3.491 \(\int x^{3/2} (a+b x)^{3/2} (A+B x) \, dx\)

Optimal. Leaf size=192 \[ \frac{a^2 x^{3/2} \sqrt{a+b x} (2 A b-a B)}{64 b^2}-\frac{3 a^3 \sqrt{x} \sqrt{a+b x} (2 A b-a B)}{128 b^3}+\frac{3 a^4 (2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{128 b^{7/2}}+\frac{a x^{5/2} \sqrt{a+b x} (2 A b-a B)}{16 b}+\frac{x^{5/2} (a+b x)^{3/2} (2 A b-a B)}{8 b}+\frac{B x^{5/2} (a+b x)^{5/2}}{5 b} \]

[Out]

(-3*a^3*(2*A*b - a*B)*Sqrt[x]*Sqrt[a + b*x])/(128*b^3) + (a^2*(2*A*b - a*B)*x^(3/2)*Sqrt[a + b*x])/(64*b^2) +
(a*(2*A*b - a*B)*x^(5/2)*Sqrt[a + b*x])/(16*b) + ((2*A*b - a*B)*x^(5/2)*(a + b*x)^(3/2))/(8*b) + (B*x^(5/2)*(a
+ b*x)^(5/2))/(5*b) + (3*a^4*(2*A*b - a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(128*b^(7/2))

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Rubi [A]  time = 0.0834286, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {80, 50, 63, 217, 206} \[ \frac{a^2 x^{3/2} \sqrt{a+b x} (2 A b-a B)}{64 b^2}-\frac{3 a^3 \sqrt{x} \sqrt{a+b x} (2 A b-a B)}{128 b^3}+\frac{3 a^4 (2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{128 b^{7/2}}+\frac{a x^{5/2} \sqrt{a+b x} (2 A b-a B)}{16 b}+\frac{x^{5/2} (a+b x)^{3/2} (2 A b-a B)}{8 b}+\frac{B x^{5/2} (a+b x)^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(a + b*x)^(3/2)*(A + B*x),x]

[Out]

(-3*a^3*(2*A*b - a*B)*Sqrt[x]*Sqrt[a + b*x])/(128*b^3) + (a^2*(2*A*b - a*B)*x^(3/2)*Sqrt[a + b*x])/(64*b^2) +
(a*(2*A*b - a*B)*x^(5/2)*Sqrt[a + b*x])/(16*b) + ((2*A*b - a*B)*x^(5/2)*(a + b*x)^(3/2))/(8*b) + (B*x^(5/2)*(a
+ b*x)^(5/2))/(5*b) + (3*a^4*(2*A*b - a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(128*b^(7/2))

