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

Optimal. Leaf size=272 \[ \frac{d x \left (a+b x^2\right )^{5/2} \left (5 a^2 d^2-20 a b c d+36 b^2 c^2\right )}{160 b^3}+\frac{x \left (a+b x^2\right )^{3/2} (4 b c-a d) \left (a^2 d^2-2 a b c d+8 b^2 c^2\right )}{128 b^3}+\frac{3 a x \sqrt{a+b x^2} (4 b c-a d) \left (a^2 d^2-2 a b c d+8 b^2 c^2\right )}{256 b^3}+\frac{3 a^2 (4 b c-a d) \left (a^2 d^2-2 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{256 b^{7/2}}+\frac{d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right ) (14 b c-5 a d)}{80 b^2}+\frac{d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b} \]

[Out]

(3*a*(4*b*c - a*d)*(8*b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x*Sqrt[a + b*x^2])/(256*b^3) + ((4*b*c - a*d)*(8*b^2*c^2
- 2*a*b*c*d + a^2*d^2)*x*(a + b*x^2)^(3/2))/(128*b^3) + (d*(36*b^2*c^2 - 20*a*b*c*d + 5*a^2*d^2)*x*(a + b*x^2)
^(5/2))/(160*b^3) + (d*(14*b*c - 5*a*d)*x*(a + b*x^2)^(5/2)*(c + d*x^2))/(80*b^2) + (d*x*(a + b*x^2)^(5/2)*(c
+ d*x^2)^2)/(10*b) + (3*a^2*(4*b*c - a*d)*(8*b^2*c^2 - 2*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2
]])/(256*b^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.217773, antiderivative size = 272, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {416, 528, 388, 195, 217, 206} \[ \frac{d x \left (a+b x^2\right )^{5/2} \left (5 a^2 d^2-20 a b c d+36 b^2 c^2\right )}{160 b^3}+\frac{x \left (a+b x^2\right )^{3/2} (4 b c-a d) \left (a^2 d^2-2 a b c d+8 b^2 c^2\right )}{128 b^3}+\frac{3 a x \sqrt{a+b x^2} (4 b c-a d) \left (a^2 d^2-2 a b c d+8 b^2 c^2\right )}{256 b^3}+\frac{3 a^2 (4 b c-a d) \left (a^2 d^2-2 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{256 b^{7/2}}+\frac{d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right ) (14 b c-5 a d)}{80 b^2}+\frac{d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*(4*b*c - a*d)*(8*b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x*Sqrt[a + b*x^2])/(256*b^3) + ((4*b*c - a*d)*(8*b^2*c^2
- 2*a*b*c*d + a^2*d^2)*x*(a + b*x^2)^(3/2))/(128*b^3) + (d*(36*b^2*c^2 - 20*a*b*c*d + 5*a^2*d^2)*x*(a + b*x^2)
^(5/2))/(160*b^3) + (d*(14*b*c - 5*a*d)*x*(a + b*x^2)^(5/2)*(c + d*x^2))/(80*b^2) + (d*x*(a + b*x^2)^(5/2)*(c
+ d*x^2)^2)/(10*b) + (3*a^2*(4*b*c - a*d)*(8*b^2*c^2 - 2*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2
]])/(256*b^(7/2))

