3.635 \(\int \frac{(a+b x^2)^2 (c+d x^2)^{5/2}}{x^7} \, dx\)

Optimal. Leaf size=222 \[ -\frac{a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac{\left (c+d x^2\right )^{5/2} \left (a d (a d+12 b c)+8 b^2 c^2\right )}{16 c^2 x^2}+\frac{5 d \left (c+d x^2\right )^{3/2} \left (a d (a d+12 b c)+8 b^2 c^2\right )}{48 c^2}+\frac{5 d \sqrt{c+d x^2} \left (a d (a d+12 b c)+8 b^2 c^2\right )}{16 c}-\frac{5 d \left (a d (a d+12 b c)+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{16 \sqrt{c}}-\frac{a \left (c+d x^2\right )^{7/2} (a d+12 b c)}{24 c^2 x^4} \]

[Out]

(5*d*(8*b^2*c^2 + a*d*(12*b*c + a*d))*Sqrt[c + d*x^2])/(16*c) + (5*d*(8*b^2*c^2 + a*d*(12*b*c + a*d))*(c + d*x
^2)^(3/2))/(48*c^2) - ((8*b^2*c^2 + a*d*(12*b*c + a*d))*(c + d*x^2)^(5/2))/(16*c^2*x^2) - (a^2*(c + d*x^2)^(7/
2))/(6*c*x^6) - (a*(12*b*c + a*d)*(c + d*x^2)^(7/2))/(24*c^2*x^4) - (5*d*(8*b^2*c^2 + a*d*(12*b*c + a*d))*ArcT
anh[Sqrt[c + d*x^2]/Sqrt[c]])/(16*Sqrt[c])

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Rubi [A]  time = 0.252158, antiderivative size = 219, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {446, 89, 78, 47, 50, 63, 208} \[ -\frac{a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac{\left (c+d x^2\right )^{5/2} \left (\frac{a d (a d+12 b c)}{c^2}+8 b^2\right )}{16 x^2}+\frac{5 d \left (c+d x^2\right )^{3/2} \left (a d (a d+12 b c)+8 b^2 c^2\right )}{48 c^2}+\frac{5 d \sqrt{c+d x^2} \left (a d (a d+12 b c)+8 b^2 c^2\right )}{16 c}-\frac{5 d \left (a d (a d+12 b c)+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{16 \sqrt{c}}-\frac{a \left (c+d x^2\right )^{7/2} (a d+12 b c)}{24 c^2 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*d*(8*b^2*c^2 + a*d*(12*b*c + a*d))*Sqrt[c + d*x^2])/(16*c) + (5*d*(8*b^2*c^2 + a*d*(12*b*c + a*d))*(c + d*x
^2)^(3/2))/(48*c^2) - ((8*b^2 + (a*d*(12*b*c + a*d))/c^2)*(c + d*x^2)^(5/2))/(16*x^2) - (a^2*(c + d*x^2)^(7/2)
)/(6*c*x^6) - (a*(12*b*c + a*d)*(c + d*x^2)^(7/2))/(24*c^2*x^4) - (5*d*(8*b^2*c^2 + a*d*(12*b*c + a*d))*ArcTan
h[Sqrt[c + d*x^2]/Sqrt[c]])/(16*Sqrt[c])

Rule 446

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

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^7} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^2 (c+d x)^{5/2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac{a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}+\frac{\operatorname{Subst}\left (\int \frac{\left (\frac{1}{2} a (12 b c+a d)+3 b^2 c x\right ) (c+d x)^{5/2}}{x^3} \, dx,x,x^2\right )}{6 c}\\ &=-\frac{a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac{a (12 b c+a d) \left (c+d x^2\right )^{7/2}}{24 c^2 x^4}+\frac{1}{16} \left (8 b^2+\frac{a d (12 b c+a d)}{c^2}\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^{5/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{\left (8 b^2+\frac{a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{16 x^2}-\frac{a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac{a (12 b c+a d) \left (c+d x^2\right )^{7/2}}{24 c^2 x^4}+\frac{1}{32} \left (5 d \left (8 b^2+\frac{a d (12 b c+a d)}{c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac{5}{48} d \left (8 b^2+\frac{a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac{\left (8 b^2+\frac{a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{16 x^2}-\frac{a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac{a (12 b c+a d) \left (c+d x^2\right )^{7/2}}{24 c^2 x^4}+\frac{1}{32} \left (5 c d \left (8 b^2+\frac{a d (12 b c+a d)}{c^2}\right )\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{x} \, dx,x,x^2\right )\\ &=\frac{5}{16} c d \left (8 b^2+\frac{a d (12 b c+a d)}{c^2}\right ) \sqrt{c+d x^2}+\frac{5}{48} d \left (8 b^2+\frac{a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac{\left (8 b^2+\frac{a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{16 x^2}-\frac{a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac{a (12 b c+a d) \left (c+d x^2\right )^{7/2}}{24 c^2 x^4}+\frac{1}{32} \left (5 d \left (8 b^2 c^2+12 a b c d+a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^2\right )\\ &=\frac{5}{16} c d \left (8 b^2+\frac{a d (12 b c+a d)}{c^2}\right ) \sqrt{c+d x^2}+\frac{5}{48} d \left (8 b^2+\frac{a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac{\left (8 b^2+\frac{a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{16 x^2}-\frac{a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac{a (12 b c+a d) \left (c+d x^2\right )^{7/2}}{24 c^2 x^4}+\frac{1}{16} \left (5 \left (8 b^2 c^2+12 a b c d+a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^2}\right )\\ &=\frac{5}{16} c d \left (8 b^2+\frac{a d (12 b c+a d)}{c^2}\right ) \sqrt{c+d x^2}+\frac{5}{48} d \left (8 b^2+\frac{a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac{\left (8 b^2+\frac{a d (12 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{5/2}}{16 x^2}-\frac{a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac{a (12 b c+a d) \left (c+d x^2\right )^{7/2}}{24 c^2 x^4}-\frac{5 d \left (8 b^2 c^2+12 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{16 \sqrt{c}}\\ \end{align*}

