3.1.99 \(\int \frac {a+b x^2}{(c+d x^2)^{7/2}} \, dx\) [99]

Optimal. Leaf size=91 \[ -\frac {(b c-a d) x}{5 c d \left (c+d x^2\right )^{5/2}}+\frac {(b c+4 a d) x}{15 c^2 d \left (c+d x^2\right )^{3/2}}+\frac {2 (b c+4 a d) x}{15 c^3 d \sqrt {c+d x^2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {393, 198, 197} \begin {gather*} \frac {2 x (4 a d+b c)}{15 c^3 d \sqrt {c+d x^2}}+\frac {x (4 a d+b c)}{15 c^2 d \left (c+d x^2\right )^{3/2}}-\frac {x (b c-a d)}{5 c d \left (c+d x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)/(c + d*x^2)^(7/2),x]

[Out]

-1/5*((b*c - a*d)*x)/(c*d*(c + d*x^2)^(5/2)) + ((b*c + 4*a*d)*x)/(15*c^2*d*(c + d*x^2)^(3/2)) + (2*(b*c + 4*a*
d)*x)/(15*c^3*d*Sqrt[c + d*x^2])

Rule 197

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

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 60, normalized size = 0.66 \begin {gather*} \frac {15 a c^2 x+5 b c^2 x^3+20 a c d x^3+2 b c d x^5+8 a d^2 x^5}{15 c^3 \left (c+d x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 132, normalized size = 1.45

method	result	size
gosper	\(\frac {x \left (8 a \,d^{2} x^{4}+2 b c d \,x^{4}+20 a c d \,x^{2}+5 b \,c^{2} x^{2}+15 c^{2} a \right )}{15 \left (d \,x^{2}+c \right )^{\frac {5}{2}} c^{3}}\)	\(57\)
trager	\(\frac {x \left (8 a \,d^{2} x^{4}+2 b c d \,x^{4}+20 a c d \,x^{2}+5 b \,c^{2} x^{2}+15 c^{2} a \right )}{15 \left (d \,x^{2}+c \right )^{\frac {5}{2}} c^{3}}\)	\(57\)
default	\(b \left (-\frac {x}{4 d \left (d \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {c \left (\frac {x}{5 c \left (d \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {8 x}{15 c^{2} \sqrt {d \,x^{2}+c}}}{c}\right )}{4 d}\right )+a \left (\frac {x}{5 c \left (d \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {8 x}{15 c^{2} \sqrt {d \,x^{2}+c}}}{c}\right )\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 103, normalized size = 1.13 \begin {gather*} \frac {8 \, a x}{15 \, \sqrt {d x^{2} + c} c^{3}} + \frac {4 \, a x}{15 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2}} + \frac {a x}{5 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} c} - \frac {b x}{5 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} d} + \frac {2 \, b x}{15 \, \sqrt {d x^{2} + c} c^{2} d} + \frac {b x}{15 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.18, size = 87, normalized size = 0.96 \begin {gather*} \frac {{\left (2 \, {\left (b c d + 4 \, a d^{2}\right )} x^{5} + 15 \, a c^{2} x + 5 \, {\left (b c^{2} + 4 \, a c d\right )} x^{3}\right )} \sqrt {d x^{2} + c}}{15 \, {\left (c^{3} d^{3} x^{6} + 3 \, c^{4} d^{2} x^{4} + 3 \, c^{5} d x^{2} + c^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 566 vs. \(2 (83) = 166\).
time = 11.43, size = 566, normalized size = 6.22 \begin {gather*} a \left (\frac {15 c^{5} x}{15 c^{\frac {17}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {15}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {13}{2}} d^{2} x^{4} \sqrt {1 + \frac {d x^{2}}{c}} + 15 c^{\frac {11}{2}} d^{3} x^{6} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {35 c^{4} d x^{3}}{15 c^{\frac {17}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {15}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {13}{2}} d^{2} x^{4} \sqrt {1 + \frac {d x^{2}}{c}} + 15 c^{\frac {11}{2}} d^{3} x^{6} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {28 c^{3} d^{2} x^{5}}{15 c^{\frac {17}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {15}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {13}{2}} d^{2} x^{4} \sqrt {1 + \frac {d x^{2}}{c}} + 15 c^{\frac {11}{2}} d^{3} x^{6} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {8 c^{2} d^{3} x^{7}}{15 c^{\frac {17}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {15}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {13}{2}} d^{2} x^{4} \sqrt {1 + \frac {d x^{2}}{c}} + 15 c^{\frac {11}{2}} d^{3} x^{6} \sqrt {1 + \frac {d x^{2}}{c}}}\right ) + b \left (\frac {5 c x^{3}}{15 c^{\frac {9}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 30 c^{\frac {7}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}} + 15 c^{\frac {5}{2}} d^{2} x^{4} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {2 d x^{5}}{15 c^{\frac {9}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 30 c^{\frac {7}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}} + 15 c^{\frac {5}{2}} d^{2} x^{4} \sqrt {1 + \frac {d x^{2}}{c}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.00, size = 72, normalized size = 0.79 \begin {gather*} \frac {{\left (x^{2} {\left (\frac {2 \, {\left (b c d^{3} + 4 \, a d^{4}\right )} x^{2}}{c^{3} d^{2}} + \frac {5 \, {\left (b c^{2} d^{2} + 4 \, a c d^{3}\right )}}{c^{3} d^{2}}\right )} + \frac {15 \, a}{c}\right )} x}{15 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.85, size = 87, normalized size = 0.96 \begin {gather*} \frac {8\,a\,d\,x\,{\left (d\,x^2+c\right )}^2-3\,b\,c^3\,x+2\,b\,c\,x\,{\left (d\,x^2+c\right )}^2+b\,c^2\,x\,\left (d\,x^2+c\right )+3\,a\,c^2\,d\,x+4\,a\,c\,d\,x\,\left (d\,x^2+c\right )}{15\,c^3\,d\,{\left (d\,x^2+c\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)/(c + d*x^2)^(7/2),x)

[Out]

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

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