3.615 \(\int x^2 (a+b x^2)^2 (c+d x^2)^{3/2} \, dx\)

Optimal. Leaf size=235 \[ \frac{c^2 x \sqrt{c+d x^2} \left (16 a^2 d^2+3 b c (b c-4 a d)\right )}{256 d^3}-\frac{c^3 \left (16 a^2 d^2+3 b c (b c-4 a d)\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{256 d^{7/2}}+\frac{x^3 \left (c+d x^2\right )^{3/2} \left (16 a^2 d^2+3 b c (b c-4 a d)\right )}{96 d^2}+\frac{c x^3 \sqrt{c+d x^2} \left (16 a^2 d^2+3 b c (b c-4 a d)\right )}{128 d^2}-\frac{b x^3 \left (c+d x^2\right )^{5/2} (b c-4 a d)}{16 d^2}+\frac{b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d} \]

[Out]

(c^2*(16*a^2*d^2 + 3*b*c*(b*c - 4*a*d))*x*Sqrt[c + d*x^2])/(256*d^3) + (c*(16*a^2*d^2 + 3*b*c*(b*c - 4*a*d))*x
^3*Sqrt[c + d*x^2])/(128*d^2) + ((16*a^2*d^2 + 3*b*c*(b*c - 4*a*d))*x^3*(c + d*x^2)^(3/2))/(96*d^2) - (b*(b*c
- 4*a*d)*x^3*(c + d*x^2)^(5/2))/(16*d^2) + (b^2*x^5*(c + d*x^2)^(5/2))/(10*d) - (c^3*(16*a^2*d^2 + 3*b*c*(b*c
- 4*a*d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(256*d^(7/2))

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Rubi [A]  time = 0.217567, antiderivative size = 232, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {464, 459, 279, 321, 217, 206} \[ \frac{c^2 x \sqrt{c+d x^2} \left (16 a^2 d^2+3 b c (b c-4 a d)\right )}{256 d^3}-\frac{c^3 \left (16 a^2 d^2+3 b c (b c-4 a d)\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{256 d^{7/2}}+\frac{1}{96} x^3 \left (c+d x^2\right )^{3/2} \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right )+\frac{c x^3 \sqrt{c+d x^2} \left (16 a^2 d^2+3 b c (b c-4 a d)\right )}{128 d^2}-\frac{b x^3 \left (c+d x^2\right )^{5/2} (b c-4 a d)}{16 d^2}+\frac{b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x^2)^2*(c + d*x^2)^(3/2),x]

[Out]

(c^2*(16*a^2*d^2 + 3*b*c*(b*c - 4*a*d))*x*Sqrt[c + d*x^2])/(256*d^3) + (c*(16*a^2*d^2 + 3*b*c*(b*c - 4*a*d))*x
^3*Sqrt[c + d*x^2])/(128*d^2) + ((16*a^2 + (3*b*c*(b*c - 4*a*d))/d^2)*x^3*(c + d*x^2)^(3/2))/96 - (b*(b*c - 4*
a*d)*x^3*(c + d*x^2)^(5/2))/(16*d^2) + (b^2*x^5*(c + d*x^2)^(5/2))/(10*d) - (c^3*(16*a^2*d^2 + 3*b*c*(b*c - 4*
a*d))*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(256*d^(7/2))

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(d^2*(e*x)^
(m + n + 1)*(a + b*x^n)^(p + 1))/(b*e^(n + 1)*(m + n*(p + 2) + 1)), x] + Dist[1/(b*(m + n*(p + 2) + 1)), Int[(
e*x)^m*(a + b*x^n)^p*Simp[b*c^2*(m + n*(p + 2) + 1) + d*((2*b*c - a*d)*(m + n + 1) + 2*b*c*n*(p + 1))*x^n, x],
x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && NeQ[m + n*(p + 2) + 1, 0]

