3.1.98 \(\int \frac {(a+b x^2)^2}{(c+d x^2)^{9/2}} \, dx\) [98]

Optimal. Leaf size=174 \[ -\frac {d x \left (a+b x^2\right )^3}{7 c (b c-a d) \left (c+d x^2\right )^{7/2}}+\frac {(7 b c-6 a d) x \left (a+b x^2\right )^2}{35 c^2 (b c-a d) \left (c+d x^2\right )^{5/2}}+\frac {4 a (7 b c-6 a d) x \left (a+b x^2\right )}{105 c^3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {8 a^2 (7 b c-6 a d) x}{105 c^4 (b c-a d) \sqrt {c+d x^2}} \]

[Out]

-1/7*d*x*(b*x^2+a)^3/c/(-a*d+b*c)/(d*x^2+c)^(7/2)+1/35*(-6*a*d+7*b*c)*x*(b*x^2+a)^2/c^2/(-a*d+b*c)/(d*x^2+c)^(
5/2)+4/105*a*(-6*a*d+7*b*c)*x*(b*x^2+a)/c^3/(-a*d+b*c)/(d*x^2+c)^(3/2)+8/105*a^2*(-6*a*d+7*b*c)*x/c^4/(-a*d+b*
c)/(d*x^2+c)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {390, 386, 197} \begin {gather*} \frac {8 a^2 x (7 b c-6 a d)}{105 c^4 \sqrt {c+d x^2} (b c-a d)}+\frac {4 a x \left (a+b x^2\right ) (7 b c-6 a d)}{105 c^3 \left (c+d x^2\right )^{3/2} (b c-a d)}+\frac {x \left (a+b x^2\right )^2 (7 b c-6 a d)}{35 c^2 \left (c+d x^2\right )^{5/2} (b c-a d)}-\frac {d x \left (a+b x^2\right )^3}{7 c \left (c+d x^2\right )^{7/2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/7*(d*x*(a + b*x^2)^3)/(c*(b*c - a*d)*(c + d*x^2)^(7/2)) + ((7*b*c - 6*a*d)*x*(a + b*x^2)^2)/(35*c^2*(b*c -
a*d)*(c + d*x^2)^(5/2)) + (4*a*(7*b*c - 6*a*d)*x*(a + b*x^2))/(105*c^3*(b*c - a*d)*(c + d*x^2)^(3/2)) + (8*a^2
*(7*b*c - 6*a*d)*x)/(105*c^4*(b*c - a*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 386

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

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a
*d)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && Eq
Q[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{9/2}} \, dx &=-\frac {d x \left (a+b x^2\right )^3}{7 c (b c-a d) \left (c+d x^2\right )^{7/2}}+\frac {(7 b c-6 a d) \int \frac {\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{7/2}} \, dx}{7 c (b c-a d)}\\ &=-\frac {d x \left (a+b x^2\right )^3}{7 c (b c-a d) \left (c+d x^2\right )^{7/2}}+\frac {(7 b c-6 a d) x \left (a+b x^2\right )^2}{35 c^2 (b c-a d) \left (c+d x^2\right )^{5/2}}+\frac {(4 a (7 b c-6 a d)) \int \frac {a+b x^2}{\left (c+d x^2\right )^{5/2}} \, dx}{35 c^2 (b c-a d)}\\ &=-\frac {d x \left (a+b x^2\right )^3}{7 c (b c-a d) \left (c+d x^2\right )^{7/2}}+\frac {(7 b c-6 a d) x \left (a+b x^2\right )^2}{35 c^2 (b c-a d) \left (c+d x^2\right )^{5/2}}+\frac {4 a (7 b c-6 a d) x \left (a+b x^2\right )}{105 c^3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {\left (8 a^2 (7 b c-6 a d)\right ) \int \frac {1}{\left (c+d x^2\right )^{3/2}} \, dx}{105 c^3 (b c-a d)}\\ &=-\frac {d x \left (a+b x^2\right )^3}{7 c (b c-a d) \left (c+d x^2\right )^{7/2}}+\frac {(7 b c-6 a d) x \left (a+b x^2\right )^2}{35 c^2 (b c-a d) \left (c+d x^2\right )^{5/2}}+\frac {4 a (7 b c-6 a d) x \left (a+b x^2\right )}{105 c^3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac {8 a^2 (7 b c-6 a d) x}{105 c^4 (b c-a d) \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 107, normalized size = 0.61 \begin {gather*} \frac {3 b^2 c^2 x^5 \left (7 c+2 d x^2\right )+2 a b c x^3 \left (35 c^2+28 c d x^2+8 d^2 x^4\right )+3 a^2 \left (35 c^3 x+70 c^2 d x^3+56 c d^2 x^5+16 d^3 x^7\right )}{105 c^4 \left (c+d x^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^2*c^2*x^5*(7*c + 2*d*x^2) + 2*a*b*c*x^3*(35*c^2 + 28*c*d*x^2 + 8*d^2*x^4) + 3*a^2*(35*c^3*x + 70*c^2*d*x^
3 + 56*c*d^2*x^5 + 16*d^3*x^7))/(105*c^4*(c + d*x^2)^(7/2))

