3.12.37 \(\int \frac {c+d x^2}{(e x)^{7/2} (a+b x^2)^{9/4}} \, dx\) [1137]

Optimal. Leaf size=181 \[ -\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (2 b c-a d)}{5 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{5/4}}+\frac {12 (2 b c-a d)}{5 a^3 e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}-\frac {24 \sqrt {b} (2 b c-a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{7/2} e^4 \sqrt [4]{a+b x^2}} \]

[Out]

-2/5*c/a/e/(e*x)^(5/2)/(b*x^2+a)^(5/4)-2/5*(-a*d+2*b*c)/a^2/e^3/(b*x^2+a)^(5/4)/(e*x)^(1/2)+12/5*(-a*d+2*b*c)/
a^3/e^3/(b*x^2+a)^(1/4)/(e*x)^(1/2)-24/5*(-a*d+2*b*c)*(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)))*EllipticE(sin(1/2*arccot(x*b^(1/2)/a^(1/2))),2^(1/2))*b^(1/2)*(e*x)^
(1/2)/a^(7/2)/e^4/(b*x^2+a)^(1/4)

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Rubi [A]
time = 0.06, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {464, 296, 292, 290, 342, 202} \begin {gather*} -\frac {24 \sqrt {b} \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (2 b c-a d) E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{7/2} e^4 \sqrt [4]{a+b x^2}}+\frac {12 (2 b c-a d)}{5 a^3 e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}-\frac {2 (2 b c-a d)}{5 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{5/4}}-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)/((e*x)^(7/2)*(a + b*x^2)^(9/4)),x]

[Out]

(-2*c)/(5*a*e*(e*x)^(5/2)*(a + b*x^2)^(5/4)) - (2*(2*b*c - a*d))/(5*a^2*e^3*Sqrt[e*x]*(a + b*x^2)^(5/4)) + (12
*(2*b*c - a*d))/(5*a^3*e^3*Sqrt[e*x]*(a + b*x^2)^(1/4)) - (24*Sqrt[b]*(2*b*c - a*d)*(1 + a/(b*x^2))^(1/4)*Sqrt
[e*x]*EllipticE[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*a^(7/2)*e^4*(a + b*x^2)^(1/4))

Rule 202

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

Rule 290

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 292

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

Rule 296

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

Rule 342

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 464

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]

Rubi steps

\begin {align*} \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{9/4}} \, dx &=-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{5/4}}-\frac {(2 b c-a d) \int \frac {1}{(e x)^{3/2} \left (a+b x^2\right )^{9/4}} \, dx}{a e^2}\\ &=-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (2 b c-a d)}{5 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{5/4}}-\frac {(6 (2 b c-a d)) \int \frac {1}{(e x)^{3/2} \left (a+b x^2\right )^{5/4}} \, dx}{5 a^2 e^2}\\ &=-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (2 b c-a d)}{5 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{5/4}}+\frac {12 (2 b c-a d)}{5 a^3 e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}+\frac {(12 b (2 b c-a d)) \int \frac {\sqrt {e x}}{\left (a+b x^2\right )^{5/4}} \, dx}{5 a^3 e^4}\\ &=-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (2 b c-a d)}{5 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{5/4}}+\frac {12 (2 b c-a d)}{5 a^3 e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}+\frac {\left (12 (2 b c-a d) \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}{5 a^3 e^4 \sqrt [4]{a+b x^2}}\\ &=-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (2 b c-a d)}{5 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{5/4}}+\frac {12 (2 b c-a d)}{5 a^3 e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}-\frac {\left (12 (2 b c-a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{5 a^3 e^4 \sqrt [4]{a+b x^2}}\\ &=-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (2 b c-a d)}{5 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{5/4}}+\frac {12 (2 b c-a d)}{5 a^3 e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}-\frac {24 \sqrt {b} (2 b c-a d) \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 )}{5 a^{7/2} e^4 \sqrt [4]{a+b x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.04, size = 86, normalized size = 0.48 \begin {gather*} \frac {2 x \left (-a^2 c-5 (-2 b c+a d) x^2 \left (a+b x^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} \, _2F_1\left (-\frac {1}{4},\frac {9}{4};\frac {3}{4};-\frac {b x^2}{a}\right )\right )}{5 a^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)/((e*x)^(7/2)*(a + b*x^2)^(9/4)),x]

[Out]

(2*x*(-(a^2*c) - 5*(-2*b*c + a*d)*x^2*(a + b*x^2)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[-1/4, 9/4, 3/4, -((b
*x^2)/a)]))/(5*a^3*(e*x)^(7/2)*(a + b*x^2)^(5/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {d \,x^{2}+c}{\left (e x \right )^{\frac {7}{2}} \left (b \,x^{2}+a \right )^{\frac {9}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)/(e*x)^(7/2)/(b*x^2+a)^(9/4),x)

[Out]

int((d*x^2+c)/(e*x)^(7/2)/(b*x^2+a)^(9/4),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((d*x^2+c)/(e*x)^(7/2)/(b*x^2+a)^(9/4),x, algorithm="maxima")

[Out]

e^(-7/2)*integrate((d*x^2 + c)/((b*x^2 + a)^(9/4)*x^(7/2)), 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((d*x^2+c)/(e*x)^(7/2)/(b*x^2+a)^(9/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/4)*(d*x^2 + c)*sqrt(x)*e^(-7/2)/(b^3*x^10 + 3*a*b^2*x^8 + 3*a^2*b*x^6 + a^3*x^4), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)/(e*x)**(7/2)/(b*x**2+a)**(9/4),x)

[Out]

Timed out

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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((d*x^2+c)/(e*x)^(7/2)/(b*x^2+a)^(9/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*e^(-7/2)/((b*x^2 + a)^(9/4)*x^(7/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {d\,x^2+c}{{\left (e\,x\right )}^{7/2}\,{\left (b\,x^2+a\right )}^{9/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)/((e*x)^(7/2)*(a + b*x^2)^(9/4)),x)

[Out]

int((c + d*x^2)/((e*x)^(7/2)*(a + b*x^2)^(9/4)), x)

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