Rule 459

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

Rule 279

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

Rule 321

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int x^2 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx &=\frac{b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}+\frac{\int x^2 \left (c+d x^2\right )^{3/2} \left (10 a^2 d-5 b (b c-4 a d) x^2\right ) \, dx}{10 d}\\ &=-\frac{b (b c-4 a d) x^3 \left (c+d x^2\right )^{5/2}}{16 d^2}+\frac{b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}+\frac{1}{16} \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right ) \int x^2 \left (c+d x^2\right )^{3/2} \, dx\\ &=\frac{1}{96} \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right ) x^3 \left (c+d x^2\right )^{3/2}-\frac{b (b c-4 a d) x^3 \left (c+d x^2\right )^{5/2}}{16 d^2}+\frac{b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}+\frac{1}{32} \left (c \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right )\right ) \int x^2 \sqrt{c+d x^2} \, dx\\ &=\frac{1}{128} c \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right ) x^3 \sqrt{c+d x^2}+\frac{1}{96} \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right ) x^3 \left (c+d x^2\right )^{3/2}-\frac{b (b c-4 a d) x^3 \left (c+d x^2\right )^{5/2}}{16 d^2}+\frac{b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}+\frac{1}{128} \left (c^2 \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right )\right ) \int \frac{x^2}{\sqrt{c+d x^2}} \, dx\\ &=\frac{c^2 \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right ) x \sqrt{c+d x^2}}{256 d}+\frac{1}{128} c \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right ) x^3 \sqrt{c+d x^2}+\frac{1}{96} \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right ) x^3 \left (c+d x^2\right )^{3/2}-\frac{b (b c-4 a d) x^3 \left (c+d x^2\right )^{5/2}}{16 d^2}+\frac{b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}-\frac{\left (c^3 \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right )\right ) \int \frac{1}{\sqrt{c+d x^2}} \, dx}{256 d}\\ &=\frac{c^2 \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right ) x \sqrt{c+d x^2}}{256 d}+\frac{1}{128} c \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right ) x^3 \sqrt{c+d x^2}+\frac{1}{96} \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right ) x^3 \left (c+d x^2\right )^{3/2}-\frac{b (b c-4 a d) x^3 \left (c+d x^2\right )^{5/2}}{16 d^2}+\frac{b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}-\frac{\left (c^3 \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{256 d}\\ &=\frac{c^2 \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right ) x \sqrt{c+d x^2}}{256 d}+\frac{1}{128} c \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right ) x^3 \sqrt{c+d x^2}+\frac{1}{96} \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right ) x^3 \left (c+d x^2\right )^{3/2}-\frac{b (b c-4 a d) x^3 \left (c+d x^2\right )^{5/2}}{16 d^2}+\frac{b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}-\frac{c^3 \left (16 a^2+\frac{3 b c (b c-4 a d)}{d^2}\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{256 d^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.116866, size = 193, normalized size = 0.82 \[ \frac{\sqrt{d} x \sqrt{c+d x^2} \left (80 a^2 d^2 \left (3 c^2+14 c d x^2+8 d^2 x^4\right )+60 a b d \left (2 c^2 d x^2-3 c^3+24 c d^2 x^4+16 d^3 x^6\right )+3 b^2 \left (8 c^2 d^2 x^4-10 c^3 d x^2+15 c^4+176 c d^3 x^6+128 d^4 x^8\right )\right )-15 c^3 \left (16 a^2 d^2-12 a b c d+3 b^2 c^2\right ) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{3840 d^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x^2)^2*(c + d*x^2)^(3/2),x]

[Out]

(Sqrt[d]*x*Sqrt[c + d*x^2]*(80*a^2*d^2*(3*c^2 + 14*c*d*x^2 + 8*d^2*x^4) + 60*a*b*d*(-3*c^3 + 2*c^2*d*x^2 + 24*
c*d^2*x^4 + 16*d^3*x^6) + 3*b^2*(15*c^4 - 10*c^3*d*x^2 + 8*c^2*d^2*x^4 + 176*c*d^3*x^6 + 128*d^4*x^8)) - 15*c^
3*(3*b^2*c^2 - 12*a*b*c*d + 16*a^2*d^2)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/(3840*d^(7/2))