Mathematica [C]  time = 0.0598314, size = 92, normalized size = 0.41 \[ \frac{\left (c+d x^2\right )^{7/2} \left (3 d x^6 \left (a^2 d^2+12 a b c d+8 b^2 c^2\right ) \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};\frac{d x^2}{c}+1\right )-7 a c^2 \left (4 a c+a d x^2+12 b c x^2\right )\right )}{168 c^4 x^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c + d*x^2)^(7/2)*(-7*a*c^2*(4*a*c + 12*b*c*x^2 + a*d*x^2) + 3*d*(8*b^2*c^2 + 12*a*b*c*d + a^2*d^2)*x^6*Hyper
geometric2F1[2, 7/2, 9/2, 1 + (d*x^2)/c]))/(168*c^4*x^6)

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Maple [A]  time = 0.014, size = 387, normalized size = 1.7 \begin{align*} -{\frac{{a}^{2}}{6\,c{x}^{6}} \left ( d{x}^{2}+c \right ) ^{{\frac{7}{2}}}}-{\frac{{a}^{2}d}{24\,{c}^{2}{x}^{4}} \left ( d{x}^{2}+c \right ) ^{{\frac{7}{2}}}}-{\frac{{a}^{2}{d}^{2}}{16\,{c}^{3}{x}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{7}{2}}}}+{\frac{{a}^{2}{d}^{3}}{16\,{c}^{3}} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{5\,{a}^{2}{d}^{3}}{48\,{c}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{a}^{2}{d}^{3}}{16}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ){\frac{1}{\sqrt{c}}}}+{\frac{5\,{a}^{2}{d}^{3}}{16\,c}\sqrt{d{x}^{2}+c}}-{\frac{ab}{2\,c{x}^{4}} \left ( d{x}^{2}+c \right ) ^{{\frac{7}{2}}}}-{\frac{3\,abd}{4\,{c}^{2}{x}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{7}{2}}}}+{\frac{3\,ab{d}^{2}}{4\,{c}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{5\,ab{d}^{2}}{4\,c} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{15\,ab{d}^{2}}{4}\sqrt{c}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ) }+{\frac{15\,ab{d}^{2}}{4}\sqrt{d{x}^{2}+c}}-{\frac{{b}^{2}}{2\,c{x}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{7}{2}}}}+{\frac{{b}^{2}d}{2\,c} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{5\,{b}^{2}d}{6} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{b}^{2}d}{2}{c}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ) }+{\frac{5\,{b}^{2}dc}{2}\sqrt{d{x}^{2}+c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a^2*(d*x^2+c)^(7/2)/c/x^6-1/24*a^2*d/c^2/x^4*(d*x^2+c)^(7/2)-1/16*a^2*d^2/c^3/x^2*(d*x^2+c)^(7/2)+1/16*a^
2*d^3/c^3*(d*x^2+c)^(5/2)+5/48*a^2*d^3/c^2*(d*x^2+c)^(3/2)-5/16*a^2*d^3/c^(1/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1
/2))/x)+5/16*a^2*d^3/c*(d*x^2+c)^(1/2)-1/2*a*b/c/x^4*(d*x^2+c)^(7/2)-3/4*a*b*d/c^2/x^2*(d*x^2+c)^(7/2)+3/4*a*b
*d^2/c^2*(d*x^2+c)^(5/2)+5/4*a*b*d^2/c*(d*x^2+c)^(3/2)-15/4*a*b*d^2*c^(1/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))
/x)+15/4*a*b*d^2*(d*x^2+c)^(1/2)-1/2*b^2/c/x^2*(d*x^2+c)^(7/2)+1/2*b^2*d/c*(d*x^2+c)^(5/2)+5/6*b^2*d*(d*x^2+c)
