\(\int \frac {a+b x^2+c x^4}{(d+e x^2)^{7/2}} \, dx\) [282]

Optimal result

Integrand size = 24, antiderivative size = 86 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {a x}{d \left (d+e x^2\right )^{5/2}}+\frac {(b d+4 a e) x^3}{3 d^2 \left (d+e x^2\right )^{5/2}}+\frac {\left (3 c d^2+2 e (b d+4 a e)\right ) x^5}{15 d^3 \left (d+e x^2\right )^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1169, 1817, 12, 270} \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {x^5 \left (2 e (4 a e+b d)+3 c d^2\right )}{15 d^3 \left (d+e x^2\right )^{5/2}}+\frac {x^3 (4 a e+b d)}{3 d^2 \left (d+e x^2\right )^{5/2}}+\frac {a x}{d \left (d+e x^2\right )^{5/2}} \]

[In]

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

[Out]

(a*x)/(d*(d + e*x^2)^(5/2)) + ((b*d + 4*a*e)*x^3)/(3*d^2*(d + e*x^2)^(5/2)) + ((3*c*d^2 + 2*e*(b*d + 4*a*e))*x
^5)/(15*d^3*(d + e*x^2)^(5/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 1169

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[a^p*x*((d + e*x^2
)^(q + 1)/d), x] + Dist[1/d, Int[x^2*(d + e*x^2)^q*(d*PolynomialQuotient[(a + b*x^2 + c*x^4)^p - a^p, x^2, x]
- e*a^p*(2*q + 3)), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0
] && IGtQ[p, 0] && ILtQ[q + 1/2, 0] && LtQ[4*p + 2*q + 1, 0]

Rule 1817

Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{A = Coeff[Pq, x, 0], Q = PolynomialQuotient
[Pq - Coeff[Pq, x, 0], x^2, x]}, Simp[A*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] + Dist[1/(a*(m + 1)),
Int[x^(m + 2)*(a + b*x^2)^p*(a*(m + 1)*Q - A*b*(m + 2*(p + 1) + 1)), x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq,
x^2] && IntegerQ[m/2] && ILtQ[(m + 1)/2 + p, 0] && LtQ[m + Expon[Pq, x] + 2*p + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {a x}{d \left (d+e x^2\right )^{5/2}}+\frac {\int \frac {x^2 \left (4 a e+d \left (b+c x^2\right )\right )}{\left (d+e x^2\right )^{7/2}} \, dx}{d} \\ & = \frac {a x}{d \left (d+e x^2\right )^{5/2}}+\frac {(b d+4 a e) x^3}{3 d^2 \left (d+e x^2\right )^{5/2}}+\frac {\int \frac {\left (3 c d^2+2 e (b d+4 a e)\right ) x^4}{\left (d+e x^2\right )^{7/2}} \, dx}{3 d^2} \\ & = \frac {a x}{d \left (d+e x^2\right )^{5/2}}+\frac {(b d+4 a e) x^3}{3 d^2 \left (d+e x^2\right )^{5/2}}+\frac {1}{3} \left (3 c+\frac {2 e (b d+4 a e)}{d^2}\right ) \int \frac {x^4}{\left (d+e x^2\right )^{7/2}} \, dx \\ & = \frac {a x}{d \left (d+e x^2\right )^{5/2}}+\frac {(b d+4 a e) x^3}{3 d^2 \left (d+e x^2\right )^{5/2}}+\frac {\left (3 c d^2+2 e (b d+4 a e)\right ) x^5}{15 d^3 \left (d+e x^2\right )^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.80 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {15 a d^2 x+5 b d^2 x^3+20 a d e x^3+3 c d^2 x^5+2 b d e x^5+8 a e^2 x^5}{15 d^3 \left (d+e x^2\right )^{5/2}} \]

[In]

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

[Out]

(15*a*d^2*x + 5*b*d^2*x^3 + 20*a*d*e*x^3 + 3*c*d^2*x^5 + 2*b*d*e*x^5 + 8*a*e^2*x^5)/(15*d^3*(d + e*x^2)^(5/2))

