3.453 \(\int \frac {(A+B x) (a+c x^2)^{5/2}}{(e x)^{3/2}} \, dx\)

Optimal. Leaf size=379 \[ \frac {8 a^{9/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (15 \sqrt {a} B+77 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{231 \sqrt [4]{c} e \sqrt {e x} \sqrt {a+c x^2}}-\frac {16 a^{9/4} A \sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 e \sqrt {e x} \sqrt {a+c x^2}}+\frac {16 a^2 A \sqrt {c} x \sqrt {a+c x^2}}{3 e \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {20 \sqrt {e x} \left (a+c x^2\right )^{3/2} (9 a B+77 A c x)}{693 e^2}+\frac {8 a \sqrt {e x} \sqrt {a+c x^2} (15 a B+77 A c x)}{231 e^2}-\frac {2 \left (a+c x^2\right )^{5/2} (11 A-B x)}{11 e \sqrt {e x}} \]

[Out]

-2/11*(-B*x+11*A)*(c*x^2+a)^(5/2)/e/(e*x)^(1/2)+20/693*(77*A*c*x+9*B*a)*(c*x^2+a)^(3/2)*(e*x)^(1/2)/e^2+16/3*a
^2*A*x*c^(1/2)*(c*x^2+a)^(1/2)/e/(a^(1/2)+x*c^(1/2))/(e*x)^(1/2)+8/231*a*(77*A*c*x+15*B*a)*(e*x)^(1/2)*(c*x^2+
a)^(1/2)/e^2-16/3*a^(9/4)*A*c^(1/4)*(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/(e*x)^(1/2)/(c*x^2+a)^(1/2)+8/231*a^(9/4)*(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))*(15*B*a^(1/2)+77*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)/c^(1/4
)/e/(e*x)^(1/2)/(c*x^2+a)^(1/2)

