\(\int \frac {c x+d x^3+e x^5+f x^7}{\sqrt {a+b x^2}} \, dx\) [171]

Optimal result

Integrand size = 31, antiderivative size = 121 \[ \int \frac {c x+d x^3+e x^5+f x^7}{\sqrt {a+b x^2}} \, dx=\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \sqrt {a+b x^2}}{b^4}+\frac {\left (b^2 d-2 a b e+3 a^2 f\right ) \left (a+b x^2\right )^{3/2}}{3 b^4}+\frac {(b e-3 a f) \left (a+b x^2\right )^{5/2}}{5 b^4}+\frac {f \left (a+b x^2\right )^{7/2}}{7 b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {1825, 1813, 1864} \[ \int \frac {c x+d x^3+e x^5+f x^7}{\sqrt {a+b x^2}} \, dx=\frac {\left (a+b x^2\right )^{3/2} \left (3 a^2 f-2 a b e+b^2 d\right )}{3 b^4}+\frac {\sqrt {a+b x^2} \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{b^4}+\frac {\left (a+b x^2\right )^{5/2} (b e-3 a f)}{5 b^4}+\frac {f \left (a+b x^2\right )^{7/2}}{7 b^4} \]

[In]

Int[(c*x + d*x^3 + e*x^5 + f*x^7)/Sqrt[a + b*x^2],x]

[Out]

((b^3*c - a*b^2*d + a^2*b*e - a^3*f)*Sqrt[a + b*x^2])/b^4 + ((b^2*d - 2*a*b*e + 3*a^2*f)*(a + b*x^2)^(3/2))/(3
*b^4) + ((b*e - 3*a*f)*(a + b*x^2)^(5/2))/(5*b^4) + (f*(a + b*x^2)^(7/2))/(7*b^4)

Rule 1813

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]

Rule 1825

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[x*PolynomialQuotient[Pq, x, x]*(a + b*x^2)^p, x] /; Fre
eQ[{a, b, p}, x] && PolyQ[Pq, x] && EqQ[Coeff[Pq, x, 0], 0] &&  !MatchQ[Pq, x^(m_.)*(u_.) /; IntegerQ[m]]

Rule 1864

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (c+d x^2+e x^4+f x^6\right )}{\sqrt {a+b x^2}} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {c+d x+e x^2+f x^3}{\sqrt {a+b x}} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {b^3 c-a b^2 d+a^2 b e-a^3 f}{b^3 \sqrt {a+b x}}+\frac {\left (b^2 d-2 a b e+3 a^2 f\right ) \sqrt {a+b x}}{b^3}+\frac {(b e-3 a f) (a+b x)^{3/2}}{b^3}+\frac {f (a+b x)^{5/2}}{b^3}\right ) \, dx,x,x^2\right ) \\ & = \frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) \sqrt {a+b x^2}}{b^4}+\frac {\left (b^2 d-2 a b e+3 a^2 f\right ) \left (a+b x^2\right )^{3/2}}{3 b^4}+\frac {(b e-3 a f) \left (a+b x^2\right )^{5/2}}{5 b^4}+\frac {f \left (a+b x^2\right )^{7/2}}{7 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.74 \[ \int \frac {c x+d x^3+e x^5+f x^7}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (-48 a^3 f+8 a^2 b \left (7 e+3 f x^2\right )-2 a b^2 \left (35 d+14 e x^2+9 f x^4\right )+b^3 \left (105 c+35 d x^2+21 e x^4+15 f x^6\right )\right )}{105 b^4} \]

[In]

Integrate[(c*x + d*x^3 + e*x^5 + f*x^7)/Sqrt[a + b*x^2],x]

[Out]

(Sqrt[a + b*x^2]*(-48*a^3*f + 8*a^2*b*(7*e + 3*f*x^2) - 2*a*b^2*(35*d + 14*e*x^2 + 9*f*x^4) + b^3*(105*c + 35*
d*x^2 + 21*e*x^4 + 15*f*x^6)))/(105*b^4)

Maple [A] (verified)