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 528

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

Rule 388

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3 \, dx &=\frac{d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b}+\frac{\int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \left (c (10 b c-a d)+d (14 b c-5 a d) x^2\right ) \, dx}{10 b}\\ &=\frac{d (14 b c-5 a d) x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )}{80 b^2}+\frac{d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b}+\frac{\int \left (a+b x^2\right )^{3/2} \left (c \left (80 b^2 c^2-22 a b c d+5 a^2 d^2\right )+3 d \left (36 b^2 c^2-20 a b c d+5 a^2 d^2\right ) x^2\right ) \, dx}{80 b^2}\\ &=\frac{d \left (36 b^2 c^2-20 a b c d+5 a^2 d^2\right ) x \left (a+b x^2\right )^{5/2}}{160 b^3}+\frac{d (14 b c-5 a d) x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )}{80 b^2}+\frac{d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b}+\frac{\left ((4 b c-a d) \left (8 b^2 c^2-2 a b c d+a^2 d^2\right )\right ) \int \left (a+b x^2\right )^{3/2} \, dx}{32 b^3}\\ &=\frac{(4 b c-a d) \left (8 b^2 c^2-2 a b c d+a^2 d^2\right ) x \left (a+b x^2\right )^{3/2}}{128 b^3}+\frac{d \left (36 b^2 c^2-20 a b c d+5 a^2 d^2\right ) x \left (a+b x^2\right )^{5/2}}{160 b^3}+\frac{d (14 b c-5 a d) x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )}{80 b^2}+\frac{d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b}+\frac{\left (3 a (4 b c-a d) \left (8 b^2 c^2-2 a b c d+a^2 d^2\right )\right ) \int \sqrt{a+b x^2} \, dx}{128 b^3}\\ &=\frac{3 a (4 b c-a d) \left (8 b^2 c^2-2 a b c d+a^2 d^2\right ) x \sqrt{a+b x^2}}{256 b^3}+\frac{(4 b c-a d) \left (8 b^2 c^2-2 a b c d+a^2 d^2\right ) x \left (a+b x^2\right )^{3/2}}{128 b^3}+\frac{d \left (36 b^2 c^2-20 a b c d+5 a^2 d^2\right ) x \left (a+b x^2\right )^{5/2}}{160 b^3}+\frac{d (14 b c-5 a d) x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )}{80 b^2}+\frac{d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b}+\frac{\left (3 a^2 (4 b c-a d) \left (8 b^2 c^2-2 a b c d+a^2 d^2\right )\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{256 b^3}\\ &=\frac{3 a (4 b c-a d) \left (8 b^2 c^2-2 a b c d+a^2 d^2\right ) x \sqrt{a+b x^2}}{256 b^3}+\frac{(4 b c-a d) \left (8 b^2 c^2-2 a b c d+a^2 d^2\right ) x \left (a+b x^2\right )^{3/2}}{128 b^3}+\frac{d \left (36 b^2 c^2-20 a b c d+5 a^2 d^2\right ) x \left (a+b x^2\right )^{5/2}}{160 b^3}+\frac{d (14 b c-5 a d) x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )}{80 b^2}+\frac{d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b}+\frac{\left (3 a^2 (4 b c-a d) \left (8 b^2 c^2-2 a b c d+a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{256 b^3}\\ &=\frac{3 a (4 b c-a d) \left (8 b^2 c^2-2 a b c d+a^2 d^2\right ) x \sqrt{a+b x^2}}{256 b^3}+\frac{(4 b c-a d) \left (8 b^2 c^2-2 a b c d+a^2 d^2\right ) x \left (a+b x^2\right )^{3/2}}{128 b^3}+\frac{d \left (36 b^2 c^2-20 a b c d+5 a^2 d^2\right ) x \left (a+b x^2\right )^{5/2}}{160 b^3}+\frac{d (14 b c-5 a d) x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )}{80 b^2}+\frac{d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b}+\frac{3 a^2 (4 b c-a d) \left (8 b^2 c^2-2 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{256 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 5.12393, size = 220, normalized size = 0.81 \[ \frac{\sqrt{b} x \sqrt{a+b x^2} \left (4 a^2 b^2 d \left (60 c^2+15 c d x^2+2 d^2 x^4\right )-10 a^3 b d^2 \left (9 c+d x^2\right )+15 a^4 d^3+16 a b^3 \left (70 c^2 d x^2+50 c^3+45 c d^2 x^4+11 d^3 x^6\right )+32 b^4 x^2 \left (20 c^2 d x^2+10 c^3+15 c d^2 x^4+4 d^3 x^6\right )\right )-15 a^2 (a d-4 b c) \left (a^2 d^2-2 a b c d+8 b^2 c^2\right ) \log \left (\sqrt{b} \sqrt{a+b x^2}+b x\right )}{1280 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*x*Sqrt[a + b*x^2]*(15*a^4*d^3 - 10*a^3*b*d^2*(9*c + d*x^2) + 4*a^2*b^2*d*(60*c^2 + 15*c*d*x^2 + 2*d^2
*x^4) + 32*b^4*x^2*(10*c^3 + 20*c^2*d*x^2 + 15*c*d^2*x^4 + 4*d^3*x^6) + 16*a*b^3*(50*c^3 + 70*c^2*d*x^2 + 45*c
*d^2*x^4 + 11*d^3*x^6)) - 15*a^2*(-4*b*c + a*d)*(8*b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Log[b*x + Sqrt[b]*Sqrt[a + b
*x^2]])/(1280*b^(7/2))

