3.1119 \(\int \frac {c+d x^2}{(e x)^{7/2} (a+b x^2)^{7/4}} \, dx\)

Optimal. Leaf size=104 \[ \frac {8 \sqrt [4]{a+b x^2} (8 b c-5 a d)}{15 a^3 e^3 \sqrt {e x}}-\frac {2 (8 b c-5 a d)}{15 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{3/4}}-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{3/4}} \]

[Out]

-2/5*c/a/e/(e*x)^(5/2)/(b*x^2+a)^(3/4)-2/15*(-5*a*d+8*b*c)/a^2/e^3/(b*x^2+a)^(3/4)/(e*x)^(1/2)+8/15*(-5*a*d+8*
b*c)*(b*x^2+a)^(1/4)/a^3/e^3/(e*x)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {453, 273, 264} \[ \frac {8 \sqrt [4]{a+b x^2} (8 b c-5 a d)}{15 a^3 e^3 \sqrt {e x}}-\frac {2 (8 b c-5 a d)}{15 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{3/4}}-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c)/(5*a*e*(e*x)^(5/2)*(a + b*x^2)^(3/4)) - (2*(8*b*c - 5*a*d))/(15*a^2*e^3*Sqrt[e*x]*(a + b*x^2)^(3/4)) +
(8*(8*b*c - 5*a*d)*(a + b*x^2)^(1/4))/(15*a^3*e^3*Sqrt[e*x])

Rule 264

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

Rule 273

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, n, p}, x] && ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[p, -1]

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 )^{7/4}} \, dx &=-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{3/4}}-\frac {(8 b c-5 a d) \int \frac {1}{(e x)^{3/2} \left (a+b x^2\right )^{7/4}} \, dx}{5 a e^2}\\ &=-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (8 b c-5 a d)}{15 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{3/4}}-\frac {(4 (8 b c-5 a d)) \int \frac {1}{(e x)^{3/2} \left (a+b x^2\right )^{3/4}} \, dx}{15 a^2 e^2}\\ &=-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (8 b c-5 a d)}{15 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{3/4}}+\frac {8 (8 b c-5 a d) \sqrt [4]{a+b x^2}}{15 a^3 e^3 \sqrt {e x}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 66, normalized size = 0.63 \[ \frac {x \left (-6 a^2 \left (c+5 d x^2\right )+8 a b x^2 \left (6 c-5 d x^2\right )+64 b^2 c x^4\right )}{15 a^3 (e x)^{7/2} \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(64*b^2*c*x^4 + 8*a*b*x^2*(6*c - 5*d*x^2) - 6*a^2*(c + 5*d*x^2)))/(15*a^3*(e*x)^(7/2)*(a + b*x^2)^(3/4))

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fricas [A]  time = 0.95, size = 81, normalized size = 0.78 \[ \frac {2 \, {\left (4 \, {\left (8 \, b^{2} c - 5 \, a b d\right )} x^{4} - 3 \, a^{2} c + 3 \, {\left (8 \, a b c - 5 \, a^{2} d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {e x}}{15 \, {\left (a^{3} b e^{4} x^{5} + a^{4} e^{4} x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(4*(8*b^2*c - 5*a*b*d)*x^4 - 3*a^2*c + 3*(8*a*b*c - 5*a^2*d)*x^2)*(b*x^2 + a)^(1/4)*sqrt(e*x)/(a^3*b*e^4*
x^5 + a^4*e^4*x^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 62, normalized size = 0.60 \[ -\frac {2 \left (20 a b d \,x^{4}-32 b^{2} c \,x^{4}+15 a^{2} d \,x^{2}-24 a b c \,x^{2}+3 c \,a^{2}\right ) x}{15 \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (e x \right )^{\frac {7}{2}} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*x*(20*a*b*d*x^4-32*b^2*c*x^4+15*a^2*d*x^2-24*a*b*c*x^2+3*a^2*c)/(b*x^2+a)^(3/4)/a^3/(e*x)^(7/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.21, size = 101, normalized size = 0.97 \[ -\frac {{\left (b\,x^2+a\right )}^{1/4}\,\left (\frac {2\,c}{5\,a\,b\,e^3}+\frac {x^2\,\left (30\,a^2\,d-48\,a\,b\,c\right )}{15\,a^3\,b\,e^3}-\frac {x^4\,\left (64\,b^2\,c-40\,a\,b\,d\right )}{15\,a^3\,b\,e^3}\right )}{x^4\,\sqrt {e\,x}+\frac {a\,x^2\,\sqrt {e\,x}}{b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a + b*x^2)^(1/4)*((2*c)/(5*a*b*e^3) + (x^2*(30*a^2*d - 48*a*b*c))/(15*a^3*b*e^3) - (x^4*(64*b^2*c - 40*a*b*
d))/(15*a^3*b*e^3)))/(x^4*(e*x)^(1/2) + (a*x^2*(e*x)^(1/2))/b)

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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)**(7/4),x)

[Out]

Timed out

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