3.3.10 \(\int \frac {(d+e x^2) (a+b x^2+c x^4)^{3/2}}{\sqrt {f x}} \, dx\) [210]

Optimal. Leaf size=297 \[ \frac {2 a d \sqrt {f x} \sqrt {a+b x^2+c x^4} F_1\left (\frac {1}{4};-\frac {3}{2},-\frac {3}{2};\frac {5}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}}+\frac {2 a e (f x)^{5/2} \sqrt {a+b x^2+c x^4} F_1\left (\frac {5}{4};-\frac {3}{2},-\frac {3}{2};\frac {9}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{5 f^3 \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}} \]

[Out]

2/5*a*e*(f*x)^(5/2)*AppellF1(5/4,-3/2,-3/2,9/4,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^2/(b+(-4*a*c+b^2)^(1/2))
)*(c*x^4+b*x^2+a)^(1/2)/f^3/(1+2*c*x^2/(b-(-4*a*c+b^2)^(1/2)))^(1/2)/(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^(1/2)+
2*a*d*AppellF1(1/4,-3/2,-3/2,5/4,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))*(f*x)^(1/2)*
(c*x^4+b*x^2+a)^(1/2)/f/(1+2*c*x^2/(b-(-4*a*c+b^2)^(1/2)))^(1/2)/(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^(1/2)

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Rubi [A]
time = 0.24, antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {1349, 1155, 524} \begin {gather*} \frac {2 a d \sqrt {f x} \sqrt {a+b x^2+c x^4} F_1\left (\frac {1}{4};-\frac {3}{2},-\frac {3}{2};\frac {5}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}}+\frac {2 a e (f x)^{5/2} \sqrt {a+b x^2+c x^4} F_1\left (\frac {5}{4};-\frac {3}{2},-\frac {3}{2};\frac {9}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{5 f^3 \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)*(a + b*x^2 + c*x^4)^(3/2))/Sqrt[f*x],x]

[Out]

(2*a*d*Sqrt[f*x]*Sqrt[a + b*x^2 + c*x^4]*AppellF1[1/4, -3/2, -3/2, 5/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-
2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(f*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sqrt
[b^2 - 4*a*c])]) + (2*a*e*(f*x)^(5/2)*Sqrt[a + b*x^2 + c*x^4]*AppellF1[5/4, -3/2, -3/2, 9/4, (-2*c*x^2)/(b - S
qrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(5*f^3*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqr
t[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 1155

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^2 +
c*x^4)^FracPart[p]/((1 + 2*c*(x^2/(b + Rt[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^2/(b - Rt[b^2 - 4*a*c, 2
])))^FracPart[p])), Int[(d*x)^m*(1 + 2*c*(x^2/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2*c*(x^2/(b - Sqrt[b^2 - 4*a*c]
)))^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x]

Rule 1349

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x]
&& NeQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0] || IntegersQ[m, q])

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{\sqrt {f x}} \, dx &=\int \left (\frac {d \left (a+b x^2+c x^4\right )^{3/2}}{\sqrt {f x}}+\frac {e (f x)^{3/2} \left (a+b x^2+c x^4\right )^{3/2}}{f^2}\right ) \, dx\\ &=d \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{\sqrt {f x}} \, dx+\frac {e \int (f x)^{3/2} \left (a+b x^2+c x^4\right )^{3/2} \, dx}{f^2}\\ &=\frac {\left (a d \sqrt {a+b x^2+c x^4}\right ) \int \frac {\left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{3/2} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{3/2}}{\sqrt {f x}} \, dx}{\sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}}+\frac {\left (a e \sqrt {a+b x^2+c x^4}\right ) \int (f x)^{3/2} \left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{3/2} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{3/2} \, dx}{f^2 \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}}\\ &=\frac {2 a d \sqrt {f x} \sqrt {a+b x^2+c x^4} F_1\left (\frac {1}{4};-\frac {3}{2},-\frac {3}{2};\frac {5}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}}+\frac {2 a e (f x)^{5/2} \sqrt {a+b x^2+c x^4} F_1\left (\frac {5}{4};-\frac {3}{2},-\frac {3}{2};\frac {9}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{5 f^3 \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}}}\\ \end {align*}

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Mathematica [A]
time = 10.80, size = 487, normalized size = 1.64 \begin {gather*} \frac {2 x \left (5 \left (a+b x^2+c x^4\right ) \left (-28 b^3 e+4 b^2 c \left (17 d+5 e x^2\right )+c^2 \left (867 a d+455 a e x^2+255 c d x^4+195 c e x^6\right )+b c \left (176 a e+5 c x^2 \left (85 d+57 e x^2\right )\right )\right )+20 a \left (-17 b^2 c d+612 a c^2 d+7 b^3 e-44 a b c e\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} F_1\left (\frac {1}{4};\frac {1}{2},\frac {1}{2};\frac {5}{4};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+4 \left (-51 b^3 c d+476 a b c^2 d+21 b^4 e-157 a b^2 c e+260 a^2 c^2 e\right ) x^2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} F_1\left (\frac {5}{4};\frac {1}{2},\frac {1}{2};\frac {9}{4};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )}{16575 c^2 \sqrt {f x} \sqrt {a+b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)*(a + b*x^2 + c*x^4)^(3/2))/Sqrt[f*x],x]

[Out]

(2*x*(5*(a + b*x^2 + c*x^4)*(-28*b^3*e + 4*b^2*c*(17*d + 5*e*x^2) + c^2*(867*a*d + 455*a*e*x^2 + 255*c*d*x^4 +
195*c*e*x^6) + b*c*(176*a*e + 5*c*x^2*(85*d + 57*e*x^2))) + 20*a*(-17*b^2*c*d + 612*a*c^2*d + 7*b^3*e - 44*a*
b*c*e)*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/
(b + Sqrt[b^2 - 4*a*c])]*AppellF1[1/4, 1/2, 1/2, 5/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt
[b^2 - 4*a*c])] + 4*(-51*b^3*c*d + 476*a*b*c^2*d + 21*b^4*e - 157*a*b^2*c*e + 260*a^2*c^2*e)*x^2*Sqrt[(b - Sqr
t[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*
c])]*AppellF1[5/4, 1/2, 1/2, 9/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]))/(1
6575*c^2*Sqrt[f*x]*Sqrt[a + b*x^2 + c*x^4])

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (e \,x^{2}+d \right ) \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{\sqrt {f x}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)*(c*x^4+b*x^2+a)^(3/2)/(f*x)^(1/2),x)

[Out]

int((e*x^2+d)*(c*x^4+b*x^2+a)^(3/2)/(f*x)^(1/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(c*x^4+b*x^2+a)^(3/2)/(f*x)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(c*x^4+b*x^2+a)^(3/2)/(f*x)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x^{2}\right ) \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}{\sqrt {f x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*(c*x**4+b*x**2+a)**(3/2)/(f*x)**(1/2),x)

[Out]

Integral((d + e*x**2)*(a + b*x**2 + c*x**4)**(3/2)/sqrt(f*x), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(c*x^4+b*x^2+a)^(3/2)/(f*x)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (e\,x^2+d\right )\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{\sqrt {f\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d + e*x^2)*(a + b*x^2 + c*x^4)^(3/2))/(f*x)^(1/2),x)

[Out]

int(((d + e*x^2)*(a + b*x^2 + c*x^4)^(3/2))/(f*x)^(1/2), x)

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