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Maple [A]  time = 0.012, size = 393, normalized size = 1.4 \begin{align*}{\frac{{d}^{3}{x}^{5}}{10\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{a{d}^{3}{x}^{3}}{16\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{{a}^{2}{d}^{3}x}{32\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{{a}^{3}{d}^{3}x}{128\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{d}^{3}{a}^{4}x}{256\,{b}^{3}}\sqrt{b{x}^{2}+a}}-{\frac{3\,{d}^{3}{a}^{5}}{256}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{7}{2}}}}+{\frac{3\,c{d}^{2}{x}^{3}}{8\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{3\,ac{d}^{2}x}{16\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{3\,{a}^{2}c{d}^{2}x}{64\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{9\,c{d}^{2}{a}^{3}x}{128\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{9\,c{d}^{2}{a}^{4}}{128}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}}+{\frac{{c}^{2}dx}{2\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{a{c}^{2}dx}{8\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{a}^{2}{c}^{2}dx}{16\,b}\sqrt{b{x}^{2}+a}}-{\frac{3\,{c}^{2}d{a}^{3}}{16}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{3}{2}}}}+{\frac{{c}^{3}x}{4} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,a{c}^{3}x}{8}\sqrt{b{x}^{2}+a}}+{\frac{3\,{c}^{3}{a}^{2}}{8}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*d^3*x^5*(b*x^2+a)^(5/2)/b-1/16*d^3/b^2*a*x^3*(b*x^2+a)^(5/2)+1/32*d^3/b^3*a^2*x*(b*x^2+a)^(5/2)-1/128*d^3
/b^3*a^3*x*(b*x^2+a)^(3/2)-3/256*d^3/b^3*a^4*x*(b*x^2+a)^(1/2)-3/256*d^3/b^(7/2)*a^5*ln(x*b^(1/2)+(b*x^2+a)^(1
/2))+3/8*c*d^2*x^3*(b*x^2+a)^(5/2)/b-3/16*c*d^2/b^2*a*x*(b*x^2+a)^(5/2)+3/64*c*d^2/b^2*a^2*x*(b*x^2+a)^(3/2)+9
/128*c*d^2/b^2*a^3*x*(b*x^2+a)^(1/2)+9/128*c*d^2/b^(5/2)*a^4*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+1/2*c^2*d*x*(b*x^2+
a)^(5/2)/b-1/8*c^2*d/b*a*x*(b*x^2+a)^(3/2)-3/16*c^2*d/b*a^2*x*(b*x^2+a)^(1/2)-3/16*c^2*d/b^(3/2)*a^3*ln(x*b^(1
/2)+(b*x^2+a)^(1/2))+1/4*c^3*x*(b*x^2+a)^(3/2)+3/8*c^3*a*x*(b*x^2+a)^(1/2)+3/8*c^3*a^2/b^(1/2)*ln(x*b^(1/2)+(b