Time = 3.55 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.68

method	result	size
pseudoelliptic	\(-\frac {16 \left (\frac {\left (-5 f \,x^{6}-7 e \,x^{4}-\frac {35}{3} d \,x^{2}-35 c \right ) b^{3}}{16}+\frac {35 \left (\frac {9}{35} f \,x^{4}+\frac {2}{5} e \,x^{2}+d \right ) a \,b^{2}}{24}-\frac {7 \left (\frac {3 f \,x^{2}}{7}+e \right ) a^{2} b}{6}+f \,a^{3}\right ) \sqrt {b \,x^{2}+a}}{35 b^{4}}\)	\(82\)
gosper	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-15 f \,x^{6} b^{3}+18 a \,b^{2} f \,x^{4}-21 b^{3} e \,x^{4}-24 a^{2} b f \,x^{2}+28 a \,b^{2} e \,x^{2}-35 b^{3} d \,x^{2}+48 f \,a^{3}-56 a^{2} b e +70 a \,b^{2} d -105 b^{3} c \right )}{105 b^{4}}\)	\(99\)
trager	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-15 f \,x^{6} b^{3}+18 a \,b^{2} f \,x^{4}-21 b^{3} e \,x^{4}-24 a^{2} b f \,x^{2}+28 a \,b^{2} e \,x^{2}-35 b^{3} d \,x^{2}+48 f \,a^{3}-56 a^{2} b e +70 a \,b^{2} d -105 b^{3} c \right )}{105 b^{4}}\)	\(99\)
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-15 f \,x^{6} b^{3}+18 a \,b^{2} f \,x^{4}-21 b^{3} e \,x^{4}-24 a^{2} b f \,x^{2}+28 a \,b^{2} e \,x^{2}-35 b^{3} d \,x^{2}+48 f \,a^{3}-56 a^{2} b e +70 a \,b^{2} d -105 b^{3} c \right )}{105 b^{4}}\)	\(99\)
default	\(f \left (\frac {x^{6} \sqrt {b \,x^{2}+a}}{7 b}-\frac {6 a \left (\frac {x^{4} \sqrt {b \,x^{2}+a}}{5 b}-\frac {4 a \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )}{5 b}\right )}{7 b}\right )+e \left (\frac {x^{4} \sqrt {b \,x^{2}+a}}{5 b}-\frac {4 a \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )}{5 b}\right )+d \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )+\frac {c \sqrt {b \,x^{2}+a}}{b}\)	\(193\)

[In]

int((f*x^7+e*x^5+d*x^3+c*x)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-16/35*(1/16*(-5*f*x^6-7*e*x^4-35/3*d*x^2-35*c)*b^3+35/24*(9/35*f*x^4+2/5*e*x^2+d)*a*b^2-7/6*(3/7*f*x^2+e)*a^2
*b+f*a^3)*(b*x^2+a)^(1/2)/b^4

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.78 \[ \int \frac {c x+d x^3+e x^5+f x^7}{\sqrt {a+b x^2}} \, dx=\frac {{\left (15 \, b^{3} f x^{6} + 3 \, {\left (7 \, b^{3} e - 6 \, a b^{2} f\right )} x^{4} + 105 \, b^{3} c - 70 \, a b^{2} d + 56 \, a^{2} b e - 48 \, a^{3} f + {\left (35 \, b^{3} d - 28 \, a b^{2} e + 24 \, a^{2} b f\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{105 \, b^{4}} \]

[In]

integrate((f*x^7+e*x^5+d*x^3+c*x)/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

1/105*(15*b^3*f*x^6 + 3*(7*b^3*e - 6*a*b^2*f)*x^4 + 105*b^3*c - 70*a*b^2*d + 56*a^2*b*e - 48*a^3*f + (35*b^3*d
- 28*a*b^2*e + 24*a^2*b*f)*x^2)*sqrt(b*x^2 + a)/b^4

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (112) = 224\).

Time = 0.29 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.97 \[ \int \frac {c x+d x^3+e x^5+f x^7}{\sqrt {a+b x^2}} \, dx=\begin {cases} - \frac {16 a^{3} f \sqrt {a + b x^{2}}}{35 b^{4}} + \frac {8 a^{2} e \sqrt {a + b x^{2}}}{15 b^{3}} + \frac {8 a^{2} f x^{2} \sqrt {a + b x^{2}}}{35 b^{3}} - \frac {2 a d \sqrt {a + b x^{2}}}{3 b^{2}} - \frac {4 a e x^{2} \sqrt {a + b x^{2}}}{15 b^{2}} - \frac {6 a f x^{4} \sqrt {a + b x^{2}}}{35 b^{2}} + \frac {c \sqrt {a + b x^{2}}}{b} + \frac {d x^{2} \sqrt {a + b x^{2}}}{3 b} + \frac {e x^{4} \sqrt {a + b x^{2}}}{5 b} + \frac {f x^{6} \sqrt {a + b x^{2}}}{7 b} & \text {for}\: b \neq 0 \\\frac {\frac {c x^{2}}{2} + \frac {d x^{4}}{4} + \frac {e x^{6}}{6} + \frac {f x^{8}}{8}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

integrate((f*x**7+e*x**5+d*x**3+c*x)/(b*x**2+a)**(1/2),x)

[Out]

