3.12.19 \(\int \frac {c+d x^2}{(e x)^{7/2} (a+b x^2)^{7/4}} \, dx\) [1119]

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

[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 = 0.03, 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 \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}} \end {gather*}

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 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)^{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 = 0.40, size = 67, normalized size = 0.64 \begin {gather*} -\frac {2 x \left (3 a^2 c-24 a b c x^2+15 a^2 d x^2-32 b^2 c x^4+20 a b d x^4\right )}{15 a^3 (e x)^{7/2} \left (a+b x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

method	result	size
gosper	\(-\frac {2 x \left (20 a b d \,x^{4}-32 b^{2} c \,x^{4}+15 a^{2} d \,x^{2}-24 a b c \,x^{2}+3 a^{2} c \right )}{15 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{3} \left (e x \right )^{\frac {7}{2}}}\)	\(62\)
risch	\(-\frac {2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (5 a d \,x^{2}-9 c \,x^{2} b +a c \right )}{5 a^{3} x^{2} e^{3} \sqrt {e x}}-\frac {2 b \left (a d -b c \right ) x^{2}}{3 a^{3} e^{3} \sqrt {e x}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}}}\)	\(79\)

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,method=_RETURNVERBOSE)

[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 [A]
time = 0.27, size = 102, normalized size = 0.98 \begin {gather*} -\frac {2}{15} \, {\left (5 \, d {\left (\frac {b x^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} a^{2}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {1}{4}}}{a^{2} \sqrt {x}}\right )} - c {\left (\frac {5 \, b^{2} x^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} a^{3}} + \frac {3 \, {\left (\frac {10 \, {\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^{3}}\right )}\right )} e^{\left (-\frac {7}{2}\right )} \end {gather*}

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]

-2/15*(5*d*(b*x^(3/2)/((b*x^2 + a)^(3/4)*a^2) + 3*(b*x^2 + a)^(1/4)/(a^2*sqrt(x))) - c*(5*b^2*x^(3/2)/((b*x^2
+ a)^(3/4)*a^3) + 3*(10*(b*x^2 + a)^(1/4)*b/sqrt(x) - (b*x^2 + a)^(5/4)/x^(5/2))/a^3))*e^(-7/2)

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Fricas [A]
time = 0.86, size = 75, normalized size = 0.72 \begin {gather*} \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 {x} e^{\left (-\frac {7}{2}\right )}}{15 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}} \end {gather*}

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(x)*e^(-7/2)/(a^3
*b*x^5 + a^4*x^3)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 469 vs. \(2 (99) = 198\).
time = 166.52, size = 469, normalized size = 4.51 \begin {gather*} c \left (- \frac {3 a^{3} b^{\frac {17}{4}} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (- \frac {5}{4}\right )}{32 a^{5} b^{4} e^{\frac {7}{2}} x^{2} \Gamma \left (\frac {7}{4}\right ) + 64 a^{4} b^{5} e^{\frac {7}{2}} x^{4} \Gamma \left (\frac {7}{4}\right ) + 32 a^{3} b^{6} e^{\frac {7}{2}} x^{6} \Gamma \left (\frac {7}{4}\right )} + \frac {21 a^{2} b^{\frac {21}{4}} x^{2} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (- \frac {5}{4}\right )}{32 a^{5} b^{4} e^{\frac {7}{2}} x^{2} \Gamma \left (\frac {7}{4}\right ) + 64 a^{4} b^{5} e^{\frac {7}{2}} x^{4} \Gamma \left (\frac {7}{4}\right ) + 32 a^{3} b^{6} e^{\frac {7}{2}} x^{6} \Gamma \left (\frac {7}{4}\right )} + \frac {56 a b^{\frac {25}{4}} x^{4} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (- \frac {5}{4}\right )}{32 a^{5} b^{4} e^{\frac {7}{2}} x^{2} \Gamma \left (\frac {7}{4}\right ) + 64 a^{4} b^{5} e^{\frac {7}{2}} x^{4} \Gamma \left (\frac {7}{4}\right ) + 32 a^{3} b^{6} e^{\frac {7}{2}} x^{6} \Gamma \left (\frac {7}{4}\right )} + \frac {32 b^{\frac {29}{4}} x^{6} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (- \frac {5}{4}\right )}{32 a^{5} b^{4} e^{\frac {7}{2}} x^{2} \Gamma \left (\frac {7}{4}\right ) + 64 a^{4} b^{5} e^{\frac {7}{2}} x^{4} \Gamma \left (\frac {7}{4}\right ) + 32 a^{3} b^{6} e^{\frac {7}{2}} x^{6} \Gamma \left (\frac {7}{4}\right )}\right ) + d \left (\frac {3 \Gamma \left (- \frac {1}{4}\right )}{8 a b^{\frac {3}{4}} e^{\frac {7}{2}} x^{2} \left (\frac {a}{b x^{2}} + 1\right )^{\frac {3}{4}} \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt [4]{b} \Gamma \left (- \frac {1}{4}\right )}{2 a^{2} e^{\frac {7}{2}} \left (\frac {a}{b x^{2}} + 1\right )^{\frac {3}{4}} \Gamma \left (\frac {7}{4}\right )}\right ) \end {gather*}

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]

c*(-3*a**3*b**(17/4)*(a/(b*x**2) + 1)**(1/4)*gamma(-5/4)/(32*a**5*b**4*e**(7/2)*x**2*gamma(7/4) + 64*a**4*b**5
*e**(7/2)*x**4*gamma(7/4) + 32*a**3*b**6*e**(7/2)*x**6*gamma(7/4)) + 21*a**2*b**(21/4)*x**2*(a/(b*x**2) + 1)**
(1/4)*gamma(-5/4)/(32*a**5*b**4*e**(7/2)*x**2*gamma(7/4) + 64*a**4*b**5*e**(7/2)*x**4*gamma(7/4) + 32*a**3*b**
6*e**(7/2)*x**6*gamma(7/4)) + 56*a*b**(25/4)*x**4*(a/(b*x**2) + 1)**(1/4)*gamma(-5/4)/(32*a**5*b**4*e**(7/2)*x
**2*gamma(7/4) + 64*a**4*b**5*e**(7/2)*x**4*gamma(7/4) + 32*a**3*b**6*e**(7/2)*x**6*gamma(7/4)) + 32*b**(29/4)
*x**6*(a/(b*x**2) + 1)**(1/4)*gamma(-5/4)/(32*a**5*b**4*e**(7/2)*x**2*gamma(7/4) + 64*a**4*b**5*e**(7/2)*x**4*
gamma(7/4) + 32*a**3*b**6*e**(7/2)*x**6*gamma(7/4))) + d*(3*gamma(-1/4)/(8*a*b**(3/4)*e**(7/2)*x**2*(a/(b*x**2
) + 1)**(3/4)*gamma(7/4)) + b**(1/4)*gamma(-1/4)/(2*a**2*e**(7/2)*(a/(b*x**2) + 1)**(3/4)*gamma(7/4)))

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

[Out]

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

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Mupad [B]
time = 0.67, size = 101, normalized size = 0.97 \begin {gather*} -\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}} \end {gather*}

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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