3.9.84 \(\int \frac {c+d x^2}{(e x)^{13/2} (a+b x^2)^{9/4}} \, dx\)

Optimal. Leaf size=178 \[ -\frac {256 \left (a+b x^2\right )^{7/4} (16 b c-11 a d)}{385 a^5 e^3 (e x)^{7/2}}+\frac {64 \left (a+b x^2\right )^{3/4} (16 b c-11 a d)}{55 a^4 e^3 (e x)^{7/2}}-\frac {24 (16 b c-11 a d)}{55 a^3 e^3 (e x)^{7/2} \sqrt [4]{a+b x^2}}-\frac {2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac {2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}} \]

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Rubi [A]  time = 0.09, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {453, 273, 264} \begin {gather*} -\frac {256 \left (a+b x^2\right )^{7/4} (16 b c-11 a d)}{385 a^5 e^3 (e x)^{7/2}}+\frac {64 \left (a+b x^2\right )^{3/4} (16 b c-11 a d)}{55 a^4 e^3 (e x)^{7/2}}-\frac {24 (16 b c-11 a d)}{55 a^3 e^3 (e x)^{7/2} \sqrt [4]{a+b x^2}}-\frac {2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac {2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)/((e*x)^(13/2)*(a + b*x^2)^(9/4)),x]

[Out]

(-2*c)/(11*a*e*(e*x)^(11/2)*(a + b*x^2)^(5/4)) - (2*(16*b*c - 11*a*d))/(55*a^2*e^3*(e*x)^(7/2)*(a + b*x^2)^(5/
4)) - (24*(16*b*c - 11*a*d))/(55*a^3*e^3*(e*x)^(7/2)*(a + b*x^2)^(1/4)) + (64*(16*b*c - 11*a*d)*(a + b*x^2)^(3
/4))/(55*a^4*e^3*(e*x)^(7/2)) - (256*(16*b*c - 11*a*d)*(a + b*x^2)^(7/4))/(385*a^5*e^3*(e*x)^(7/2))

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)^{13/2} \left (a+b x^2\right )^{9/4}} \, dx &=-\frac {2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac {(16 b c-11 a d) \int \frac {1}{(e x)^{9/2} \left (a+b x^2\right )^{9/4}} \, dx}{11 a e^2}\\ &=-\frac {2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac {(12 (16 b c-11 a d)) \int \frac {1}{(e x)^{9/2} \left (a+b x^2\right )^{5/4}} \, dx}{55 a^2 e^2}\\ &=-\frac {2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac {24 (16 b c-11 a d)}{55 a^3 e^3 (e x)^{7/2} \sqrt [4]{a+b x^2}}-\frac {(96 (16 b c-11 a d)) \int \frac {1}{(e x)^{9/2} \sqrt [4]{a+b x^2}} \, dx}{55 a^3 e^2}\\ &=-\frac {2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac {24 (16 b c-11 a d)}{55 a^3 e^3 (e x)^{7/2} \sqrt [4]{a+b x^2}}+\frac {64 (16 b c-11 a d) \left (a+b x^2\right )^{3/4}}{55 a^4 e^3 (e x)^{7/2}}+\frac {(128 (16 b c-11 a d)) \int \frac {\left (a+b x^2\right )^{3/4}}{(e x)^{9/2}} \, dx}{55 a^4 e^2}\\ &=-\frac {2 c}{11 a e (e x)^{11/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (16 b c-11 a d)}{55 a^2 e^3 (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac {24 (16 b c-11 a d)}{55 a^3 e^3 (e x)^{7/2} \sqrt [4]{a+b x^2}}+\frac {64 (16 b c-11 a d) \left (a+b x^2\right )^{3/4}}{55 a^4 e^3 (e x)^{7/2}}-\frac {256 (16 b c-11 a d) \left (a+b x^2\right )^{7/4}}{385 a^5 e^3 (e x)^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 79, normalized size = 0.44 \begin {gather*} \frac {2 x \left (a x^2 \left (-5 a^3+20 a^2 b x^2+160 a b^2 x^4+128 b^3 x^6\right ) (11 a d-16 b c)-35 a^5 c\right )}{385 a^6 (e x)^{13/2} \left (a+b x^2\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)/((e*x)^(13/2)*(a + b*x^2)^(9/4)),x]

[Out]

(2*x*(-35*a^5*c + a*(-16*b*c + 11*a*d)*x^2*(-5*a^3 + 20*a^2*b*x^2 + 160*a*b^2*x^4 + 128*b^3*x^6)))/(385*a^6*(e
*x)^(13/2)*(a + b*x^2)^(5/4))

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IntegrateAlgebraic [A]  time = 45.91, size = 160, normalized size = 0.90 \begin {gather*} \frac {2 \left (a+b x^2\right )^{3/4} \left (-35 a^4 c e^8-55 a^4 d e^8 x^2+80 a^3 b c e^8 x^2+220 a^3 b d e^8 x^4-320 a^2 b^2 c e^8 x^4+1760 a^2 b^2 d e^8 x^6-2560 a b^3 c e^8 x^6+1408 a b^3 d e^8 x^8-2048 b^4 c e^8 x^8\right )}{385 a^5 e^5 (e x)^{11/2} \left (a e^2+b e^2 x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + d*x^2)/((e*x)^(13/2)*(a + b*x^2)^(9/4)),x]