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Maple [A]
time = 0.07, size = 301, normalized size = 1.73

method	result	size
gosper	\(\frac {x \left (48 a^{2} d^{3} x^{6}+16 a b c \,d^{2} x^{6}+6 b^{2} c^{2} d \,x^{6}+168 a^{2} c \,d^{2} x^{4}+56 a b \,c^{2} d \,x^{4}+21 b^{2} c^{3} x^{4}+210 a^{2} c^{2} d \,x^{2}+70 a b \,c^{3} x^{2}+105 a^{2} c^{3}\right )}{105 \left (d \,x^{2}+c \right )^{\frac {7}{2}} c^{4}}\)	\(115\)
trager	\(\frac {x \left (48 a^{2} d^{3} x^{6}+16 a b c \,d^{2} x^{6}+6 b^{2} c^{2} d \,x^{6}+168 a^{2} c \,d^{2} x^{4}+56 a b \,c^{2} d \,x^{4}+21 b^{2} c^{3} x^{4}+210 a^{2} c^{2} d \,x^{2}+70 a b \,c^{3} x^{2}+105 a^{2} c^{3}\right )}{105 \left (d \,x^{2}+c \right )^{\frac {7}{2}} c^{4}}\)	\(115\)
default	\(b^{2} \left (-\frac {x^{3}}{4 d \left (d \,x^{2}+c \right )^{\frac {7}{2}}}+\frac {3 c \left (-\frac {x}{6 d \left (d \,x^{2}+c \right )^{\frac {7}{2}}}+\frac {c \left (\frac {x}{7 c \left (d \,x^{2}+c \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 c \left (d \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {8 x}{15 c^{2} \sqrt {d \,x^{2}+c}}\right )}{7 c}}{c}\right )}{6 d}\right )}{4 d}\right )+2 a b \left (-\frac {x}{6 d \left (d \,x^{2}+c \right )^{\frac {7}{2}}}+\frac {c \left (\frac {x}{7 c \left (d \,x^{2}+c \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 c \left (d \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {8 x}{15 c^{2} \sqrt {d \,x^{2}+c}}\right )}{7 c}}{c}\right )}{6 d}\right )+a^{2} \left (\frac {x}{7 c \left (d \,x^{2}+c \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 c \left (d \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {8 x}{15 c^{2} \sqrt {d \,x^{2}+c}}\right )}{7 c}}{c}\right )\)	\(301\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/(d*x^2+c)^(9/2),x,method=_RETURNVERBOSE)

[Out]

b^2*(-1/4*x^3/d/(d*x^2+c)^(7/2)+3/4*c/d*(-1/6*x/d/(d*x^2+c)^(7/2)+1/6*c/d*(1/7*x/c/(d*x^2+c)^(7/2)+6/7/c*(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))))))+2*a*b*(-1/6*x/d/(d*x^2+c)^(7
/2)+1/6*c/d*(1/7*x/c/(d*x^2+c)^(7/2)+6/7/c*(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^2*(1/7*x/c/(d*x^2+c)^(7/2)+6/7/c*(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 = 0.29, size = 249, normalized size = 1.43 \begin {gather*} -\frac {b^{2} x^{3}}{4 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} d} + \frac {16 \, a^{2} x}{35 \, \sqrt {d x^{2} + c} c^{4}} + \frac {8 \, a^{2} x}{35 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{3}} + \frac {6 \, a^{2} x}{35 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} c^{2}} + \frac {a^{2} x}{7 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} c} + \frac {3 \, b^{2} x}{140 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} d^{2}} + \frac {2 \, b^{2} x}{35 \, \sqrt {d x^{2} + c} c^{2} d^{2}} + \frac {b^{2} x}{35 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c d^{2}} - \frac {3 \, b^{2} c x}{28 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} d^{2}} - \frac {2 \, a b x}{7 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} d} + \frac {16 \, a b x}{105 \, \sqrt {d x^{2} + c} c^{3} d} + \frac {8 \, a b x}{105 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2} d} + \frac {2 \, a b x}{35 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b^2*x^3/((d*x^2 + c)^(7/2)*d) + 16/35*a^2*x/(sqrt(d*x^2 + c)*c^4) + 8/35*a^2*x/((d*x^2 + c)^(3/2)*c^3) +
6/35*a^2*x/((d*x^2 + c)^(5/2)*c^2) + 1/7*a^2*x/((d*x^2 + c)^(7/2)*c) + 3/140*b^2*x/((d*x^2 + c)^(5/2)*d^2) + 2
/35*b^2*x/(sqrt(d*x^2 + c)*c^2*d^2) + 1/35*b^2*x/((d*x^2 + c)^(3/2)*c*d^2) - 3/28*b^2*c*x/((d*x^2 + c)^(7/2)*d
^2) - 2/7*a*b*x/((d*x^2 + c)^(7/2)*d) + 16/105*a*b*x/(sqrt(d*x^2 + c)*c^3*d) + 8/105*a*b*x/((d*x^2 + c)^(3/2)*
c^2*d) + 2/35*a*b*x/((d*x^2 + c)^(5/2)*c*d)