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Maple [A]  time = 0.007, size = 321, normalized size = 1.4 \begin{align*}{\frac{{b}^{2}{x}^{5}}{10\,d} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{{b}^{2}c{x}^{3}}{16\,{d}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{{b}^{2}{c}^{2}x}{32\,{d}^{3}} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{{b}^{2}{c}^{3}x}{128\,{d}^{3}} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{b}^{2}{c}^{4}x}{256\,{d}^{3}}\sqrt{d{x}^{2}+c}}-{\frac{3\,{b}^{2}{c}^{5}}{256}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{7}{2}}}}+{\frac{ab{x}^{3}}{4\,d} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{abcx}{8\,{d}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}+{\frac{ab{c}^{2}x}{32\,{d}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{\frac{3\,ab{c}^{3}x}{64\,{d}^{2}}\sqrt{d{x}^{2}+c}}+{\frac{3\,ab{c}^{4}}{64}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{5}{2}}}}+{\frac{{a}^{2}x}{6\,d} \left ( d{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{{a}^{2}cx}{24\,d} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{{a}^{2}{c}^{2}x}{16\,d}\sqrt{d{x}^{2}+c}}-{\frac{{a}^{2}{c}^{3}}{16}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x^2+a)^2*(d*x^2+c)^(3/2),x)

[Out]

1/10*b^2*x^5*(d*x^2+c)^(5/2)/d-1/16*b^2*c/d^2*x^3*(d*x^2+c)^(5/2)+1/32*b^2*c^2/d^3*x*(d*x^2+c)^(5/2)-1/128*b^2
*c^3/d^3*x*(d*x^2+c)^(3/2)-3/256*b^2*c^4/d^3*x*(d*x^2+c)^(1/2)-3/256*b^2*c^5/d^(7/2)*ln(x*d^(1/2)+(d*x^2+c)^(1
/2))+1/4*a*b*x^3*(d*x^2+c)^(5/2)/d-1/8*a*b*c/d^2*x*(d*x^2+c)^(5/2)+1/32*a*b*c^2/d^2*x*(d*x^2+c)^(3/2)+3/64*a*b
*c^3/d^2*x*(d*x^2+c)^(1/2)+3/64*a*b*c^4/d^(5/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))+1/6*a^2*x*(d*x^2+c)^(5/2)/d-1/24
*a^2*c/d*x*(d*x^2+c)^(3/2)-1/16*a^2*c^2/d*x*(d*x^2+c)^(1/2)-1/16*a^2*c^3/d^(3/2)*ln(x*d^(1/2)+(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(x^2*(b*x^2+a)^2*(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.96273, size = 953, normalized size = 4.06 \begin{align*} \left [\frac{15 \,{\left (3 \, b^{2} c^{5} - 12 \, a b c^{4} d + 16 \, a^{2} c^{3} d^{2}\right )} \sqrt{d} \log \left (-2 \, d x^{2} + 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) + 2 \,{\left (384 \, b^{2} d^{5} x^{9} + 48 \,{\left (11 \, b^{2} c d^{4} + 20 \, a b d^{5}\right )} x^{7} + 8 \,{\left (3 \, b^{2} c^{2} d^{3} + 180 \, a b c d^{4} + 80 \, a^{2} d^{5}\right )} x^{5} - 10 \,{\left (3 \, b^{2} c^{3} d^{2} - 12 \, a b c^{2} d^{3} - 112 \, a^{2} c d^{4}\right )} x^{3} + 15 \,{\left (3 \, b^{2} c^{4} d - 12 \, a b c^{3} d^{2} + 16 \, a^{2} c^{2} d^{3}\right )} x\right )} \sqrt{d x^{2} + c}}{7680 \, d^{4}}, \frac{15 \,{\left (3 \, b^{2} c^{5} - 12 \, a b c^{4} d + 16 \, a^{2} c^{3} d^{2}\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) +{\left (384 \, b^{2} d^{5} x^{9} + 48 \,{\left (11 \, b^{2} c d^{4} + 20 \, a b d^{5}\right )} x^{7} + 8 \,{\left (3 \, b^{2} c^{2} d^{3} + 180 \, a b c d^{4} + 80 \, a^{2} d^{5}\right )} x^{5} - 10 \,{\left (3 \, b^{2} c^{3} d^{2} - 12 \, a b c^{2} d^{3} - 112 \, a^{2} c d^{4}\right )} x^{3} + 15 \,{\left (3 \, b^{2} c^{4} d - 12 \, a b c^{3} d^{2} + 16 \, a^{2} c^{2} d^{3}\right )} x\right )} \sqrt{d x^{2} + c}}{3840 \, d^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+a)^2*(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[1/7680*(15*(3*b^2*c^5 - 12*a*b*c^4*d + 16*a^2*c^3*d^2)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c
) + 2*(384*b^2*d^5*x^9 + 48*(11*b^2*c*d^4 + 20*a*b*d^5)*x^7 + 8*(3*b^2*c^2*d^3 + 180*a*b*c*d^4 + 80*a^2*d^5)*x
^5 - 10*(3*b^2*c^3*d^2 - 12*a*b*c^2*d^3 - 112*a^2*c*d^4)*x^3 + 15*(3*b^2*c^4*d - 12*a*b*c^3*d^2 + 16*a^2*c^2*d
^3)*x)*sqrt(d*x^2 + c))/d^4, 1/3840*(15*(3*b^2*c^5 - 12*a*b*c^4*d + 16*a^2*c^3*d^2)*sqrt(-d)*arctan(sqrt(-d)*x
/sqrt(d*x^2 + c)) + (384*b^2*d^5*x^9 + 48*(11*b^2*c*d^4 + 20*a*b*d^5)*x^7 + 8*(3*b^2*c^2*d^3 + 180*a*b*c*d^4 +
80*a^2*d^5)*x^5 - 10*(3*b^2*c^3*d^2 - 12*a*b*c^2*d^3 - 112*a^2*c*d^4)*x^3 + 15*(3*b^2*c^4*d - 12*a*b*c^3*d^2
+ 16*a^2*c^2*d^3)*x)*sqrt(d*x^2 + c))/d^4]

