3.11.97 \(\int \frac {c+d x^2}{(e x)^{11/2} (a+b x^2)^{3/4}} \, dx\) [1097]

Optimal. Leaf size=104 \[ -\frac {2 c \sqrt [4]{a+b x^2}}{9 a e (e x)^{9/2}}+\frac {2 (8 b c-9 a d) \sqrt [4]{a+b x^2}}{9 a^2 e^3 (e x)^{5/2}}-\frac {8 (8 b c-9 a d) \left (a+b x^2\right )^{5/4}}{45 a^3 e^3 (e x)^{5/2}} \]

[Out]

-2/9*c*(b*x^2+a)^(1/4)/a/e/(e*x)^(9/2)+2/9*(-9*a*d+8*b*c)*(b*x^2+a)^(1/4)/a^2/e^3/(e*x)^(5/2)-8/45*(-9*a*d+8*b
*c)*(b*x^2+a)^(5/4)/a^3/e^3/(e*x)^(5/2)

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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 )^{5/4} (8 b c-9 a d)}{45 a^3 e^3 (e x)^{5/2}}+\frac {2 \sqrt [4]{a+b x^2} (8 b c-9 a d)}{9 a^2 e^3 (e x)^{5/2}}-\frac {2 c \sqrt [4]{a+b x^2}}{9 a e (e x)^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(3/4)),x]

[Out]

(-2*c*(a + b*x^2)^(1/4))/(9*a*e*(e*x)^(9/2)) + (2*(8*b*c - 9*a*d)*(a + b*x^2)^(1/4))/(9*a^2*e^3*(e*x)^(5/2)) -
(8*(8*b*c - 9*a*d)*(a + b*x^2)^(5/4))/(45*a^3*e^3*(e*x)^(5/2))

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)^{11/2} \left (a+b x^2\right )^{3/4}} \, dx &=-\frac {2 c \sqrt [4]{a+b x^2}}{9 a e (e x)^{9/2}}-\frac {(8 b c-9 a d) \int \frac {1}{(e x)^{7/2} \left (a+b x^2\right )^{3/4}} \, dx}{9 a e^2}\\ &=-\frac {2 c \sqrt [4]{a+b x^2}}{9 a e (e x)^{9/2}}+\frac {2 (8 b c-9 a d) \sqrt [4]{a+b x^2}}{9 a^2 e^3 (e x)^{5/2}}+\frac {(4 (8 b c-9 a d)) \int \frac {\sqrt [4]{a+b x^2}}{(e x)^{7/2}} \, dx}{9 a^2 e^2}\\ &=-\frac {2 c \sqrt [4]{a+b x^2}}{9 a e (e x)^{9/2}}+\frac {2 (8 b c-9 a d) \sqrt [4]{a+b x^2}}{9 a^2 e^3 (e x)^{5/2}}-\frac {8 (8 b c-9 a d) \left (a+b x^2\right )^{5/4}}{45 a^3 e^3 (e x)^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.37, size = 67, normalized size = 0.64 \begin {gather*} -\frac {2 x \sqrt [4]{a+b x^2} \left (5 a^2 c-8 a b c x^2+9 a^2 d x^2+32 b^2 c x^4-36 a b d x^4\right )}{45 a^3 (e x)^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(3/4)),x]

[Out]

(-2*x*(a + b*x^2)^(1/4)*(5*a^2*c - 8*a*b*c*x^2 + 9*a^2*d*x^2 + 32*b^2*c*x^4 - 36*a*b*d*x^4))/(45*a^3*(e*x)^(11
/2))

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

method	result	size
gosper	\(-\frac {2 x \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (-36 a b d \,x^{4}+32 b^{2} c \,x^{4}+9 a^{2} d \,x^{2}-8 a b c \,x^{2}+5 a^{2} c \right )}{45 a^{3} \left (e x \right )^{\frac {11}{2}}}\)	\(62\)
risch	\(-\frac {2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (-36 a b d \,x^{4}+32 b^{2} c \,x^{4}+9 a^{2} d \,x^{2}-8 a b c \,x^{2}+5 a^{2} c \right )}{45 e^{5} \sqrt {e x}\, a^{3} x^{4}}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(3/4),x,method=_RETURNVERBOSE)

[Out]

-2/45*x*(b*x^2+a)^(1/4)*(-36*a*b*d*x^4+32*b^2*c*x^4+9*a^2*d*x^2-8*a*b*c*x^2+5*a^2*c)/a^3/(e*x)^(11/2)

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Maxima [A]
time = 0.27, size = 94, normalized size = 0.90 \begin {gather*} \frac {2}{45} \, {\left (\frac {9 \, d {\left (\frac {5 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}} b}{\sqrt {x}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{4}}}{x^{\frac {5}{2}}}\right )}}{a^{2}} - \frac {{\left (\frac {45 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}} b^{2}}{\sqrt {x}} - \frac {18 \, {\left (b x^{2} + a\right )}^{\frac {5}{4}} b}{x^{\frac {5}{2}}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {9}{4}}}{x^{\frac {9}{2}}}\right )} c}{a^{3}}\right )} e^{\left (-\frac {11}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

2/45*(9*d*(5*(b*x^2 + a)^(1/4)*b/sqrt(x) - (b*x^2 + a)^(5/4)/x^(5/2))/a^2 - (45*(b*x^2 + a)^(1/4)*b^2/sqrt(x)
- 18*(b*x^2 + a)^(5/4)*b/x^(5/2) + 5*(b*x^2 + a)^(9/4)/x^(9/2))*c/a^3)*e^(-11/2)

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Fricas [A]
time = 1.21, size = 60, normalized size = 0.58 \begin {gather*} -\frac {2 \, {\left (4 \, {\left (8 \, b^{2} c - 9 \, a b d\right )} x^{4} + 5 \, a^{2} c - {\left (8 \, a b c - 9 \, a^{2} d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}} e^{\left (-\frac {11}{2}\right )}}{45 \, a^{3} x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

-2/45*(4*(8*b^2*c - 9*a*b*d)*x^4 + 5*a^2*c - (8*a*b*c - 9*a^2*d)*x^2)*(b*x^2 + a)^(1/4)*e^(-11/2)/(a^3*x^(9/2)
)

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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)**(11/2)/(b*x**2+a)**(3/4),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)^(11/2)/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*e^(-11/2)/((b*x^2 + a)^(3/4)*x^(11/2)), x)

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Mupad [B]
time = 0.69, size = 75, normalized size = 0.72 \begin {gather*} -\frac {{\left (b\,x^2+a\right )}^{1/4}\,\left (\frac {2\,c}{9\,a\,e^5}+\frac {x^2\,\left (18\,a^2\,d-16\,a\,b\,c\right )}{45\,a^3\,e^5}+\frac {x^4\,\left (64\,b^2\,c-72\,a\,b\,d\right )}{45\,a^3\,e^5}\right )}{x^4\,\sqrt {e\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(3/4)),x)

[Out]

-((a + b*x^2)^(1/4)*((2*c)/(9*a*e^5) + (x^2*(18*a^2*d - 16*a*b*c))/(45*a^3*e^5) + (x^4*(64*b^2*c - 72*a*b*d))/
(45*a^3*e^5)))/(x^4*(e*x)^(1/2))

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