3.9.48 \(\int \frac {(e x)^m (a+b x^4)}{(c+d x^4)^{3/2}} \, dx\) [848]

Optimal. Leaf size=132 \[ -\frac {(b c-a d) (e x)^{1+m}}{2 c d e \sqrt {c+d x^4}}+\frac {(a d (1-m)+b c (1+m)) (e x)^{1+m} \sqrt {1+\frac {d x^4}{c}} \, _2F_1\left (\frac {1}{2},\frac {1+m}{4};\frac {5+m}{4};-\frac {d x^4}{c}\right )}{2 c d e (1+m) \sqrt {c+d x^4}} \]

[Out]

-1/2*(-a*d+b*c)*(e*x)^(1+m)/c/d/e/(d*x^4+c)^(1/2)+1/2*(a*d*(1-m)+b*c*(1+m))*(e*x)^(1+m)*hypergeom([1/2, 1/4+1/
4*m],[5/4+1/4*m],-d*x^4/c)*(1+d*x^4/c)^(1/2)/c/d/e/(1+m)/(d*x^4+c)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {468, 372, 371} \begin {gather*} \frac {\sqrt {\frac {d x^4}{c}+1} (e x)^{m+1} (a d (1-m)+b c (m+1)) \, _2F_1\left (\frac {1}{2},\frac {m+1}{4};\frac {m+5}{4};-\frac {d x^4}{c}\right )}{2 c d e (m+1) \sqrt {c+d x^4}}-\frac {(e x)^{m+1} (b c-a d)}{2 c d e \sqrt {c+d x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d
))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b*e*n*(p + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a
*b*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0]
&& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0]
&& LeQ[-1, m, (-n)*(p + 1)]))

Rubi steps

\begin {align*} \int \frac {(e x)^m \left (a+b x^4\right )}{\left (c+d x^4\right )^{3/2}} \, dx &=-\frac {(b c-a d) (e x)^{1+m}}{2 c d e \sqrt {c+d x^4}}+\frac {(-a d (-1+m)+b c (1+m)) \int \frac {(e x)^m}{\sqrt {c+d x^4}} \, dx}{2 c d}\\ &=-\frac {(b c-a d) (e x)^{1+m}}{2 c d e \sqrt {c+d x^4}}+\frac {\left ((-a d (-1+m)+b c (1+m)) \sqrt {1+\frac {d x^4}{c}}\right ) \int \frac {(e x)^m}{\sqrt {1+\frac {d x^4}{c}}} \, dx}{2 c d \sqrt {c+d x^4}}\\ &=-\frac {(b c-a d) (e x)^{1+m}}{2 c d e \sqrt {c+d x^4}}+\frac {(a d (1-m)+b c (1+m)) (e x)^{1+m} \sqrt {1+\frac {d x^4}{c}} \, _2F_1\left (\frac {1}{2},\frac {1+m}{4};\frac {5+m}{4};-\frac {d x^4}{c}\right )}{2 c d e (1+m) \sqrt {c+d x^4}}\\ \end {align*}

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Mathematica [A]
time = 3.51, size = 110, normalized size = 0.83 \begin {gather*} \frac {x (e x)^m \sqrt {1+\frac {d x^4}{c}} \left (b c \, _2F_1\left (\frac {1}{2},\frac {1+m}{4};\frac {5+m}{4};-\frac {d x^4}{c}\right )+(-b c+a d) \, _2F_1\left (\frac {3}{2},\frac {1+m}{4};\frac {5+m}{4};-\frac {d x^4}{c}\right )\right )}{c d (1+m) \sqrt {c+d x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (b \,x^{4}+a \right )}{\left (d \,x^{4}+c \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((b*x^4 + a)*sqrt(d*x^4 + c)*(x*e)^m/(d^2*x^8 + 2*c*d*x^4 + c^2), x)

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Sympy [C] Result contains complex when optimal does not.
time = 22.15, size = 119, normalized size = 0.90 \begin {gather*} \frac {a e^{m} x x^{m} \Gamma \left (\frac {m}{4} + \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{4} + \frac {1}{4} \\ \frac {m}{4} + \frac {5}{4} \end {matrix}\middle | {\frac {d x^{4} e^{i \pi }}{c}} \right )}}{4 c^{\frac {3}{2}} \Gamma \left (\frac {m}{4} + \frac {5}{4}\right )} + \frac {b e^{m} x^{5} x^{m} \Gamma \left (\frac {m}{4} + \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{4} + \frac {5}{4} \\ \frac {m}{4} + \frac {9}{4} \end {matrix}\middle | {\frac {d x^{4} e^{i \pi }}{c}} \right )}}{4 c^{\frac {3}{2}} \Gamma \left (\frac {m}{4} + \frac {9}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*e**m*x*x**m*gamma(m/4 + 1/4)*hyper((3/2, m/4 + 1/4), (m/4 + 5/4,), d*x**4*exp_polar(I*pi)/c)/(4*c**(3/2)*gam
ma(m/4 + 5/4)) + b*e**m*x**5*x**m*gamma(m/4 + 5/4)*hyper((3/2, m/4 + 5/4), (m/4 + 9/4,), d*x**4*exp_polar(I*pi
)/c)/(4*c**(3/2)*gamma(m/4 + 9/4))

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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)^m*(b*x^4+a)/(d*x^4+c)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (e\,x\right )}^m\,\left (b\,x^4+a\right )}{{\left (d\,x^4+c\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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