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Sympy [B]  time = 41.3064, size = 505, normalized size = 2.15 \begin{align*} \frac{a^{2} c^{\frac{5}{2}} x}{16 d \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{17 a^{2} c^{\frac{3}{2}} x^{3}}{48 \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{11 a^{2} \sqrt{c} d x^{5}}{24 \sqrt{1 + \frac{d x^{2}}{c}}} - \frac{a^{2} c^{3} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )}}{16 d^{\frac{3}{2}}} + \frac{a^{2} d^{2} x^{7}}{6 \sqrt{c} \sqrt{1 + \frac{d x^{2}}{c}}} - \frac{3 a b c^{\frac{7}{2}} x}{64 d^{2} \sqrt{1 + \frac{d x^{2}}{c}}} - \frac{a b c^{\frac{5}{2}} x^{3}}{64 d \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{13 a b c^{\frac{3}{2}} x^{5}}{32 \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{5 a b \sqrt{c} d x^{7}}{8 \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{3 a b c^{4} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )}}{64 d^{\frac{5}{2}}} + \frac{a b d^{2} x^{9}}{4 \sqrt{c} \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{3 b^{2} c^{\frac{9}{2}} x}{256 d^{3} \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{b^{2} c^{\frac{7}{2}} x^{3}}{256 d^{2} \sqrt{1 + \frac{d x^{2}}{c}}} - \frac{b^{2} c^{\frac{5}{2}} x^{5}}{640 d \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{23 b^{2} c^{\frac{3}{2}} x^{7}}{160 \sqrt{1 + \frac{d x^{2}}{c}}} + \frac{19 b^{2} \sqrt{c} d x^{9}}{80 \sqrt{1 + \frac{d x^{2}}{c}}} - \frac{3 b^{2} c^{5} \operatorname{asinh}{\left (\frac{\sqrt{d} x}{\sqrt{c}} \right )}}{256 d^{\frac{7}{2}}} + \frac{b^{2} d^{2} x^{11}}{10 \sqrt{c} \sqrt{1 + \frac{d x^{2}}{c}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x**2+a)**2*(d*x**2+c)**(3/2),x)

