3.845 \(\int \frac {(a+b x^2)^2}{(e x)^{7/2} \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=387 \[ \frac {2 \sqrt {e x} \sqrt {c+d x^2} \left (-3 a^2 d^2+10 a b c d+5 b^2 c^2\right )}{5 c^2 \sqrt {d} e^4 \left (\sqrt {c}+\sqrt {d} x\right )}+\frac {\left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (-3 a^2 d^2+10 a b c d+5 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 c^{7/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}}-\frac {2 \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (-3 a^2 d^2+10 a b c d+5 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 c^{7/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}}-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}-\frac {2 a \sqrt {c+d x^2} (10 b c-3 a d)}{5 c^2 e^3 \sqrt {e x}} \]

[Out]

-2/5*a^2*(d*x^2+c)^(1/2)/c/e/(e*x)^(5/2)-2/5*a*(-3*a*d+10*b*c)*(d*x^2+c)^(1/2)/c^2/e^3/(e*x)^(1/2)+2/5*(-3*a^2
*d^2+10*a*b*c*d+5*b^2*c^2)*(e*x)^(1/2)*(d*x^2+c)^(1/2)/c^2/e^4/d^(1/2)/(c^(1/2)+x*d^(1/2))-2/5*(-3*a^2*d^2+10*
a*b*c*d+5*b^2*c^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)))*EllipticE(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)/c^(7/4)/d^(3/4)/e^(7/2)/(d*x^2+c)^(1/2)+1/5*(-3*a^2*d^2+10*a*b*c*
d+5*b^2*c^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)/c^(7/4)/d^(3/4)/e^(7/2)/(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.35, antiderivative size = 387, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {462, 453, 329, 305, 220, 1196} \[ \frac {2 \sqrt {e x} \sqrt {c+d x^2} \left (-3 a^2 d^2+10 a b c d+5 b^2 c^2\right )}{5 c^2 \sqrt {d} e^4 \left (\sqrt {c}+\sqrt {d} x\right )}+\frac {\left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (-3 a^2 d^2+10 a b c d+5 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 c^{7/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}}-\frac {2 \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (-3 a^2 d^2+10 a b c d+5 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 c^{7/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}}-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}-\frac {2 a \sqrt {c+d x^2} (10 b c-3 a d)}{5 c^2 e^3 \sqrt {e x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/((e*x)^(7/2)*Sqrt[c + d*x^2]),x]

[Out]

(-2*a^2*Sqrt[c + d*x^2])/(5*c*e*(e*x)^(5/2)) - (2*a*(10*b*c - 3*a*d)*Sqrt[c + d*x^2])/(5*c^2*e^3*Sqrt[e*x]) +
(2*(5*b^2*c^2 + 10*a*b*c*d - 3*a^2*d^2)*Sqrt[e*x]*Sqrt[c + d*x^2])/(5*c^2*Sqrt[d]*e^4*(Sqrt[c] + Sqrt[d]*x)) -
(2*(5*b^2*c^2 + 10*a*b*c*d - 3*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*Ellip
ticE[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(5*c^(7/4)*d^(3/4)*e^(7/2)*Sqrt[c + d*x^2]) + ((5*
b^2*c^2 + 10*a*b*c*d - 3*a^2*d^2)*(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])/(5*c^(7/4)*d^(3/4)*e^(7/2)*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 305

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

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 453

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

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]