Rule 80

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

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 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^{3/2} (a+b x)^{3/2} (A+B x) \, dx &=\frac{B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac{\left (5 A b-\frac{5 a B}{2}\right ) \int x^{3/2} (a+b x)^{3/2} \, dx}{5 b}\\ &=\frac{(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac{B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac{(3 a (2 A b-a B)) \int x^{3/2} \sqrt{a+b x} \, dx}{16 b}\\ &=\frac{a (2 A b-a B) x^{5/2} \sqrt{a+b x}}{16 b}+\frac{(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac{B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac{\left (a^2 (2 A b-a B)\right ) \int \frac{x^{3/2}}{\sqrt{a+b x}} \, dx}{32 b}\\ &=\frac{a^2 (2 A b-a B) x^{3/2} \sqrt{a+b x}}{64 b^2}+\frac{a (2 A b-a B) x^{5/2} \sqrt{a+b x}}{16 b}+\frac{(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac{B x^{5/2} (a+b x)^{5/2}}{5 b}-\frac{\left (3 a^3 (2 A b-a B)\right ) \int \frac{\sqrt{x}}{\sqrt{a+b x}} \, dx}{128 b^2}\\ &=-\frac{3 a^3 (2 A b-a B) \sqrt{x} \sqrt{a+b x}}{128 b^3}+\frac{a^2 (2 A b-a B) x^{3/2} \sqrt{a+b x}}{64 b^2}+\frac{a (2 A b-a B) x^{5/2} \sqrt{a+b x}}{16 b}+\frac{(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac{B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac{\left (3 a^4 (2 A b-a B)\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx}{256 b^3}\\ &=-\frac{3 a^3 (2 A b-a B) \sqrt{x} \sqrt{a+b x}}{128 b^3}+\frac{a^2 (2 A b-a B) x^{3/2} \sqrt{a+b x}}{64 b^2}+\frac{a (2 A b-a B) x^{5/2} \sqrt{a+b x}}{16 b}+\frac{(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac{B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac{\left (3 a^4 (2 A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )}{128 b^3}\\ &=-\frac{3 a^3 (2 A b-a B) \sqrt{x} \sqrt{a+b x}}{128 b^3}+\frac{a^2 (2 A b-a B) x^{3/2} \sqrt{a+b x}}{64 b^2}+\frac{a (2 A b-a B) x^{5/2} \sqrt{a+b x}}{16 b}+\frac{(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac{B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac{\left (3 a^4 (2 A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a+b x}}\right )}{128 b^3}\\ &=-\frac{3 a^3 (2 A b-a B) \sqrt{x} \sqrt{a+b x}}{128 b^3}+\frac{a^2 (2 A b-a B) x^{3/2} \sqrt{a+b x}}{64 b^2}+\frac{a (2 A b-a B) x^{5/2} \sqrt{a+b x}}{16 b}+\frac{(2 A b-a B) x^{5/2} (a+b x)^{3/2}}{8 b}+\frac{B x^{5/2} (a+b x)^{5/2}}{5 b}+\frac{3 a^4 (2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{128 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.256709, size = 144, normalized size = 0.75 \[ \frac{\sqrt{a+b x} \left (\sqrt{b} \sqrt{x} \left (4 a^2 b^2 x (5 A+2 B x)-10 a^3 b (3 A+B x)+15 a^4 B+16 a b^3 x^2 (15 A+11 B x)+32 b^4 x^3 (5 A+4 B x)\right )-\frac{15 a^{7/2} (a B-2 A b) \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{\frac{b x}{a}+1}}\right )}{640 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*(a + b*x)^(3/2)*(A + B*x),x]

[Out]

(Sqrt[a + b*x]*(Sqrt[b]*Sqrt[x]*(15*a^4*B - 10*a^3*b*(3*A + B*x) + 4*a^2*b^2*x*(5*A + 2*B*x) + 32*b^4*x^3*(5*A
+ 4*B*x) + 16*a*b^3*x^2*(15*A + 11*B*x)) - (15*a^(7/2)*(-2*A*b + a*B)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/Sqr
t[1 + (b*x)/a]))/(640*b^(7/2))

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Maple [A]  time = 0.01, size = 260, normalized size = 1.4 \begin{align*}{\frac{1}{1280}\sqrt{x}\sqrt{bx+a} \left ( 256\,B{x}^{4}{b}^{9/2}\sqrt{x \left ( bx+a \right ) }+320\,A{x}^{3}{b}^{9/2}\sqrt{x \left ( bx+a \right ) }+352\,B{x}^{3}a{b}^{7/2}\sqrt{x \left ( bx+a \right ) }+480\,A{x}^{2}a{b}^{7/2}\sqrt{x \left ( bx+a \right ) }+16\,B{x}^{2}{a}^{2}{b}^{5/2}\sqrt{x \left ( bx+a \right ) }+40\,A\sqrt{x \left ( bx+a \right ) }{b}^{5/2}x{a}^{2}-20\,B\sqrt{x \left ( bx+a \right ) }{b}^{3/2}x{a}^{3}+30\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{4}b-60\,A\sqrt{x \left ( bx+a \right ) }{b}^{3/2}{a}^{3}-15\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{5}+30\,B\sqrt{x \left ( bx+a \right ) }\sqrt{b}{a}^{4} \right ){b}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{x \left ( bx+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(b*x+a)^(3/2)*(B*x+A),x)