*x^2+a)^(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)^(3/2)*(d*x^2+c)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.01576, size = 1102, normalized size = 4.05 \begin{align*} \left [-\frac{15 \,{\left (32 \, a^{2} b^{3} c^{3} - 16 \, a^{3} b^{2} c^{2} d + 6 \, a^{4} b c d^{2} - a^{5} d^{3}\right )} \sqrt{b} \log \left (-2 \, b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) - 2 \,{\left (128 \, b^{5} d^{3} x^{9} + 16 \,{\left (30 \, b^{5} c d^{2} + 11 \, a b^{4} d^{3}\right )} x^{7} + 8 \,{\left (80 \, b^{5} c^{2} d + 90 \, a b^{4} c d^{2} + a^{2} b^{3} d^{3}\right )} x^{5} + 10 \,{\left (32 \, b^{5} c^{3} + 112 \, a b^{4} c^{2} d + 6 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}\right )} x^{3} + 5 \,{\left (160 \, a b^{4} c^{3} + 48 \, a^{2} b^{3} c^{2} d - 18 \, a^{3} b^{2} c d^{2} + 3 \, a^{4} b d^{3}\right )} x\right )} \sqrt{b x^{2} + a}}{2560 \, b^{4}}, -\frac{15 \,{\left (32 \, a^{2} b^{3} c^{3} - 16 \, a^{3} b^{2} c^{2} d + 6 \, a^{4} b c d^{2} - a^{5} d^{3}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (128 \, b^{5} d^{3} x^{9} + 16 \,{\left (30 \, b^{5} c d^{2} + 11 \, a b^{4} d^{3}\right )} x^{7} + 8 \,{\left (80 \, b^{5} c^{2} d + 90 \, a b^{4} c d^{2} + a^{2} b^{3} d^{3}\right )} x^{5} + 10 \,{\left (32 \, b^{5} c^{3} + 112 \, a b^{4} c^{2} d + 6 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}\right )} x^{3} + 5 \,{\left (160 \, a b^{4} c^{3} + 48 \, a^{2} b^{3} c^{2} d - 18 \, a^{3} b^{2} c d^{2} + 3 \, a^{4} b d^{3}\right )} x\right )} \sqrt{b x^{2} + a}}{1280 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2560*(15*(32*a^2*b^3*c^3 - 16*a^3*b^2*c^2*d + 6*a^4*b*c*d^2 - a^5*d^3)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2
+ a)*sqrt(b)*x - a) - 2*(128*b^5*d^3*x^9 + 16*(30*b^5*c*d^2 + 11*a*b^4*d^3)*x^7 + 8*(80*b^5*c^2*d + 90*a*b^4*
c*d^2 + a^2*b^3*d^3)*x^5 + 10*(32*b^5*c^3 + 112*a*b^4*c^2*d + 6*a^2*b^3*c*d^2 - a^3*b^2*d^3)*x^3 + 5*(160*a*b^
4*c^3 + 48*a^2*b^3*c^2*d - 18*a^3*b^2*c*d^2 + 3*a^4*b*d^3)*x)*sqrt(b*x^2 + a))/b^4, -1/1280*(15*(32*a^2*b^3*c^
3 - 16*a^3*b^2*c^2*d + 6*a^4*b*c*d^2 - a^5*d^3)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (128*b^5*d^3*x^9
+ 16*(30*b^5*c*d^2 + 11*a*b^4*d^3)*x^7 + 8*(80*b^5*c^2*d + 90*a*b^4*c*d^2 + a^2*b^3*d^3)*x^5 + 10*(32*b^5*c^3
+ 112*a*b^4*c^2*d + 6*a^2*b^3*c*d^2 - a^3*b^2*d^3)*x^3 + 5*(160*a*b^4*c^3 + 48*a^2*b^3*c^2*d - 18*a^3*b^2*c*d
^2 + 3*a^4*b*d^3)*x)*sqrt(b*x^2 + a))/b^4]