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Rubi [A]  time = 0.41, antiderivative size = 379, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {813, 815, 842, 840, 1198, 220, 1196} \[ \frac {8 a^{9/4} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} \left (15 \sqrt {a} B+77 A \sqrt {c}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{231 \sqrt [4]{c} e \sqrt {e x} \sqrt {a+c x^2}}+\frac {16 a^2 A \sqrt {c} x \sqrt {a+c x^2}}{3 e \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}-\frac {16 a^{9/4} A \sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x\right ) \sqrt {\frac {a+c x^2}{\left (\sqrt {a}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 e \sqrt {e x} \sqrt {a+c x^2}}+\frac {20 \sqrt {e x} \left (a+c x^2\right )^{3/2} (9 a B+77 A c x)}{693 e^2}+\frac {8 a \sqrt {e x} \sqrt {a+c x^2} (15 a B+77 A c x)}{231 e^2}-\frac {2 \left (a+c x^2\right )^{5/2} (11 A-B x)}{11 e \sqrt {e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*a^2*A*Sqrt[c]*x*Sqrt[a + c*x^2])/(3*e*Sqrt[e*x]*(Sqrt[a] + Sqrt[c]*x)) + (8*a*Sqrt[e*x]*(15*a*B + 77*A*c*x
)*Sqrt[a + c*x^2])/(231*e^2) + (20*Sqrt[e*x]*(9*a*B + 77*A*c*x)*(a + c*x^2)^(3/2))/(693*e^2) - (2*(11*A - B*x)
*(a + c*x^2)^(5/2))/(11*e*Sqrt[e*x]) - (16*a^(9/4)*A*c^(1/4)*Sqrt[x]*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + c*x^2)/(S
qrt[a] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(3*e*Sqrt[e*x]*Sqrt[a + c*x^2]) +
(8*a^(9/4)*(15*Sqrt[a]*B + 77*A*Sqrt[c])*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])/(231*c^(1/4)*e*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 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 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 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 )^{5/2}}{(e x)^{3/2}} \, dx &=-\frac {2 (11 A-B x) \left (a+c x^2\right )^{5/2}}{11 e \sqrt {e x}}-\frac {10 \int \frac {(-a B e-11 A c e x) \left (a+c x^2\right )^{3/2}}{\sqrt {e x}} \, dx}{11 e^2}\\ &=\frac {20 \sqrt {e x} (9 a B+77 A c x) \left (a+c x^2\right )^{3/2}}{693 e^2}-\frac {2 (11 A-B x) \left (a+c x^2\right )^{5/2}}{11 e \sqrt {e x}}-\frac {40 \int \frac {\left (-\frac {9}{2} a^2 B c e^3-\frac {77}{2} a A c^2 e^3 x\right ) \sqrt {a+c x^2}}{\sqrt {e x}} \, dx}{231 c e^4}\\ &=\frac {8 a \sqrt {e x} (15 a B+77 A c x) \sqrt {a+c x^2}}{231 e^2}+\frac {20 \sqrt {e x} (9 a B+77 A c x) \left (a+c x^2\right )^{3/2}}{693 e^2}-\frac {2 (11 A-B x) \left (a+c x^2\right )^{5/2}}{11 e \sqrt {e x}}-\frac {32 \int \frac {-\frac {45}{4} a^3 B c^2 e^5-\frac {231}{4} a^2 A c^3 e^5 x}{\sqrt {e x} \sqrt {a+c x^2}} \, dx}{693 c^2 e^6}\\ &=\frac {8 a \sqrt {e x} (15 a B+77 A c x) \sqrt {a+c x^2}}{231 e^2}+\frac {20 \sqrt {e x} (9 a B+77 A c x) \left (a+c x^2\right )^{3/2}}{693 e^2}-\frac {2 (11 A-B x) \left (a+c x^2\right )^{5/2}}{11 e \sqrt {e x}}-\frac {\left (32 \sqrt {x}\right ) \int \frac {-\frac {45}{4} a^3 B c^2 e^5-\frac {231}{4} a^2 A c^3 e^5 x}{\sqrt {x} \sqrt {a+c x^2}} \, dx}{693 c^2 e^6 \sqrt {e x}}\\ &=\frac {8 a \sqrt {e x} (15 a B+77 A c x) \sqrt {a+c x^2}}{231 e^2}+\frac {20 \sqrt {e x} (9 a B+77 A c x) \left (a+c x^2\right )^{3/2}}{693 e^2}-\frac {2 (11 A-B x) \left (a+c x^2\right )^{5/2}}{11 e \sqrt {e x}}-\frac {\left (64 \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {-\frac {45}{4} a^3 B c^2 e^5-\frac {231}{4} a^2 A c^3 e^5 x^2}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{693 c^2 e^6 \sqrt {e x}}\\ &=\frac {8 a \sqrt {e x} (15 a B+77 A c x) \sqrt {a+c x^2}}{231 e^2}+\frac {20 \sqrt {e x} (9 a B+77 A c x) \left (a+c x^2\right )^{3/2}}{693 e^2}-\frac {2 (11 A-B x) \left (a+c x^2\right )^{5/2}}{11 e \sqrt {e x}}+\frac {\left (16 a^{5/2} \left (15 \sqrt {a} B+77 A \sqrt {c}\right ) \sqrt {x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^4}} \, dx,x,\sqrt {x}\right )}{231 e \sqrt {e x}}-\frac {\left (16 a^{5/2} A \sqrt {c} \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 )}{3 e \sqrt {e x}}\\ &=\frac {16 a^2 A \sqrt {c} x \sqrt {a+c x^2}}{3 e \sqrt {e x} \left (\sqrt {a}+\sqrt {c} x\right )}+\frac {8 a \sqrt {e x} (15 a B+77 A c x) \sqrt {a+c x^2}}{231 e^2}+\frac {20 \sqrt {e x} (9 a B+77 A c x) \left (a+c x^2\right )^{3/2}}{693 e^2}-\frac {2 (11 A-B x) \left (a+c x^2\right )^{5/2}}{11 e \sqrt {e x}}-\frac {16 a^{9/4} A \sqrt [4]{c} \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 )}{3 e \sqrt {e x} \sqrt {a+c x^2}}+\frac {8 a^{9/4} \left (15 \sqrt {a} B+77 A \sqrt {c}\right ) \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 )}{231 \sqrt [4]{c} e \sqrt {e x} \sqrt {a+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 83, normalized size = 0.22 \[ \frac {2 a^2 x \sqrt {a+c x^2} \left (B x \, _2F_1\left (-\frac {5}{2},\frac {1}{4};\frac {5}{4};-\frac {c x^2}{a}\right )-A \, _2F_1\left (-\frac {5}{2},-\frac {1}{4};\frac {3}{4};-\frac {c x^2}{a}\right )\right )}{(e x)^{3/2} \sqrt {\frac {c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B c^{2} x^{5} + A c^{2} x^{4} + 2 \, B a c x^{3} + 2 \, A a c x^{2} + B a^{2} x + A a^{2}\right )} \sqrt {c x^{2} + a} \sqrt {e x}}{e^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*c^2*x^5 + A*c^2*x^4 + 2*B*a*c*x^3 + 2*A*a*c*x^2 + B*a^2*x + A*a^2)*sqrt(c*x^2 + a)*sqrt(e*x)/(e^2*
x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + a)^(5/2)*(B*x + A)/(e*x)^(3/2), x)