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.67

method	result	size
pseudoelliptic	\(\frac {x \left (\left (\frac {1}{5} c \,x^{4}+\frac {1}{3} b \,x^{2}+a \right ) d^{2}+\frac {4 e \left (\frac {b \,x^{2}}{10}+a \right ) x^{2} d}{3}+\frac {8 a \,e^{2} x^{4}}{15}\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}} d^{3}}\)	\(58\)
gosper	\(\frac {x \left (8 a \,e^{2} x^{4}+2 b d e \,x^{4}+3 c \,d^{2} x^{4}+20 a d e \,x^{2}+5 b \,d^{2} x^{2}+15 a \,d^{2}\right )}{15 \left (e \,x^{2}+d \right )^{\frac {5}{2}} d^{3}}\)	\(66\)
trager	\(\frac {x \left (8 a \,e^{2} x^{4}+2 b d e \,x^{4}+3 c \,d^{2} x^{4}+20 a d e \,x^{2}+5 b \,d^{2} x^{2}+15 a \,d^{2}\right )}{15 \left (e \,x^{2}+d \right )^{\frac {5}{2}} d^{3}}\)	\(66\)
default	\(a \left (\frac {x}{5 d \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{2} \sqrt {e \,x^{2}+d}}}{d}\right )+c \left (-\frac {x^{3}}{2 e \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {3 d \left (-\frac {x}{4 e \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {d \left (\frac {x}{5 d \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{2} \sqrt {e \,x^{2}+d}}}{d}\right )}{4 e}\right )}{2 e}\right )+b \left (-\frac {x}{4 e \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {d \left (\frac {x}{5 d \left (e \,x^{2}+d \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{2} \sqrt {e \,x^{2}+d}}}{d}\right )}{4 e}\right )\)	\(232\)

[In]

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

[Out]

1/(e*x^2+d)^(5/2)*x*((1/5*c*x^4+1/3*b*x^2+a)*d^2+4/3*e*(1/10*b*x^2+a)*x^2*d+8/15*a*e^2*x^4)/d^3

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {{\left ({\left (3 \, c d^{2} + 2 \, b d e + 8 \, a e^{2}\right )} x^{5} + 15 \, a d^{2} x + 5 \, {\left (b d^{2} + 4 \, a d e\right )} x^{3}\right )} \sqrt {e x^{2} + d}}{15 \, {\left (d^{3} e^{3} x^{6} + 3 \, d^{4} e^{2} x^{4} + 3 \, d^{5} e x^{2} + d^{6}\right )}} \]

[In]

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

[Out]

1/15*((3*c*d^2 + 2*b*d*e + 8*a*e^2)*x^5 + 15*a*d^2*x + 5*(b*d^2 + 4*a*d*e)*x^3)*sqrt(e*x^2 + d)/(d^3*e^3*x^6 +
3*d^4*e^2*x^4 + 3*d^5*e*x^2 + d^6)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 639 vs. \(2 (80) = 160\).

Time = 14.75 (sec) , antiderivative size = 639, normalized size of antiderivative = 7.43 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=a \left (\frac {15 d^{5} x}{15 d^{\frac {17}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {15}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {13}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {11}{2}} e^{3} x^{6} \sqrt {1 + \frac {e x^{2}}{d}}} + \frac {35 d^{4} e x^{3}}{15 d^{\frac {17}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {15}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {13}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {11}{2}} e^{3} x^{6} \sqrt {1 + \frac {e x^{2}}{d}}} + \frac {28 d^{3} e^{2} x^{5}}{15 d^{\frac {17}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {15}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {13}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {11}{2}} e^{3} x^{6} \sqrt {1 + \frac {e x^{2}}{d}}} + \frac {8 d^{2} e^{3} x^{7}}{15 d^{\frac {17}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {15}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 45 d^{\frac {13}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {11}{2}} e^{3} x^{6} \sqrt {1 + \frac {e x^{2}}{d}}}\right ) + b \left (\frac {5 d x^{3}}{15 d^{\frac {9}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 30 d^{\frac {7}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {5}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}}} + \frac {2 e x^{5}}{15 d^{\frac {9}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 30 d^{\frac {7}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 15 d^{\frac {5}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}}}\right ) + \frac {c x^{5}}{5 d^{\frac {7}{2}} \sqrt {1 + \frac {e x^{2}}{d}} + 10 d^{\frac {5}{2}} e x^{2} \sqrt {1 + \frac {e x^{2}}{d}} + 5 d^{\frac {3}{2}} e^{2} x^{4} \sqrt {1 + \frac {e x^{2}}{d}}} \]

[In]

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

[Out]

