3.1120 \(\int \frac {c+d x^2}{(e x)^{11/2} (a+b x^2)^{7/4}} \, dx\)

Optimal. Leaf size=141 \[ -\frac {64 \left (a+b x^2\right )^{5/4} (4 b c-3 a d)}{45 a^4 e^3 (e x)^{5/2}}+\frac {16 \sqrt [4]{a+b x^2} (4 b c-3 a d)}{9 a^3 e^3 (e x)^{5/2}}-\frac {2 (4 b c-3 a d)}{9 a^2 e^3 (e x)^{5/2} \left (a+b x^2\right )^{3/4}}-\frac {2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{3/4}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {453, 273, 264} \[ -\frac {64 \left (a+b x^2\right )^{5/4} (4 b c-3 a d)}{45 a^4 e^3 (e x)^{5/2}}+\frac {16 \sqrt [4]{a+b x^2} (4 b c-3 a d)}{9 a^3 e^3 (e x)^{5/2}}-\frac {2 (4 b c-3 a d)}{9 a^2 e^3 (e x)^{5/2} \left (a+b x^2\right )^{3/4}}-\frac {2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(7/4)),x]

[Out]

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

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)^{11/2} \left (a+b x^2\right )^{7/4}} \, dx &=-\frac {2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{3/4}}-\frac {(4 b c-3 a d) \int \frac {1}{(e x)^{7/2} \left (a+b x^2\right )^{7/4}} \, dx}{3 a e^2}\\ &=-\frac {2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (4 b c-3 a d)}{9 a^2 e^3 (e x)^{5/2} \left (a+b x^2\right )^{3/4}}-\frac {(8 (4 b c-3 a d)) \int \frac {1}{(e x)^{7/2} \left (a+b x^2\right )^{3/4}} \, dx}{9 a^2 e^2}\\ &=-\frac {2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (4 b c-3 a d)}{9 a^2 e^3 (e x)^{5/2} \left (a+b x^2\right )^{3/4}}+\frac {16 (4 b c-3 a d) \sqrt [4]{a+b x^2}}{9 a^3 e^3 (e x)^{5/2}}+\frac {(32 (4 b c-3 a d)) \int \frac {\sqrt [4]{a+b x^2}}{(e x)^{7/2}} \, dx}{9 a^3 e^2}\\ &=-\frac {2 c}{9 a e (e x)^{9/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (4 b c-3 a d)}{9 a^2 e^3 (e x)^{5/2} \left (a+b x^2\right )^{3/4}}+\frac {16 (4 b c-3 a d) \sqrt [4]{a+b x^2}}{9 a^3 e^3 (e x)^{5/2}}-\frac {64 (4 b c-3 a d) \left (a+b x^2\right )^{5/4}}{45 a^4 e^3 (e x)^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 88, normalized size = 0.62 \[ \frac {4 x^3 \left (3 a^2-24 a b x^2-32 b^2 x^4\right ) \left (6 b c-\frac {9 a d}{2}\right )}{135 a^4 (e x)^{11/2} \left (a+b x^2\right )^{3/4}}-\frac {2 c x}{9 a (e x)^{11/2} \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(7/4)),x]

[Out]

(-2*c*x)/(9*a*(e*x)^(11/2)*(a + b*x^2)^(3/4)) + (4*(6*b*c - (9*a*d)/2)*x^3*(3*a^2 - 24*a*b*x^2 - 32*b^2*x^4))/
(135*a^4*(e*x)^(11/2)*(a + b*x^2)^(3/4))

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fricas [A]  time = 0.91, size = 105, normalized size = 0.74 \[ -\frac {2 \, {\left (32 \, {\left (4 \, b^{3} c - 3 \, a b^{2} d\right )} x^{6} + 24 \, {\left (4 \, a b^{2} c - 3 \, a^{2} b d\right )} x^{4} + 5 \, a^{3} c - 3 \, {\left (4 \, a^{2} b c - 3 \, a^{3} d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {e x}}{45 \, {\left (a^{4} b e^{6} x^{7} + a^{5} e^{6} x^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(7/4),x, algorithm="fricas")

[Out]

-2/45*(32*(4*b^3*c - 3*a*b^2*d)*x^6 + 24*(4*a*b^2*c - 3*a^2*b*d)*x^4 + 5*a^3*c - 3*(4*a^2*b*c - 3*a^3*d)*x^2)*
(b*x^2 + a)^(1/4)*sqrt(e*x)/(a^4*b*e^6*x^7 + a^5*e^6*x^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} \left (e x\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(7/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(7/4)*(e*x)^(11/2)), x)

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maple [A]  time = 0.01, size = 86, normalized size = 0.61 \[ -\frac {2 \left (-96 a \,b^{2} d \,x^{6}+128 b^{3} c \,x^{6}-72 a^{2} b d \,x^{4}+96 a \,b^{2} c \,x^{4}+9 a^{3} d \,x^{2}-12 a^{2} b c \,x^{2}+5 c \,a^{3}\right ) x}{45 \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (e x \right )^{\frac {11}{2}} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(7/4),x)

[Out]

-2/45*x*(-96*a*b^2*d*x^6+128*b^3*c*x^6-72*a^2*b*d*x^4+96*a*b^2*c*x^4+9*a^3*d*x^2-12*a^2*b*c*x^2+5*a^3*c)/(b*x^
2+a)^(3/4)/a^4/(e*x)^(11/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} \left (e x\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(7/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(7/4)*(e*x)^(11/2)), x)

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mupad [B]  time = 1.26, size = 125, normalized size = 0.89 \[ -\frac {{\left (b\,x^2+a\right )}^{1/4}\,\left (\frac {2\,c}{9\,a\,b\,e^5}-\frac {16\,x^4\,\left (3\,a\,d-4\,b\,c\right )}{15\,a^3\,e^5}+\frac {x^2\,\left (18\,a^3\,d-24\,a^2\,b\,c\right )}{45\,a^4\,b\,e^5}+\frac {x^6\,\left (256\,b^3\,c-192\,a\,b^2\,d\right )}{45\,a^4\,b\,e^5}\right )}{x^6\,\sqrt {e\,x}+\frac {a\,x^4\,\sqrt {e\,x}}{b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(7/4)),x)

[Out]

-((a + b*x^2)^(1/4)*((2*c)/(9*a*b*e^5) - (16*x^4*(3*a*d - 4*b*c))/(15*a^3*e^5) + (x^2*(18*a^3*d - 24*a^2*b*c))
/(45*a^4*b*e^5) + (x^6*(256*b^3*c - 192*a*b^2*d))/(45*a^4*b*e^5)))/(x^6*(e*x)^(1/2) + (a*x^4*(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)**(11/2)/(b*x**2+a)**(7/4),x)

[Out]

Timed out

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