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Sympy [B]  time = 45.9015, size = 665, normalized size = 2.44 \begin{align*} \frac{3 a^{\frac{9}{2}} d^{3} x}{256 b^{3} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{9 a^{\frac{7}{2}} c d^{2} x}{128 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{a^{\frac{7}{2}} d^{3} x^{3}}{256 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 a^{\frac{5}{2}} c^{2} d x}{16 b \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{3 a^{\frac{5}{2}} c d^{2} x^{3}}{128 b \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{a^{\frac{5}{2}} d^{3} x^{5}}{640 b \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{a^{\frac{3}{2}} c^{3} x \sqrt{1 + \frac{b x^{2}}{a}}}{2} + \frac{a^{\frac{3}{2}} c^{3} x}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{17 a^{\frac{3}{2}} c^{2} d x^{3}}{16 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{39 a^{\frac{3}{2}} c d^{2} x^{5}}{64 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{23 a^{\frac{3}{2}} d^{3} x^{7}}{160 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 \sqrt{a} b c^{3} x^{3}}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{11 \sqrt{a} b c^{2} d x^{5}}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{15 \sqrt{a} b c d^{2} x^{7}}{16 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{19 \sqrt{a} b d^{3} x^{9}}{80 \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{3 a^{5} d^{3} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{256 b^{\frac{7}{2}}} + \frac{9 a^{4} c d^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{128 b^{\frac{5}{2}}} - \frac{3 a^{3} c^{2} d \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{16 b^{\frac{3}{2}}} + \frac{3 a^{2} c^{3} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{8 \sqrt{b}} + \frac{b^{2} c^{3} x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{b^{2} c^{2} d x^{7}}{2 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 b^{2} c d^{2} x^{9}}{8 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{b^{2} d^{3} x^{11}}{10 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a**(9/2)*d**3*x/(256*b**3*sqrt(1 + b*x**2/a)) - 9*a**(7/2)*c*d**2*x/(128*b**2*sqrt(1 + b*x**2/a)) + a**(7/2)
*d**3*x**3/(256*b**2*sqrt(1 + b*x**2/a)) + 3*a**(5/2)*c**2*d*x/(16*b*sqrt(1 + b*x**2/a)) - 3*a**(5/2)*c*d**2*x
**3/(128*b*sqrt(1 + b*x**2/a)) - a**(5/2)*d**3*x**5/(640*b*sqrt(1 + b*x**2/a)) + a**(3/2)*c**3*x*sqrt(1 + b*x*
*2/a)/2 + a**(3/2)*c**3*x/(8*sqrt(1 + b*x**2/a)) + 17*a**(3/2)*c**2*d*x**3/(16*sqrt(1 + b*x**2/a)) + 39*a**(3/
2)*c*d**2*x**5/(64*sqrt(1 + b*x**2/a)) + 23*a**(3/2)*d**3*x**7/(160*sqrt(1 + b*x**2/a)) + 3*sqrt(a)*b*c**3*x**
3/(8*sqrt(1 + b*x**2/a)) + 11*sqrt(a)*b*c**2*d*x**5/(8*sqrt(1 + b*x**2/a)) + 15*sqrt(a)*b*c*d**2*x**7/(16*sqrt
(1 + b*x**2/a)) + 19*sqrt(a)*b*d**3*x**9/(80*sqrt(1 + b*x**2/a)) - 3*a**5*d**3*asinh(sqrt(b)*x/sqrt(a))/(256*b
**(7/2)) + 9*a**4*c*d**2*asinh(sqrt(b)*x/sqrt(a))/(128*b**(5/2)) - 3*a**3*c**2*d*asinh(sqrt(b)*x/sqrt(a))/(16*
b**(3/2)) + 3*a**2*c**3*asinh(sqrt(b)*x/sqrt(a))/(8*sqrt(b)) + b**2*c**3*x**5/(4*sqrt(a)*sqrt(1 + b*x**2/a)) +
b**2*c**2*d*x**7/(2*sqrt(a)*sqrt(1 + b*x**2/a)) + 3*b**2*c*d**2*x**9/(8*sqrt(a)*sqrt(1 + b*x**2/a)) + b**2*d*
*3*x**11/(10*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]  time = 1.11162, size = 351, normalized size = 1.29 \begin{align*} \frac{1}{1280} \,{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \, b d^{3} x^{2} + \frac{30 \, b^{9} c d^{2} + 11 \, a b^{8} d^{3}}{b^{8}}\right )} x^{2} + \frac{80 \, b^{9} c^{2} d + 90 \, a b^{8} c d^{2} + a^{2} b^{7} d^{3}}{b^{8}}\right )} x^{2} + \frac{5 \,{\left (32 \, b^{9} c^{3} + 112 \, a b^{8} c^{2} d + 6 \, a^{2} b^{7} c d^{2} - a^{3} b^{6} d^{3}\right )}}{b^{8}}\right )} x^{2} + \frac{5 \,{\left (160 \, a b^{8} c^{3} + 48 \, a^{2} b^{7} c^{2} d - 18 \, a^{3} b^{6} c d^{2} + 3 \, a^{4} b^{5} d^{3}\right )}}{b^{8}}\right )} \sqrt{b x^{2} + a} x - \frac{3 \,{\left (32 \, a^{2} b^{3} c^{3} - 16 \, a^{3} b^{2} c^{2} d + 6 \, a^{4} b c d^{2} - a^{5} d^{3}\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{256 \, b^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1280*(2*(4*(2*(8*b*d^3*x^2 + (30*b^9*c*d^2 + 11*a*b^8*d^3)/b^8)*x^2 + (80*b^9*c^2*d + 90*a*b^8*c*d^2 + a^2*b
^7*d^3)/b^8)*x^2 + 5*(32*b^9*c^3 + 112*a*b^8*c^2*d + 6*a^2*b^7*c*d^2 - a^3*b^6*d^3)/b^8)*x^2 + 5*(160*a*b^8*c^
3 + 48*a^2*b^7*c^2*d - 18*a^3*b^6*c*d^2 + 3*a^4*b^5*d^3)/b^8)*sqrt(b*x^2 + a)*x - 3/256*(32*a^2*b^3*c^3 - 16*a
^3*b^2*c^2*d + 6*a^4*b*c*d^2 - a^5*d^3)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)