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Fricas [A]
time = 2.37, size = 151, normalized size = 0.87 \begin {gather*} \frac {{\left (2 \, {\left (3 \, b^{2} c^{2} d + 8 \, a b c d^{2} + 24 \, a^{2} d^{3}\right )} x^{7} + 105 \, a^{2} c^{3} x + 7 \, {\left (3 \, b^{2} c^{3} + 8 \, a b c^{2} d + 24 \, a^{2} c d^{2}\right )} x^{5} + 70 \, {\left (a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{3}\right )} \sqrt {d x^{2} + c}}{105 \, {\left (c^{4} d^{4} x^{8} + 4 \, c^{5} d^{3} x^{6} + 6 \, c^{6} d^{2} x^{4} + 4 \, c^{7} d x^{2} + c^{8}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(2*(3*b^2*c^2*d + 8*a*b*c*d^2 + 24*a^2*d^3)*x^7 + 105*a^2*c^3*x + 7*(3*b^2*c^3 + 8*a*b*c^2*d + 24*a^2*c*
d^2)*x^5 + 70*(a*b*c^3 + 3*a^2*c^2*d)*x^3)*sqrt(d*x^2 + c)/(c^4*d^4*x^8 + 4*c^5*d^3*x^6 + 6*c^6*d^2*x^4 + 4*c^
7*d*x^2 + c^8)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/(d*x**2+c)**(9/2),x)

[Out]

Integral((a + b*x**2)**2/(c + d*x**2)**(9/2), x)

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Giac [A]
time = 0.88, size = 138, normalized size = 0.79 \begin {gather*} \frac {{\left ({\left (x^{2} {\left (\frac {2 \, {\left (3 \, b^{2} c^{2} d^{4} + 8 \, a b c d^{5} + 24 \, a^{2} d^{6}\right )} x^{2}}{c^{4} d^{3}} + \frac {7 \, {\left (3 \, b^{2} c^{3} d^{3} + 8 \, a b c^{2} d^{4} + 24 \, a^{2} c d^{5}\right )}}{c^{4} d^{3}}\right )} + \frac {70 \, {\left (a b c^{3} d^{3} + 3 \, a^{2} c^{2} d^{4}\right )}}{c^{4} d^{3}}\right )} x^{2} + \frac {105 \, a^{2}}{c}\right )} x}{105 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*((x^2*(2*(3*b^2*c^2*d^4 + 8*a*b*c*d^5 + 24*a^2*d^6)*x^2/(c^4*d^3) + 7*(3*b^2*c^3*d^3 + 8*a*b*c^2*d^4 + 2
4*a^2*c*d^5)/(c^4*d^3)) + 70*(a*b*c^3*d^3 + 3*a^2*c^2*d^4)/(c^4*d^3))*x^2 + 105*a^2/c)*x/(d*x^2 + c)^(7/2)

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Mupad [B]
time = 4.99, size = 176, normalized size = 1.01 \begin {gather*} \frac {x\,\left (\frac {a^2}{7\,c}+\frac {c\,\left (\frac {b^2}{7\,d}-\frac {2\,a\,b}{7\,c}\right )}{d}\right )}{{\left (d\,x^2+c\right )}^{7/2}}-\frac {x\,\left (\frac {b^2}{5\,d^2}-\frac {6\,a^2\,d^2+2\,a\,b\,c\,d-b^2\,c^2}{35\,c^2\,d^2}\right )}{{\left (d\,x^2+c\right )}^{5/2}}+\frac {x\,\left (24\,a^2\,d^2+8\,a\,b\,c\,d+3\,b^2\,c^2\right )}{105\,c^3\,d^2\,{\left (d\,x^2+c\right )}^{3/2}}+\frac {x\,\left (48\,a^2\,d^2+16\,a\,b\,c\,d+6\,b^2\,c^2\right )}{105\,c^4\,d^2\,\sqrt {d\,x^2+c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*(a^2/(7*c) + (c*(b^2/(7*d) - (2*a*b)/(7*c)))/d))/(c + d*x^2)^(7/2) - (x*(b^2/(5*d^2) - (6*a^2*d^2 - b^2*c^2
+ 2*a*b*c*d)/(35*c^2*d^2)))/(c + d*x^2)^(5/2) + (x*(24*a^2*d^2 + 3*b^2*c^2 + 8*a*b*c*d))/(105*c^3*d^2*(c + d*
x^2)^(3/2)) + (x*(48*a^2*d^2 + 6*b^2*c^2 + 16*a*b*c*d))/(105*c^4*d^2*(c + d*x^2)^(1/2))

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