^(3/2)-5/2*b^2*d*c^(3/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x)+5/2*b^2*d*c*(d*x^2+c)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.48991, size = 778, normalized size = 3.5 \begin{align*} \left [\frac{15 \,{\left (8 \, b^{2} c^{2} d + 12 \, a b c d^{2} + a^{2} d^{3}\right )} \sqrt{c} x^{6} \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) + 2 \,{\left (16 \, b^{2} c d^{2} x^{8} + 16 \,{\left (7 \, b^{2} c^{2} d + 6 \, a b c d^{2}\right )} x^{6} - 8 \, a^{2} c^{3} - 3 \,{\left (8 \, b^{2} c^{3} + 36 \, a b c^{2} d + 11 \, a^{2} c d^{2}\right )} x^{4} - 2 \,{\left (12 \, a b c^{3} + 13 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{96 \, c x^{6}}, \frac{15 \,{\left (8 \, b^{2} c^{2} d + 12 \, a b c d^{2} + a^{2} d^{3}\right )} \sqrt{-c} x^{6} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) +{\left (16 \, b^{2} c d^{2} x^{8} + 16 \,{\left (7 \, b^{2} c^{2} d + 6 \, a b c d^{2}\right )} x^{6} - 8 \, a^{2} c^{3} - 3 \,{\left (8 \, b^{2} c^{3} + 36 \, a b c^{2} d + 11 \, a^{2} c d^{2}\right )} x^{4} - 2 \,{\left (12 \, a b c^{3} + 13 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{48 \, c x^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(15*(8*b^2*c^2*d + 12*a*b*c*d^2 + a^2*d^3)*sqrt(c)*x^6*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^
2) + 2*(16*b^2*c*d^2*x^8 + 16*(7*b^2*c^2*d + 6*a*b*c*d^2)*x^6 - 8*a^2*c^3 - 3*(8*b^2*c^3 + 36*a*b*c^2*d + 11*a
^2*c*d^2)*x^4 - 2*(12*a*b*c^3 + 13*a^2*c^2*d)*x^2)*sqrt(d*x^2 + c))/(c*x^6), 1/48*(15*(8*b^2*c^2*d + 12*a*b*c*
d^2 + a^2*d^3)*sqrt(-c)*x^6*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (16*b^2*c*d^2*x^8 + 16*(7*b^2*c^2*d + 6*a*b*c*d
^2)*x^6 - 8*a^2*c^3 - 3*(8*b^2*c^3 + 36*a*b*c^2*d + 11*a^2*c*d^2)*x^4 - 2*(12*a*b*c^3 + 13*a^2*c^2*d)*x^2)*sqr
t(d*x^2 + c))/(c*x^6)]

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Sympy [A]  time = 131.077, size = 468, normalized size = 2.11 \begin{align*} - \frac{a^{2} c^{3}}{6 \sqrt{d} x^{7} \sqrt{\frac{c}{d x^{2}} + 1}} - \frac{17 a^{2} c^{2} \sqrt{d}}{24 x^{5} \sqrt{\frac{c}{d x^{2}} + 1}} - \frac{35 a^{2} c d^{\frac{3}{2}}}{48 x^{3} \sqrt{\frac{c}{d x^{2}} + 1}} - \frac{a^{2} d^{\frac{5}{2}} \sqrt{\frac{c}{d x^{2}} + 1}}{2 x} - \frac{3 a^{2} d^{\frac{5}{2}}}{16 x \sqrt{\frac{c}{d x^{2}} + 1}} - \frac{5 a^{2} d^{3} \operatorname{asinh}{\left (\frac{\sqrt{c}}{\sqrt{d} x} \right )}}{16 \sqrt{c}} - \frac{15 a b \sqrt{c} d^{2} \operatorname{asinh}{\left (\frac{\sqrt{c}}{\sqrt{d} x} \right )}}{4} - \frac{a b c^{3}}{2 \sqrt{d} x^{5} \sqrt{\frac{c}{d x^{2}} + 1}} - \frac{3 a b c^{2} \sqrt{d}}{4 x^{3} \sqrt{\frac{c}{d x^{2}} + 1}} - \frac{2 a b c d^{\frac{3}{2}} \sqrt{\frac{c}{d x^{2}} + 1}}{x} + \frac{7 a b c d^{\frac{3}{2}}}{4 x \sqrt{\frac{c}{d x^{2}} + 1}} + \frac{2 a b d^{\frac{5}{2}} x}{\sqrt{\frac{c}{d x^{2}} + 1}} - \frac{5 b^{2} c^{\frac{3}{2}} d \operatorname{asinh}{\left (\frac{\sqrt{c}}{\sqrt{d} x} \right )}}{2} - \frac{b^{2} c^{2} \sqrt{d} \sqrt{\frac{c}{d x^{2}} + 1}}{2 x} + \frac{2 b^{2} c^{2} \sqrt{d}}{x \sqrt{\frac{c}{d x^{2}} + 1}} + \frac{2 b^{2} c d^{\frac{3}{2}} x}{\sqrt{\frac{c}{d x^{2}} + 1}} + b^{2} d^{2} \left (\begin{cases} \frac{\sqrt{c} x^{2}}{2} & \text{for}\: d = 0 \\\frac{\left (c + d x^{2}\right )^{\frac{3}{2}}}{3 d} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(5/2)/x**7,x)