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maple [A]  time = 0.08, size = 364, normalized size = 0.96 \[ -\frac {2 \left (-63 B \,c^{4} x^{7}-77 A \,c^{4} x^{6}-279 B a \,c^{3} x^{5}-385 A a \,c^{3} x^{4}-549 B \,a^{2} c^{2} x^{3}+385 A \,a^{2} c^{2} x^{2}-1848 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, A \,a^{3} c \EllipticE \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )+924 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, A \,a^{3} c \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )-333 B \,a^{3} c x +693 A \,a^{3} c -180 \sqrt {2}\, \sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {\frac {-c x +\sqrt {-a c}}{\sqrt {-a c}}}\, \sqrt {-\frac {c x}{\sqrt {-a c}}}\, \sqrt {-a c}\, B \,a^{3} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-a c}}{\sqrt {-a c}}}, \frac {\sqrt {2}}{2}\right )\right )}{693 \sqrt {c \,x^{2}+a}\, \sqrt {e x}\, c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(5/2)/(e*x)^(3/2),x)

[Out]

-2/693*(-63*B*c^4*x^7-77*A*c^4*x^6+924*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)*EllipticF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))
*a^3*c-1848*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))*a^3*c-180*B*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)*Ellipti
cF(((c*x+(-a*c)^(1/2))/(-a*c)^(1/2))^(1/2),1/2*2^(1/2))*(-a*c)^(1/2)*a^3-279*B*a*c^3*x^5-385*A*a*c^3*x^4-549*B
*a^2*c^2*x^3+385*A*a^2*c^2*x^2-333*B*a^3*c*x+693*a^3*A*c)/(c*x^2+a)^(1/2)/c/e/(e*x)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + a)^(5/2)*(B*x + A)/(e*x)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^(5/2)*(A + B*x))/(e*x)^(3/2),x)

[Out]

int(((a + c*x^2)^(5/2)*(A + B*x))/(e*x)^(3/2), x)

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sympy [C]  time = 18.55, size = 304, normalized size = 0.80 \[ \frac {A a^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} + \frac {A a^{\frac {3}{2}} c x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{e^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} + \frac {A \sqrt {a} c^{2} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} + \frac {B a^{\frac {5}{2}} \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {3}{2}} \Gamma \left (\frac {5}{4}\right )} + \frac {B a^{\frac {3}{2}} c x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{e^{\frac {3}{2}} \Gamma \left (\frac {9}{4}\right )} + \frac {B \sqrt {a} c^{2} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {3}{2}} \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(5/2)/(e*x)**(3/2),x)

[Out]

A*a**(5/2)*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), c*x**2*exp_polar(I*pi)/a)/(2*e**(3/2)*sqrt(x)*gamma(3/4)) +
A*a**(3/2)*c*x**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), c*x**2*exp_polar(I*pi)/a)/(e**(3/2)*gamma(7/4)) +
A*sqrt(a)*c**2*x**(7/2)*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), c*x**2*exp_polar(I*pi)/a)/(2*e**(3/2)*gamma(11
/4)) + B*a**(5/2)*sqrt(x)*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), c*x**2*exp_polar(I*pi)/a)/(2*e**(3/2)*gamma(5/
4)) + B*a**(3/2)*c*x**(5/2)*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), c*x**2*exp_polar(I*pi)/a)/(e**(3/2)*gamma(9/
4)) + B*sqrt(a)*c**2*x**(9/2)*gamma(9/4)*hyper((-1/2, 9/4), (13/4,), c*x**2*exp_polar(I*pi)/a)/(2*e**(3/2)*gam
ma(13/4))

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