3.1122 \(\int \frac {(e x)^{3/2} (c+d x^2)}{(a+b x^2)^{7/4}} \, dx\)

Optimal. Leaf size=152 \[ -\frac {(e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} (2 b c-5 a d) F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 \sqrt {a} b^{3/2} \left (a+b x^2\right )^{3/4}}-\frac {e \sqrt {e x} \sqrt [4]{a+b x^2} (2 b c-5 a d)}{3 a b^2}+\frac {2 (e x)^{5/2} (b c-a d)}{3 a b e \left (a+b x^2\right )^{3/4}} \]

[Out]

2/3*(-a*d+b*c)*(e*x)^(5/2)/a/b/e/(b*x^2+a)^(3/4)-1/3*(-5*a*d+2*b*c)*(1+a/b/x^2)^(3/4)*(e*x)^(3/2)*(cos(1/2*arc
cot(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arccot(x*b^(1/2)/a^(1/2)
)),2^(1/2))/b^(3/2)/(b*x^2+a)^(3/4)/a^(1/2)-1/3*(-5*a*d+2*b*c)*e*(b*x^2+a)^(1/4)*(e*x)^(1/2)/a/b^2

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Rubi [A]  time = 0.11, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {457, 321, 329, 237, 335, 275, 231} \[ -\frac {e \sqrt {e x} \sqrt [4]{a+b x^2} (2 b c-5 a d)}{3 a b^2}-\frac {(e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} (2 b c-5 a d) F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 \sqrt {a} b^{3/2} \left (a+b x^2\right )^{3/4}}+\frac {2 (e x)^{5/2} (b c-a d)}{3 a b e \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(7/4),x]

[Out]

(2*(b*c - a*d)*(e*x)^(5/2))/(3*a*b*e*(a + b*x^2)^(3/4)) - ((2*b*c - 5*a*d)*e*Sqrt[e*x]*(a + b*x^2)^(1/4))/(3*a
*b^2) - ((2*b*c - 5*a*d)*(1 + a/(b*x^2))^(3/4)*(e*x)^(3/2)*EllipticF[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(3*Sqr
t[a]*b^(3/2)*(a + b*x^2)^(3/4))

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 237

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[(x^3*(1 + a/(b*x^4))^(3/4))/(a + b*x^4)^(3/4), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 321

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

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 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 457

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)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx &=\frac {2 (b c-a d) (e x)^{5/2}}{3 a b e \left (a+b x^2\right )^{3/4}}+\frac {\left (2 \left (-b c+\frac {5 a d}{2}\right )\right ) \int \frac {(e x)^{3/2}}{\left (a+b x^2\right )^{3/4}} \, dx}{3 a b}\\ &=\frac {2 (b c-a d) (e x)^{5/2}}{3 a b e \left (a+b x^2\right )^{3/4}}-\frac {(2 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{3 a b^2}+\frac {\left ((2 b c-5 a d) e^2\right ) \int \frac {1}{\sqrt {e x} \left (a+b x^2\right )^{3/4}} \, dx}{6 b^2}\\ &=\frac {2 (b c-a d) (e x)^{5/2}}{3 a b e \left (a+b x^2\right )^{3/4}}-\frac {(2 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{3 a b^2}+\frac {((2 b c-5 a d) e) \operatorname {Subst}\left (\int \frac {1}{\left (a+\frac {b x^4}{e^2}\right )^{3/4}} \, dx,x,\sqrt {e x}\right )}{3 b^2}\\ &=\frac {2 (b c-a d) (e x)^{5/2}}{3 a b e \left (a+b x^2\right )^{3/4}}-\frac {(2 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{3 a b^2}+\frac {\left ((2 b c-5 a d) e \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {a e^2}{b x^4}\right )^{3/4} x^3} \, dx,x,\sqrt {e x}\right )}{3 b^2 \left (a+b x^2\right )^{3/4}}\\ &=\frac {2 (b c-a d) (e x)^{5/2}}{3 a b e \left (a+b x^2\right )^{3/4}}-\frac {(2 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{3 a b^2}-\frac {\left ((2 b c-5 a d) e \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1+\frac {a e^2 x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{\sqrt {e x}}\right )}{3 b^2 \left (a+b x^2\right )^{3/4}}\\ &=\frac {2 (b c-a d) (e x)^{5/2}}{3 a b e \left (a+b x^2\right )^{3/4}}-\frac {(2 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{3 a b^2}-\frac {\left ((2 b c-5 a d) e \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {a e^2 x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{e x}\right )}{6 b^2 \left (a+b x^2\right )^{3/4}}\\ &=\frac {2 (b c-a d) (e x)^{5/2}}{3 a b e \left (a+b x^2\right )^{3/4}}-\frac {(2 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{3 a b^2}-\frac {(2 b c-5 a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 \sqrt {a} b^{3/2} \left (a+b x^2\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.13, size = 85, normalized size = 0.56 \[ \frac {e \sqrt {e x} \left (\left (\frac {b x^2}{a}+1\right )^{3/4} (2 b c-5 a d) \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {5}{4};-\frac {b x^2}{a}\right )+5 a d-2 b c+3 b d x^2\right )}{3 b^2 \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(7/4),x]

[Out]

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

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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d e x^{3} + c e x\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {e x}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {7}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*(e*x)^(3/2)/(b*x^2 + a)^(7/4), x)

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x \right )^{\frac {3}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {7}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x)

[Out]

int((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {7}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)*(e*x)^(3/2)/(b*x^2 + a)^(7/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e\,x\right )}^{3/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{7/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(7/4),x)

[Out]

int(((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(7/4), x)

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sympy [C]  time = 70.52, size = 94, normalized size = 0.62 \[ \frac {c e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {7}{4}} \Gamma \left (\frac {9}{4}\right )} + \frac {d e^{\frac {3}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {7}{4}} \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(3/2)*(d*x**2+c)/(b*x**2+a)**(7/4),x)

[Out]

c*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((5/4, 7/4), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(7/4)*gamma(9/4)) + d
*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((7/4, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(7/4)*gamma(13/4))

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