3.1111 \(\int \frac {(e x)^{5/2} (c+d x^2)}{(a+b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=142 \[ \frac {\sqrt {a} e^2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (6 b c-7 a d) E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 b^{5/2} \sqrt [4]{a+b x^2}}+\frac {e (e x)^{3/2} (6 b c-7 a d)}{6 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{7/2}}{3 b e \sqrt [4]{a+b x^2}} \]

[Out]

1/6*(-7*a*d+6*b*c)*e*(e*x)^(3/2)/b^2/(b*x^2+a)^(1/4)+1/3*d*(e*x)^(7/2)/b/e/(b*x^2+a)^(1/4)+1/2*(-7*a*d+6*b*c)*
e^2*(1+a/b/x^2)^(1/4)*(cos(1/2*arccot(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x*b^(1/2)/a^(1/2)))*Elliptic
E(sin(1/2*arccot(x*b^(1/2)/a^(1/2))),2^(1/2))*a^(1/2)*(e*x)^(1/2)/b^(5/2)/(b*x^2+a)^(1/4)

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Rubi [A]  time = 0.07, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {459, 285, 284, 335, 196} \[ \frac {\sqrt {a} e^2 \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (6 b c-7 a d) E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 b^{5/2} \sqrt [4]{a+b x^2}}+\frac {e (e x)^{3/2} (6 b c-7 a d)}{6 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{7/2}}{3 b e \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x]

[Out]

((6*b*c - 7*a*d)*e*(e*x)^(3/2))/(6*b^2*(a + b*x^2)^(1/4)) + (d*(e*x)^(7/2))/(3*b*e*(a + b*x^2)^(1/4)) + (Sqrt[
a]*(6*b*c - 7*a*d)*e^2*(1 + a/(b*x^2))^(1/4)*Sqrt[e*x]*EllipticE[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(2*b^(5/2)
*(a + b*x^2)^(1/4))

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 284

Int[Sqrt[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Dist[(Sqrt[c*x]*(1 + a/(b*x^2))^(1/4))/(b*(a +
b*x^2)^(1/4)), Int[1/(x^2*(1 + a/(b*x^2))^(5/4)), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]

Rule 285

Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[(2*c*(c*x)^(m - 1))/(b*(2*m - 3)*(a + b*x
^2)^(1/4)), x] - Dist[(2*a*c^2*(m - 1))/(b*(2*m - 3)), Int[(c*x)^(m - 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a
, b, c}, x] && PosQ[b/a] && IntegerQ[2*m] && GtQ[m, 3/2]

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 459

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

Rubi steps

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

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Mathematica [C]  time = 0.12, size = 85, normalized size = 0.60 \[ \frac {e (e x)^{3/2} \left (\sqrt [4]{\frac {b x^2}{a}+1} (7 a d-6 b c) \, _2F_1\left (\frac {3}{4},\frac {5}{4};\frac {7}{4};-\frac {b x^2}{a}\right )-7 a d+6 b c+2 b d x^2\right )}{6 b^2 \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x]

[Out]

(e*(e*x)^(3/2)*(6*b*c - 7*a*d + 2*b*d*x^2 + (-6*b*c + 7*a*d)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[3/4, 5/4,
7/4, -((b*x^2)/a)]))/(6*b^2*(a + b*x^2)^(1/4))

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fricas [F]  time = 1.30, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d e^{2} x^{4} + c e^{2} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

integral((d*e^2*x^4 + c*e^2*x^2)*(b*x^2 + a)^(3/4)*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 \[ \int \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*(e*x)^(5/2)/(b*x^2 + a)^(5/4), x)

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x \right )^{\frac {5}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)

[Out]

int((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)*(e*x)^(5/2)/(b*x^2 + a)^(5/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e\,x\right )}^{5/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x)

[Out]

int(((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(5/4), x)

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sympy [C]  time = 155.67, size = 94, normalized size = 0.66 \[ \frac {c e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {11}{4}\right )} + \frac {d e^{\frac {5}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {15}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(5/2)*(d*x**2+c)/(b*x**2+a)**(5/4),x)

[Out]

c*e**(5/2)*x**(7/2)*gamma(7/4)*hyper((5/4, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*gamma(11/4)) +
d*e**(5/2)*x**(11/2)*gamma(11/4)*hyper((5/4, 11/4), (15/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*gamma(15/4
))

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