[Out]

-a**2*c**3/(6*sqrt(d)*x**7*sqrt(c/(d*x**2) + 1)) - 17*a**2*c**2*sqrt(d)/(24*x**5*sqrt(c/(d*x**2) + 1)) - 35*a*
*2*c*d**(3/2)/(48*x**3*sqrt(c/(d*x**2) + 1)) - a**2*d**(5/2)*sqrt(c/(d*x**2) + 1)/(2*x) - 3*a**2*d**(5/2)/(16*
x*sqrt(c/(d*x**2) + 1)) - 5*a**2*d**3*asinh(sqrt(c)/(sqrt(d)*x))/(16*sqrt(c)) - 15*a*b*sqrt(c)*d**2*asinh(sqrt
(c)/(sqrt(d)*x))/4 - a*b*c**3/(2*sqrt(d)*x**5*sqrt(c/(d*x**2) + 1)) - 3*a*b*c**2*sqrt(d)/(4*x**3*sqrt(c/(d*x**
2) + 1)) - 2*a*b*c*d**(3/2)*sqrt(c/(d*x**2) + 1)/x + 7*a*b*c*d**(3/2)/(4*x*sqrt(c/(d*x**2) + 1)) + 2*a*b*d**(5
/2)*x/sqrt(c/(d*x**2) + 1) - 5*b**2*c**(3/2)*d*asinh(sqrt(c)/(sqrt(d)*x))/2 - b**2*c**2*sqrt(d)*sqrt(c/(d*x**2
) + 1)/(2*x) + 2*b**2*c**2*sqrt(d)/(x*sqrt(c/(d*x**2) + 1)) + 2*b**2*c*d**(3/2)*x/sqrt(c/(d*x**2) + 1) + b**2*
d**2*Piecewise((sqrt(c)*x**2/2, Eq(d, 0)), ((c + d*x**2)**(3/2)/(3*d), True))

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Giac [A]  time = 1.18783, size = 386, normalized size = 1.74 \begin{align*} \frac{16 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} b^{2} d^{2} + 96 \, \sqrt{d x^{2} + c} b^{2} c d^{2} + 96 \, \sqrt{d x^{2} + c} a b d^{3} + \frac{15 \,{\left (8 \, b^{2} c^{2} d^{2} + 12 \, a b c d^{3} + a^{2} d^{4}\right )} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{\sqrt{-c}} - \frac{24 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} b^{2} c^{2} d^{2} - 48 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} b^{2} c^{3} d^{2} + 24 \, \sqrt{d x^{2} + c} b^{2} c^{4} d^{2} + 108 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} a b c d^{3} - 192 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a b c^{2} d^{3} + 84 \, \sqrt{d x^{2} + c} a b c^{3} d^{3} + 33 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} a^{2} d^{4} - 40 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a^{2} c d^{4} + 15 \, \sqrt{d x^{2} + c} a^{2} c^{2} d^{4}}{d^{3} x^{6}}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*(d*x^2 + c)^(3/2)*b^2*d^2 + 96*sqrt(d*x^2 + c)*b^2*c*d^2 + 96*sqrt(d*x^2 + c)*a*b*d^3 + 15*(8*b^2*c^2
*d^2 + 12*a*b*c*d^3 + a^2*d^4)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) - (24*(d*x^2 + c)^(5/2)*b^2*c^2*d^2 -
48*(d*x^2 + c)^(3/2)*b^2*c^3*d^2 + 24*sqrt(d*x^2 + c)*b^2*c^4*d^2 + 108*(d*x^2 + c)^(5/2)*a*b*c*d^3 - 192*(d*
x^2 + c)^(3/2)*a*b*c^2*d^3 + 84*sqrt(d*x^2 + c)*a*b*c^3*d^3 + 33*(d*x^2 + c)^(5/2)*a^2*d^4 - 40*(d*x^2 + c)^(3
/2)*a^2*c*d^4 + 15*sqrt(d*x^2 + c)*a^2*c^2*d^4)/(d^3*x^6))/d