3.1137 \(\int \frac {c+d x^2}{(e x)^{7/2} (a+b x^2)^{9/4}} \, dx\)

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

[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 = 0.09, 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 = {453, 290, 286, 284, 335, 196} \[ \frac {12 (2 b c-a d)}{5 a^3 e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}-\frac {2 (2 b c-a d)}{5 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{5/4}}-\frac {24 \sqrt {b} \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (2 b c-a d) E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{7/2} e^4 \sqrt [4]{a+b x^2}}-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{5/4}} \]

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 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 286

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 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 453

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 ) \operatorname {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]  time = 0.04, size = 86, normalized size = 0.48 \[ \frac {2 x \left (a^2 (-c)-5 x^2 \left (a+b x^2\right ) \sqrt [4]{\frac {b x^2}{a}+1} (a d-2 b c) \, _2F_1\left (-\frac {1}{4},\frac {9}{4};\frac {3}{4};-\frac {b x^2}{a}\right )\right )}{5 a^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}} \]

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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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (d x^{2} + c\right )} \sqrt {e x}}{b^{3} e^{4} x^{10} + 3 \, a b^{2} e^{4} x^{8} + 3 \, a^{2} b e^{4} x^{6} + a^{3} e^{4} x^{4}}, x\right ) \]

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(e*x)/(b^3*e^4*x^10 + 3*a*b^2*e^4*x^8 + 3*a^2*b*e^4*x^6 + a^3*e^4*x
^4), x)

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

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)/((b*x^2 + a)^(9/4)*(e*x)^(7/2)), x)

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maple [F]  time = 0.13, 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 \[ \int \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} \left (e x\right )^{\frac {7}{2}}}\,{d x} \]

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]

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

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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