[Out]

a**2*c**(5/2)*x/(16*d*sqrt(1 + d*x**2/c)) + 17*a**2*c**(3/2)*x**3/(48*sqrt(1 + d*x**2/c)) + 11*a**2*sqrt(c)*d*
x**5/(24*sqrt(1 + d*x**2/c)) - a**2*c**3*asinh(sqrt(d)*x/sqrt(c))/(16*d**(3/2)) + a**2*d**2*x**7/(6*sqrt(c)*sq
rt(1 + d*x**2/c)) - 3*a*b*c**(7/2)*x/(64*d**2*sqrt(1 + d*x**2/c)) - a*b*c**(5/2)*x**3/(64*d*sqrt(1 + d*x**2/c)
) + 13*a*b*c**(3/2)*x**5/(32*sqrt(1 + d*x**2/c)) + 5*a*b*sqrt(c)*d*x**7/(8*sqrt(1 + d*x**2/c)) + 3*a*b*c**4*as
inh(sqrt(d)*x/sqrt(c))/(64*d**(5/2)) + a*b*d**2*x**9/(4*sqrt(c)*sqrt(1 + d*x**2/c)) + 3*b**2*c**(9/2)*x/(256*d
**3*sqrt(1 + d*x**2/c)) + b**2*c**(7/2)*x**3/(256*d**2*sqrt(1 + d*x**2/c)) - b**2*c**(5/2)*x**5/(640*d*sqrt(1
+ d*x**2/c)) + 23*b**2*c**(3/2)*x**7/(160*sqrt(1 + d*x**2/c)) + 19*b**2*sqrt(c)*d*x**9/(80*sqrt(1 + d*x**2/c))
- 3*b**2*c**5*asinh(sqrt(d)*x/sqrt(c))/(256*d**(7/2)) + b**2*d**2*x**11/(10*sqrt(c)*sqrt(1 + d*x**2/c))

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Giac [A]  time = 1.11371, size = 296, normalized size = 1.26 \begin{align*} \frac{1}{3840} \,{\left (2 \,{\left (4 \,{\left (6 \,{\left (8 \, b^{2} d x^{2} + \frac{11 \, b^{2} c d^{8} + 20 \, a b d^{9}}{d^{8}}\right )} x^{2} + \frac{3 \, b^{2} c^{2} d^{7} + 180 \, a b c d^{8} + 80 \, a^{2} d^{9}}{d^{8}}\right )} x^{2} - \frac{5 \,{\left (3 \, b^{2} c^{3} d^{6} - 12 \, a b c^{2} d^{7} - 112 \, a^{2} c d^{8}\right )}}{d^{8}}\right )} x^{2} + \frac{15 \,{\left (3 \, b^{2} c^{4} d^{5} - 12 \, a b c^{3} d^{6} + 16 \, a^{2} c^{2} d^{7}\right )}}{d^{8}}\right )} \sqrt{d x^{2} + c} x + \frac{{\left (3 \, b^{2} c^{5} - 12 \, a b c^{4} d + 16 \, a^{2} c^{3} d^{2}\right )} \log \left ({\left | -\sqrt{d} x + \sqrt{d x^{2} + c} \right |}\right )}{256 \, d^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+a)^2*(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

1/3840*(2*(4*(6*(8*b^2*d*x^2 + (11*b^2*c*d^8 + 20*a*b*d^9)/d^8)*x^2 + (3*b^2*c^2*d^7 + 180*a*b*c*d^8 + 80*a^2*
d^9)/d^8)*x^2 - 5*(3*b^2*c^3*d^6 - 12*a*b*c^2*d^7 - 112*a^2*c*d^8)/d^8)*x^2 + 15*(3*b^2*c^4*d^5 - 12*a*b*c^3*d
^6 + 16*a^2*c^2*d^7)/d^8)*sqrt(d*x^2 + c)*x + 1/256*(3*b^2*c^5 - 12*a*b*c^4*d + 16*a^2*c^3*d^2)*log(abs(-sqrt(
d)*x + sqrt(d*x^2 + c)))/d^(7/2)