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]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2}{(e x)^{7/2} \sqrt {c+d x^2}} \, dx &=-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}+\frac {2 \int \frac {\frac {1}{2} a (10 b c-3 a d)+\frac {5}{2} b^2 c x^2}{(e x)^{3/2} \sqrt {c+d x^2}} \, dx}{5 c e^2}\\ &=-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}-\frac {2 a (10 b c-3 a d) \sqrt {c+d x^2}}{5 c^2 e^3 \sqrt {e x}}+\frac {\left (5 b^2 c^2+10 a b c d-3 a^2 d^2\right ) \int \frac {\sqrt {e x}}{\sqrt {c+d x^2}} \, dx}{5 c^2 e^4}\\ &=-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}-\frac {2 a (10 b c-3 a d) \sqrt {c+d x^2}}{5 c^2 e^3 \sqrt {e x}}+\frac {\left (2 \left (5 b^2 c^2+10 a b c d-3 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{5 c^2 e^5}\\ &=-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}-\frac {2 a (10 b c-3 a d) \sqrt {c+d x^2}}{5 c^2 e^3 \sqrt {e x}}+\frac {\left (2 \left (5 b^2 c^2+10 a b c d-3 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{5 c^{3/2} \sqrt {d} e^4}-\frac {\left (2 \left (5 b^2 c^2+10 a b c d-3 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {d} x^2}{\sqrt {c} e}}{\sqrt {c+\frac {d x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{5 c^{3/2} \sqrt {d} e^4}\\ &=-\frac {2 a^2 \sqrt {c+d x^2}}{5 c e (e x)^{5/2}}-\frac {2 a (10 b c-3 a d) \sqrt {c+d x^2}}{5 c^2 e^3 \sqrt {e x}}+\frac {2 \left (5 b^2 c^2+10 a b c d-3 a^2 d^2\right ) \sqrt {e x} \sqrt {c+d x^2}}{5 c^2 \sqrt {d} e^4 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 \left (5 b^2 c^2+10 a b c d-3 a^2 d^2\right ) \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 c^{7/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}}+\frac {\left (5 b^2 c^2+10 a b c d-3 a^2 d^2\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 )}{5 c^{7/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.13, size = 116, normalized size = 0.30 \[ \frac {x \left (2 x^4 \sqrt {\frac {c}{d x^2}+1} \left (-3 a^2 d^2+10 a b c d+5 b^2 c^2\right ) \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-\frac {c}{d x^2}\right )-2 a \left (c+d x^2\right ) \left (a \left (c-3 d x^2\right )+10 b c x^2\right )\right )}{5 c^2 (e x)^{7/2} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/((e*x)^(7/2)*Sqrt[c + d*x^2]),x]

[Out]

(x*(-2*a*(c + d*x^2)*(10*b*c*x^2 + a*(c - 3*d*x^2)) + 2*(5*b^2*c^2 + 10*a*b*c*d - 3*a^2*d^2)*Sqrt[1 + c/(d*x^2
)]*x^4*Hypergeometric2F1[-1/4, 1/2, 3/4, -(c/(d*x^2))]))/(5*c^2*(e*x)^(7/2)*Sqrt[c + d*x^2])

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {d x^{2} + c} \sqrt {e x}}{d e^{4} x^{6} + c e^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2/(sqrt(d*x^2 + c)*(e*x)^(7/2)), x)

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maple [A]  time = 0.03, size = 626, normalized size = 1.62 \[ -\frac {-6 a^{2} d^{3} x^{4}+20 a b c \,d^{2} x^{4}+6 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a^{2} c \,d^{2} x^{2} \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a^{2} c \,d^{2} x^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-20 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a b \,c^{2} d \,x^{2} \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+10 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a b \,c^{2} d \,x^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-10 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, b^{2} c^{3} x^{2} \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+5 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, b^{2} c^{3} x^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-4 a^{2} c \,d^{2} x^{2}+20 a b \,c^{2} d \,x^{2}+2 a^{2} c^{2} d}{5 \sqrt {d \,x^{2}+c}\, \sqrt {e x}\, c^{2} d \,e^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/(e*x)^(7/2)/(d*x^2+c)^(1/2),x)

[Out]

-1/5/x^2*(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)*EllipticE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a^2*c*d^2-20*((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)*Elli
pticE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*b*c^2*d-10*((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)*EllipticE(((d*x+(-c*d)^(1/2
))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^2*b^2*c^3-3*((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))*x^2*a^2*c*d^2+10*((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))*x^2*a*b*c^2*d
+5*((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))*x^2*b^2*c^3-6*a^2*d^3*x^4+20*a*b*c*d^
2*x^4-4*a^2*c*d^2*x^2+20*a*b*c^2*d*x^2+2*a^2*c^2*d)/(d*x^2+c)^(1/2)/d/e^3/(e*x)^(1/2)/c^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(e*x)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2/(sqrt(d*x^2 + c)*(e*x)^(7/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^2/((e*x)^(7/2)*(c + d*x^2)^(1/2)),x)

[Out]

int((a + b*x^2)^2/((e*x)^(7/2)*(c + d*x^2)^(1/2)), x)

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sympy [C]  time = 40.79, size = 155, normalized size = 0.40 \[ \frac {a^{2} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt {c} e^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right )} + \frac {a b \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{\sqrt {c} e^{\frac {7}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} + \frac {b^{2} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt {c} e^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/(e*x)**(7/2)/(d*x**2+c)**(1/2),x)

[Out]

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

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