3.12.8 \(\int \frac {c+d x^2}{(e x)^{9/2} (a+b x^2)^{5/4}} \, dx\) [1108]

Optimal. Leaf size=104 \[ -\frac {2 c}{7 a e (e x)^{7/2} \sqrt [4]{a+b x^2}}-\frac {2 (8 b c-7 a d)}{7 a^2 e^3 (e x)^{3/2} \sqrt [4]{a+b x^2}}+\frac {8 (8 b c-7 a d) \left (a+b x^2\right )^{3/4}}{21 a^3 e^3 (e x)^{3/2}} \]

[Out]

-2/7*c/a/e/(e*x)^(7/2)/(b*x^2+a)^(1/4)-2/7*(-7*a*d+8*b*c)/a^2/e^3/(e*x)^(3/2)/(b*x^2+a)^(1/4)+8/21*(-7*a*d+8*b
*c)*(b*x^2+a)^(3/4)/a^3/e^3/(e*x)^(3/2)

________________________________________________________________________________________

Rubi [A]
time = 0.03, 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 = {464, 279, 270} \begin {gather*} \frac {8 \left (a+b x^2\right )^{3/4} (8 b c-7 a d)}{21 a^3 e^3 (e x)^{3/2}}-\frac {2 (8 b c-7 a d)}{7 a^2 e^3 (e x)^{3/2} \sqrt [4]{a+b x^2}}-\frac {2 c}{7 a e (e x)^{7/2} \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)/((e*x)^(9/2)*(a + b*x^2)^(5/4)),x]

[Out]

(-2*c)/(7*a*e*(e*x)^(7/2)*(a + b*x^2)^(1/4)) - (2*(8*b*c - 7*a*d))/(7*a^2*e^3*(e*x)^(3/2)*(a + b*x^2)^(1/4)) +
(8*(8*b*c - 7*a*d)*(a + b*x^2)^(3/4))/(21*a^3*e^3*(e*x)^(3/2))

Rule 270

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 279

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

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)^{9/2} \left (a+b x^2\right )^{5/4}} \, dx &=-\frac {2 c}{7 a e (e x)^{7/2} \sqrt [4]{a+b x^2}}-\frac {(8 b c-7 a d) \int \frac {1}{(e x)^{5/2} \left (a+b x^2\right )^{5/4}} \, dx}{7 a e^2}\\ &=-\frac {2 c}{7 a e (e x)^{7/2} \sqrt [4]{a+b x^2}}-\frac {2 (8 b c-7 a d)}{7 a^2 e^3 (e x)^{3/2} \sqrt [4]{a+b x^2}}-\frac {(4 (8 b c-7 a d)) \int \frac {1}{(e x)^{5/2} \sqrt [4]{a+b x^2}} \, dx}{7 a^2 e^2}\\ &=-\frac {2 c}{7 a e (e x)^{7/2} \sqrt [4]{a+b x^2}}-\frac {2 (8 b c-7 a d)}{7 a^2 e^3 (e x)^{3/2} \sqrt [4]{a+b x^2}}+\frac {8 (8 b c-7 a d) \left (a+b x^2\right )^{3/4}}{21 a^3 e^3 (e x)^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.44, size = 67, normalized size = 0.64 \begin {gather*} -\frac {2 x \left (3 a^2 c-8 a b c x^2+7 a^2 d x^2-32 b^2 c x^4+28 a b d x^4\right )}{21 a^3 (e x)^{9/2} \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)/((e*x)^(9/2)*(a + b*x^2)^(5/4)),x]

[Out]

(-2*x*(3*a^2*c - 8*a*b*c*x^2 + 7*a^2*d*x^2 - 32*b^2*c*x^4 + 28*a*b*d*x^4))/(21*a^3*(e*x)^(9/2)*(a + b*x^2)^(1/
4))