[Out]

(2*(a + b*x^2)^(3/4)*(-35*a^4*c*e^8 + 80*a^3*b*c*e^8*x^2 - 55*a^4*d*e^8*x^2 - 320*a^2*b^2*c*e^8*x^4 + 220*a^3*
b*d*e^8*x^4 - 2560*a*b^3*c*e^8*x^6 + 1760*a^2*b^2*d*e^8*x^6 - 2048*b^4*c*e^8*x^8 + 1408*a*b^3*d*e^8*x^8))/(385
*a^5*e^5*(e*x)^(11/2)*(a*e^2 + b*e^2*x^2)^2)

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fricas [A]  time = 1.08, size = 143, normalized size = 0.80 \begin {gather*} -\frac {2 \, {\left (128 \, {\left (16 \, b^{4} c - 11 \, a b^{3} d\right )} x^{8} + 160 \, {\left (16 \, a b^{3} c - 11 \, a^{2} b^{2} d\right )} x^{6} + 35 \, a^{4} c + 20 \, {\left (16 \, a^{2} b^{2} c - 11 \, a^{3} b d\right )} x^{4} - 5 \, {\left (16 \, a^{3} b c - 11 \, a^{4} d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x}}{385 \, {\left (a^{5} b^{2} e^{7} x^{10} + 2 \, a^{6} b e^{7} x^{8} + a^{7} e^{7} x^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(9/4),x, algorithm="fricas")

[Out]

-2/385*(128*(16*b^4*c - 11*a*b^3*d)*x^8 + 160*(16*a*b^3*c - 11*a^2*b^2*d)*x^6 + 35*a^4*c + 20*(16*a^2*b^2*c -
11*a^3*b*d)*x^4 - 5*(16*a^3*b*c - 11*a^4*d)*x^2)*(b*x^2 + a)^(3/4)*sqrt(e*x)/(a^5*b^2*e^7*x^10 + 2*a^6*b*e^7*x
^8 + a^7*e^7*x^6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} \left (e x\right )^{\frac {13}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(9/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(9/4)*(e*x)^(13/2)), x)

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maple [A]  time = 0.01, size = 110, normalized size = 0.62 \begin {gather*} -\frac {2 \left (-1408 a \,b^{3} d \,x^{8}+2048 b^{4} c \,x^{8}-1760 a^{2} b^{2} d \,x^{6}+2560 a \,b^{3} c \,x^{6}-220 a^{3} b d \,x^{4}+320 a^{2} b^{2} c \,x^{4}+55 a^{4} d \,x^{2}-80 a^{3} b c \,x^{2}+35 c \,a^{4}\right ) x}{385 \left (b \,x^{2}+a \right )^{\frac {5}{4}} \left (e x \right )^{\frac {13}{2}} a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(9/4),x)

[Out]

-2/385*x*(-1408*a*b^3*d*x^8+2048*b^4*c*x^8-1760*a^2*b^2*d*x^6+2560*a*b^3*c*x^6-220*a^3*b*d*x^4+320*a^2*b^2*c*x
^4+55*a^4*d*x^2-80*a^3*b*c*x^2+35*a^4*c)/(b*x^2+a)^(5/4)/a^5/(e*x)^(13/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} \left (e x\right )^{\frac {13}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(13/2)/(b*x^2+a)^(9/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(9/4)*(e*x)^(13/2)), x)

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mupad [B]  time = 1.41, size = 156, normalized size = 0.88 \begin {gather*} \frac {{\left (b\,x^2+a\right )}^{3/4}\,\left (\frac {64\,x^6\,\left (11\,a\,d-16\,b\,c\right )}{77\,a^4\,e^6}-\frac {2\,c}{11\,a\,b^2\,e^6}+\frac {8\,x^4\,\left (11\,a\,d-16\,b\,c\right )}{77\,a^3\,b\,e^6}-\frac {x^2\,\left (110\,a^4\,d-160\,a^3\,b\,c\right )}{385\,a^5\,b^2\,e^6}+\frac {256\,b\,x^8\,\left (11\,a\,d-16\,b\,c\right )}{385\,a^5\,e^6}\right )}{x^9\,\sqrt {e\,x}+\frac {a^2\,x^5\,\sqrt {e\,x}}{b^2}+\frac {2\,a\,x^7\,\sqrt {e\,x}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2)/((e*x)^(13/2)*(a + b*x^2)^(9/4)),x)

[Out]

((a + b*x^2)^(3/4)*((64*x^6*(11*a*d - 16*b*c))/(77*a^4*e^6) - (2*c)/(11*a*b^2*e^6) + (8*x^4*(11*a*d - 16*b*c))
/(77*a^3*b*e^6) - (x^2*(110*a^4*d - 160*a^3*b*c))/(385*a^5*b^2*e^6) + (256*b*x^8*(11*a*d - 16*b*c))/(385*a^5*e
^6)))/(x^9*(e*x)^(1/2) + (a^2*x^5*(e*x)^(1/2))/b^2 + (2*a*x^7*(e*x)^(1/2))/b)

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sympy [F(-1)]  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)**(13/2)/(b*x**2+a)**(9/4),x)

[Out]

Timed out

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