a*(15*d**5*x/(15*d**(17/2)*sqrt(1 + e*x**2/d) + 45*d**(15/2)*e*x**2*sqrt(1 + e*x**2/d) + 45*d**(13/2)*e**2*x**
4*sqrt(1 + e*x**2/d) + 15*d**(11/2)*e**3*x**6*sqrt(1 + e*x**2/d)) + 35*d**4*e*x**3/(15*d**(17/2)*sqrt(1 + e*x*
*2/d) + 45*d**(15/2)*e*x**2*sqrt(1 + e*x**2/d) + 45*d**(13/2)*e**2*x**4*sqrt(1 + e*x**2/d) + 15*d**(11/2)*e**3
*x**6*sqrt(1 + e*x**2/d)) + 28*d**3*e**2*x**5/(15*d**(17/2)*sqrt(1 + e*x**2/d) + 45*d**(15/2)*e*x**2*sqrt(1 +
e*x**2/d) + 45*d**(13/2)*e**2*x**4*sqrt(1 + e*x**2/d) + 15*d**(11/2)*e**3*x**6*sqrt(1 + e*x**2/d)) + 8*d**2*e*
*3*x**7/(15*d**(17/2)*sqrt(1 + e*x**2/d) + 45*d**(15/2)*e*x**2*sqrt(1 + e*x**2/d) + 45*d**(13/2)*e**2*x**4*sqr
t(1 + e*x**2/d) + 15*d**(11/2)*e**3*x**6*sqrt(1 + e*x**2/d))) + b*(5*d*x**3/(15*d**(9/2)*sqrt(1 + e*x**2/d) +
30*d**(7/2)*e*x**2*sqrt(1 + e*x**2/d) + 15*d**(5/2)*e**2*x**4*sqrt(1 + e*x**2/d)) + 2*e*x**5/(15*d**(9/2)*sqrt
(1 + e*x**2/d) + 30*d**(7/2)*e*x**2*sqrt(1 + e*x**2/d) + 15*d**(5/2)*e**2*x**4*sqrt(1 + e*x**2/d))) + c*x**5/(
5*d**(7/2)*sqrt(1 + e*x**2/d) + 10*d**(5/2)*e*x**2*sqrt(1 + e*x**2/d) + 5*d**(3/2)*e**2*x**4*sqrt(1 + e*x**2/d
))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (76) = 152\).

Time = 0.20 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.01 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=-\frac {c x^{3}}{2 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} e} + \frac {8 \, a x}{15 \, \sqrt {e x^{2} + d} d^{3}} + \frac {4 \, a x}{15 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2}} + \frac {a x}{5 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d} + \frac {c x}{10 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} e^{2}} + \frac {c x}{5 \, \sqrt {e x^{2} + d} d e^{2}} - \frac {3 \, c d x}{10 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} e^{2}} - \frac {b x}{5 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} e} + \frac {2 \, b x}{15 \, \sqrt {e x^{2} + d} d^{2} e} + \frac {b x}{15 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d e} \]

[In]

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

[Out]

-1/2*c*x^3/((e*x^2 + d)^(5/2)*e) + 8/15*a*x/(sqrt(e*x^2 + d)*d^3) + 4/15*a*x/((e*x^2 + d)^(3/2)*d^2) + 1/5*a*x
/((e*x^2 + d)^(5/2)*d) + 1/10*c*x/((e*x^2 + d)^(3/2)*e^2) + 1/5*c*x/(sqrt(e*x^2 + d)*d*e^2) - 3/10*c*d*x/((e*x
^2 + d)^(5/2)*e^2) - 1/5*b*x/((e*x^2 + d)^(5/2)*e) + 2/15*b*x/(sqrt(e*x^2 + d)*d^2*e) + 1/15*b*x/((e*x^2 + d)^
(3/2)*d*e)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {{\left (x^{2} {\left (\frac {{\left (3 \, c d^{2} e^{2} + 2 \, b d e^{3} + 8 \, a e^{4}\right )} x^{2}}{d^{3} e^{2}} + \frac {5 \, {\left (b d^{2} e^{2} + 4 \, a d e^{3}\right )}}{d^{3} e^{2}}\right )} + \frac {15 \, a}{d}\right )} x}{15 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}}} \]

[In]

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

[Out]

1/15*(x^2*((3*c*d^2*e^2 + 2*b*d*e^3 + 8*a*e^4)*x^2/(d^3*e^2) + 5*(b*d^2*e^2 + 4*a*d*e^3)/(d^3*e^2)) + 15*a/d)*
x/(e*x^2 + d)^(5/2)

Mupad [B] (verification not implemented)

Time = 7.87 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.55 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^{7/2}} \, dx=\frac {3\,c\,d^4\,x-6\,c\,d^3\,x\,\left (e\,x^2+d\right )-3\,b\,d^3\,e\,x+8\,a\,e^2\,x\,{\left (e\,x^2+d\right )}^2+3\,c\,d^2\,x\,{\left (e\,x^2+d\right )}^2+3\,a\,d^2\,e^2\,x+4\,a\,d\,e^2\,x\,\left (e\,x^2+d\right )+2\,b\,d\,e\,x\,{\left (e\,x^2+d\right )}^2+b\,d^2\,e\,x\,\left (e\,x^2+d\right )}{15\,d^3\,e^2\,{\left (e\,x^2+d\right )}^{5/2}} \]

[In]

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

[Out]

(3*c*d^4*x - 6*c*d^3*x*(d + e*x^2) - 3*b*d^3*e*x + 8*a*e^2*x*(d + e*x^2)^2 + 3*c*d^2*x*(d + e*x^2)^2 + 3*a*d^2
*e^2*x + 4*a*d*e^2*x*(d + e*x^2) + 2*b*d*e*x*(d + e*x^2)^2 + b*d^2*e*x*(d + e*x^2))/(15*d^3*e^2*(d + e*x^2)^(5
/2))