^{2} + B a x + A a\right )} \sqrt {c x^{2} + a} \sqrt {e x}}{e^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/(e*x)^(7/2),x, algorithm="fricas")

[Out]

integral((B*c*x^3 + A*c*x^2 + B*a*x + A*a)*sqrt(c*x^2 + a)*sqrt(e*x)/(e^4*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} {\left (B x + A\right )}}{\left (e x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/(e*x)^(7/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + a)^(3/2)*(B*x + A)/(e*x)^(7/2), x)

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maple [A]  time = 0.08, size = 329, normalized size = 0.97 \[ \frac {\frac {2 B \,c^{2} x^{5}}{3}-\frac {14 A \,c^{2} x^{4}}{5}+\frac {24 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, A a c \,x^{2} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{5}-\frac {12 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, A a c \,x^{2} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{5}-\frac {16 A a c \,x^{2}}{5}+\frac {4 \sqrt {2}\, \sqrt {-a c}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, B a \,x^{2} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )}{3}-\frac {2 B \,a^{2} x}{3}-\frac {2 A \,a^{2}}{5}}{\sqrt {c \,x^{2}+a}\, \sqrt {e x}\, e^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(3/2)/(e*x)^(7/2),x)

[Out]

2/15/x^2*(36*A*2^(1/2)*((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-1/(-
a*c)^(1/2)*c*x)^(1/2)*EllipticE(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*c-18*A*2^(1/2)*((c*
x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-1/(-a*c)^(1/2)*c*x)^(1/2)*Ellip
ticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*c+10*B*2^(1/2)*(-a*c)^(1/2)*((c*x+(-a*c)^(1/2)
)/(-a*c)^(1/2))^(1/2)*((-c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2)*(-1/(-a*c)^(1/2)*c*x)^(1/2)*EllipticF(((c*x+(-a
*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a+5*B*c^2*x^5-21*A*c^2*x^4-24*A*a*c*x^2-5*B*a^2*x-3*A*a^2)/(c*
x^2+a)^(1/2)/e^3/(e*x)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} {\left (B x + A\right )}}{\left (e x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/(e*x)^(7/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + a)^(3/2)*(B*x + A)/(e*x)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,x^2+a\right )}^{3/2}\,\left (A+B\,x\right )}{{\left (e\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^(3/2)*(A + B*x))/(e*x)^(7/2),x)

[Out]

int(((a + c*x^2)^(3/2)*(A + B*x))/(e*x)^(7/2), x)

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sympy [C]  time = 37.89, size = 212, normalized size = 0.63 \[ \frac {A a^{\frac {3}{2}} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right )} + \frac {A \sqrt {a} c \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {7}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} + \frac {B a^{\frac {3}{2}} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {7}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {B \sqrt {a} c \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {7}{2}} \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(3/2)/(e*x)**(7/2),x)

[Out]

A*a**(3/2)*gamma(-5/4)*hyper((-5/4, -1/2), (-1/4,), c*x**2*exp_polar(I*pi)/a)/(2*e**(7/2)*x**(5/2)*gamma(-1/4)
) + A*sqrt(a)*c*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), c*x**2*exp_polar(I*pi)/a)/(2*e**(7/2)*sqrt(x)*gamma(3/
4)) + B*a**(3/2)*gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), c*x**2*exp_polar(I*pi)/a)/(2*e**(7/2)*x**(3/2)*gamma(
1/4)) + B*sqrt(a)*c*sqrt(x)*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), c*x**2*exp_polar(I*pi)/a)/(2*e**(7/2)*gamma(
5/4))

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