________________________________________________________________________________________

Maple [A]
time = 0.11, size = 62, normalized size = 0.60

method	result	size
gosper	\(-\frac {2 x \left (28 a b d \,x^{4}-32 b^{2} c \,x^{4}+7 a^{2} d \,x^{2}-8 a b c \,x^{2}+3 a^{2} c \right )}{21 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{3} \left (e x \right )^{\frac {9}{2}}}\)	\(62\)
risch	\(-\frac {2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (7 a d \,x^{2}-11 c \,x^{2} b +3 a c \right )}{21 a^{3} x^{3} e^{4} \sqrt {e x}}-\frac {2 b x \left (a d -b c \right )}{a^{3} e^{4} \sqrt {e x}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}}}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)/(e*x)^(9/2)/(b*x^2+a)^(5/4),x,method=_RETURNVERBOSE)

[Out]

-2/21*x*(28*a*b*d*x^4-32*b^2*c*x^4+7*a^2*d*x^2-8*a*b*c*x^2+3*a^2*c)/(b*x^2+a)^(1/4)/a^3/(e*x)^(9/2)

________________________________________________________________________________________

Maxima [A]
time = 0.28, size = 101, normalized size = 0.97 \begin {gather*} -\frac {2}{21} \, {\left (7 \, d {\left (\frac {3 \, b \sqrt {x}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} a^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}}}{a^{2} x^{\frac {3}{2}}}\right )} - c {\left (\frac {21 \, b^{2} \sqrt {x}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} a^{3}} + \frac {\frac {14 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} b}{x^{\frac {3}{2}}} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}}}{x^{\frac {7}{2}}}}{a^{3}}\right )}\right )} e^{\left (-\frac {9}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(9/2)/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

-2/21*(7*d*(3*b*sqrt(x)/((b*x^2 + a)^(1/4)*a^2) + (b*x^2 + a)^(3/4)/(a^2*x^(3/2))) - c*(21*b^2*sqrt(x)/((b*x^2
+ a)^(1/4)*a^3) + (14*(b*x^2 + a)^(3/4)*b/x^(3/2) - 3*(b*x^2 + a)^(7/4)/x^(7/2))/a^3))*e^(-9/2)

________________________________________________________________________________________

Fricas [A]
time = 1.55, size = 74, normalized size = 0.71 \begin {gather*} \frac {2 \, {\left (4 \, {\left (8 \, b^{2} c - 7 \, a b d\right )} x^{4} - 3 \, a^{2} c + {\left (8 \, a b c - 7 \, a^{2} d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {x} e^{\left (-\frac {9}{2}\right )}}{21 \, {\left (a^{3} b x^{6} + a^{4} x^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(9/2)/(b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

2/21*(4*(8*b^2*c - 7*a*b*d)*x^4 - 3*a^2*c + (8*a*b*c - 7*a^2*d)*x^2)*(b*x^2 + a)^(3/4)*sqrt(x)*e^(-9/2)/(a^3*b
*x^6 + a^4*x^4)

________________________________________________________________________________________

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)**(9/2)/(b*x**2+a)**(5/4),x)

[Out]

Timed out

________________________________________________________________________________________

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)^(9/2)/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*e^(-9/2)/((b*x^2 + a)^(5/4)*x^(9/2)), x)

________________________________________________________________________________________

Mupad [B]
time = 0.66, size = 101, normalized size = 0.97 \begin {gather*} -\frac {{\left (b\,x^2+a\right )}^{3/4}\,\left (\frac {2\,c}{7\,a\,b\,e^4}+\frac {x^2\,\left (14\,a^2\,d-16\,a\,b\,c\right )}{21\,a^3\,b\,e^4}-\frac {x^4\,\left (64\,b^2\,c-56\,a\,b\,d\right )}{21\,a^3\,b\,e^4}\right )}{x^5\,\sqrt {e\,x}+\frac {a\,x^3\,\sqrt {e\,x}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)/((e*x)^(9/2)*(a + b*x^2)^(5/4)),x)

[Out]

-((a + b*x^2)^(3/4)*((2*c)/(7*a*b*e^4) + (x^2*(14*a^2*d - 16*a*b*c))/(21*a^3*b*e^4) - (x^4*(64*b^2*c - 56*a*b*
d))/(21*a^3*b*e^4)))/(x^5*(e*x)^(1/2) + (a*x^3*(e*x)^(1/2))/b)

________________________________________________________________________________________