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

Optimal. Leaf size=421 \[ -\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{5 c e (e x)^{5/2}}+\frac {2 (e x)^{3/2} \sqrt {c+d x^2} \left (a d (a d+10 b c)+b^2 c^2\right )}{5 c^2 e^5}+\frac {4 \sqrt {e x} \sqrt {c+d x^2} \left (a d (a d+10 b c)+b^2 c^2\right )}{5 c \sqrt {d} e^4 \left (\sqrt {c}+\sqrt {d} x\right )}+\frac {2 \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (a d (a d+10 b c)+b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 c^{3/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}}-\frac {4 \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (a d (a d+10 b c)+b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 c^{3/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}}-\frac {2 a \left (c+d x^2\right )^{3/2} (a d+10 b c)}{5 c^2 e^3 \sqrt {e x}} \]

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 421, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {462, 453, 279, 329, 305, 220, 1196} \[ -\frac {2 a^2 \left (c+d x^2\right )^{3/2}}{5 c e (e x)^{5/2}}+\frac {2 \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (a d (a d+10 b c)+b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 c^{3/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}}-\frac {4 \left (\sqrt {c}+\sqrt {d} x\right ) \sqrt {\frac {c+d x^2}{\left (\sqrt {c}+\sqrt {d} x\right )^2}} \left (a d (a d+10 b c)+b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt {e x}}{\sqrt [4]{c} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 c^{3/4} d^{3/4} e^{7/2} \sqrt {c+d x^2}}+\frac {2 (e x)^{3/2} \sqrt {c+d x^2} \left (a d (a d+10 b c)+b^2 c^2\right )}{5 c^2 e^5}+\frac {4 \sqrt {e x} \sqrt {c+d x^2} \left (a d (a d+10 b c)+b^2 c^2\right )}{5 c \sqrt {d} e^4 \left (\sqrt {c}+\sqrt {d} x\right )}-\frac {2 a \left (c+d x^2\right )^{3/2} (a d+10 b c)}{5 c^2 e^3 \sqrt {e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.13, size = 125, normalized size = 0.30 \[ \frac {x \left (4 x^4 \sqrt {\frac {c}{d x^2}+1} \left (a^2 d^2+10 a b c d+b^2 c^2\right ) \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-\frac {c}{d x^2}\right )-2 \left (c+d x^2\right ) \left (a^2 \left (c+2 d x^2\right )+10 a b c x^2-b^2 c x^4\right )\right )}{5 c (e x)^{7/2} \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.08, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {d x^{2} + c} \sqrt {e x}}{e^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(d*x^2 + c)*sqrt(e*x)/(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*(d*x^2+c)^(1/2)/(e*x)^(7/2),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 = 0.05, size = 648, normalized size = 1.54 \[ \frac {\frac {2 b^{2} c \,d^{2} x^{6}}{5}-\frac {4 a^{2} d^{3} x^{4}}{5}-4 a b c \,d^{2} x^{4}+\frac {2 b^{2} c^{2} d \,x^{4}}{5}+\frac {4 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a^{2} c \,d^{2} x^{2} \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{5}-\frac {2 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a^{2} c \,d^{2} x^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{5}+8 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a b \,c^{2} d \,x^{2} \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )-4 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, a b \,c^{2} d \,x^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )+\frac {4 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, b^{2} c^{3} x^{2} \EllipticE \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{5}-\frac {2 \sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {2}\, \sqrt {\frac {-d x +\sqrt {-c d}}{\sqrt {-c d}}}\, \sqrt {-\frac {d x}{\sqrt {-c d}}}\, b^{2} c^{3} x^{2} \EllipticF \left (\sqrt {\frac {d x +\sqrt {-c d}}{\sqrt {-c d}}}, \frac {\sqrt {2}}{2}\right )}{5}-\frac {6 a^{2} c \,d^{2} x^{2}}{5}-4 a b \,c^{2} d \,x^{2}-\frac {2 a^{2} c^{2} d}{5}}{\sqrt {d \,x^{2}+c}\, \sqrt {e x}\, c d \,e^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5/x^2*(2*((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)*Ellip
ticE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*b*c^2*d+2*((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-((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-((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+x^6*b^2*c*d^2-2*x^4*a^2*d^3-10*
x^4*a*b*c*d^2+x^4*b^2*c^2*d-3*x^2*a^2*c*d^2-10*x^2*a*b*c^2*d-a^2*c^2*d)/(d*x^2+c)^(1/2)/d/e^3/(e*x)^(1/2)/c

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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*(d*x^2+c)^(1/2)/(e*x)^(7/2),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 \[ \int \frac {{\left (b\,x^2+a\right )}^2\,\sqrt {d\,x^2+c}}{{\left (e\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 30.18, size = 160, normalized size = 0.38 \[ \frac {a^{2} \sqrt {c} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right )} + \frac {a b \sqrt {c} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{e^{\frac {7}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} + \frac {b^{2} \sqrt {c} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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