Piecewise((-16*a**3*f*sqrt(a + b*x**2)/(35*b**4) + 8*a**2*e*sqrt(a + b*x**2)/(15*b**3) + 8*a**2*f*x**2*sqrt(a
+ b*x**2)/(35*b**3) - 2*a*d*sqrt(a + b*x**2)/(3*b**2) - 4*a*e*x**2*sqrt(a + b*x**2)/(15*b**2) - 6*a*f*x**4*sqr
t(a + b*x**2)/(35*b**2) + c*sqrt(a + b*x**2)/b + d*x**2*sqrt(a + b*x**2)/(3*b) + e*x**4*sqrt(a + b*x**2)/(5*b)
+ f*x**6*sqrt(a + b*x**2)/(7*b), Ne(b, 0)), ((c*x**2/2 + d*x**4/4 + e*x**6/6 + f*x**8/8)/sqrt(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.49 \[ \int \frac {c x+d x^3+e x^5+f x^7}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} f x^{6}}{7 \, b} + \frac {\sqrt {b x^{2} + a} e x^{4}}{5 \, b} - \frac {6 \, \sqrt {b x^{2} + a} a f x^{4}}{35 \, b^{2}} + \frac {\sqrt {b x^{2} + a} d x^{2}}{3 \, b} - \frac {4 \, \sqrt {b x^{2} + a} a e x^{2}}{15 \, b^{2}} + \frac {8 \, \sqrt {b x^{2} + a} a^{2} f x^{2}}{35 \, b^{3}} + \frac {\sqrt {b x^{2} + a} c}{b} - \frac {2 \, \sqrt {b x^{2} + a} a d}{3 \, b^{2}} + \frac {8 \, \sqrt {b x^{2} + a} a^{2} e}{15 \, b^{3}} - \frac {16 \, \sqrt {b x^{2} + a} a^{3} f}{35 \, b^{4}} \]

[In]

integrate((f*x^7+e*x^5+d*x^3+c*x)/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

1/7*sqrt(b*x^2 + a)*f*x^6/b + 1/5*sqrt(b*x^2 + a)*e*x^4/b - 6/35*sqrt(b*x^2 + a)*a*f*x^4/b^2 + 1/3*sqrt(b*x^2
+ a)*d*x^2/b - 4/15*sqrt(b*x^2 + a)*a*e*x^2/b^2 + 8/35*sqrt(b*x^2 + a)*a^2*f*x^2/b^3 + sqrt(b*x^2 + a)*c/b - 2
/3*sqrt(b*x^2 + a)*a*d/b^2 + 8/15*sqrt(b*x^2 + a)*a^2*e/b^3 - 16/35*sqrt(b*x^2 + a)*a^3*f/b^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.05 \[ \int \frac {c x+d x^3+e x^5+f x^7}{\sqrt {a+b x^2}} \, dx=\frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} \sqrt {b x^{2} + a}}{b^{4}} + \frac {35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2} d + 21 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b e - 70 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b e + 15 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} f - 63 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a f + 105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} f}{105 \, b^{4}} \]

[In]

integrate((f*x^7+e*x^5+d*x^3+c*x)/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*sqrt(b*x^2 + a)/b^4 + 1/105*(35*(b*x^2 + a)^(3/2)*b^2*d + 21*(b*x^2 + a)^(
5/2)*b*e - 70*(b*x^2 + a)^(3/2)*a*b*e + 15*(b*x^2 + a)^(7/2)*f - 63*(b*x^2 + a)^(5/2)*a*f + 105*(b*x^2 + a)^(3
/2)*a^2*f)/b^4

Mupad [B] (verification not implemented)

Time = 5.99 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.85 \[ \int \frac {c x+d x^3+e x^5+f x^7}{\sqrt {a+b x^2}} \, dx=\sqrt {b\,x^2+a}\,\left (\frac {-48\,f\,a^3+56\,e\,a^2\,b-70\,d\,a\,b^2+105\,c\,b^3}{105\,b^4}+\frac {f\,x^6}{7\,b}+\frac {x^2\,\left (24\,f\,a^2\,b-28\,e\,a\,b^2+35\,d\,b^3\right )}{105\,b^4}+\frac {x^4\,\left (21\,b^3\,e-18\,a\,b^2\,f\right )}{105\,b^4}\right ) \]

[In]

int((c*x + d*x^3 + e*x^5 + f*x^7)/(a + b*x^2)^(1/2),x)

[Out]

(a + b*x^2)^(1/2)*((105*b^3*c - 48*a^3*f - 70*a*b^2*d + 56*a^2*b*e)/(105*b^4) + (f*x^6)/(7*b) + (x^2*(35*b^3*d
- 28*a*b^2*e + 24*a^2*b*f))/(105*b^4) + (x^4*(21*b^3*e - 18*a*b^2*f))/(105*b^4))