[Out]

1/1280*x^(1/2)*(b*x+a)^(1/2)/b^(7/2)*(256*B*x^4*b^(9/2)*(x*(b*x+a))^(1/2)+320*A*x^3*b^(9/2)*(x*(b*x+a))^(1/2)+
352*B*x^3*a*b^(7/2)*(x*(b*x+a))^(1/2)+480*A*x^2*a*b^(7/2)*(x*(b*x+a))^(1/2)+16*B*x^2*a^2*b^(5/2)*(x*(b*x+a))^(
1/2)+40*A*(x*(b*x+a))^(1/2)*b^(5/2)*x*a^2-20*B*(x*(b*x+a))^(1/2)*b^(3/2)*x*a^3+30*A*ln(1/2*(2*(x*(b*x+a))^(1/2
)*b^(1/2)+2*b*x+a)/b^(1/2))*a^4*b-60*A*(x*(b*x+a))^(1/2)*b^(3/2)*a^3-15*B*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+
2*b*x+a)/b^(1/2))*a^5+30*B*(x*(b*x+a))^(1/2)*b^(1/2)*a^4)/(x*(b*x+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(x^(3/2)*(b*x+a)^(3/2)*(B*x+A),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.77428, size = 706, normalized size = 3.68 \begin{align*} \left [-\frac{15 \,{\left (B a^{5} - 2 \, A a^{4} b\right )} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (128 \, B b^{5} x^{4} + 15 \, B a^{4} b - 30 \, A a^{3} b^{2} + 16 \,{\left (11 \, B a b^{4} + 10 \, A b^{5}\right )} x^{3} + 8 \,{\left (B a^{2} b^{3} + 30 \, A a b^{4}\right )} x^{2} - 10 \,{\left (B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{1280 \, b^{4}}, \frac{15 \,{\left (B a^{5} - 2 \, A a^{4} b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (128 \, B b^{5} x^{4} + 15 \, B a^{4} b - 30 \, A a^{3} b^{2} + 16 \,{\left (11 \, B a b^{4} + 10 \, A b^{5}\right )} x^{3} + 8 \,{\left (B a^{2} b^{3} + 30 \, A a b^{4}\right )} x^{2} - 10 \,{\left (B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{640 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/1280*(15*(B*a^5 - 2*A*a^4*b)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(128*B*b^5*x^4 +
15*B*a^4*b - 30*A*a^3*b^2 + 16*(11*B*a*b^4 + 10*A*b^5)*x^3 + 8*(B*a^2*b^3 + 30*A*a*b^4)*x^2 - 10*(B*a^3*b^2 -
2*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^4, 1/640*(15*(B*a^5 - 2*A*a^4*b)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(
-b)/(b*sqrt(x))) + (128*B*b^5*x^4 + 15*B*a^4*b - 30*A*a^3*b^2 + 16*(11*B*a*b^4 + 10*A*b^5)*x^3 + 8*(B*a^2*b^3
+ 30*A*a*b^4)*x^2 - 10*(B*a^3*b^2 - 2*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^4]

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Sympy [C]  time = 96.8735, size = 1865, normalized size = 9.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(b*x+a)**(3/2)*(B*x+A),x)

[Out]

-2*A*a*Piecewise((a**(5/2)*sqrt(a + b*x)/(16*sqrt(b)*sqrt(b*x/a)) - a**(3/2)*(a + b*x)**(3/2)/(48*sqrt(b)*sqrt
(b*x/a)) - 5*sqrt(a)*(a + b*x)**(5/2)/(24*sqrt(b)*sqrt(b*x/a)) - a**3*acosh(sqrt(a + b*x)/sqrt(a))/(16*sqrt(b)
) + (a + b*x)**(7/2)/(6*sqrt(a)*sqrt(b)*sqrt(b*x/a)), Abs(a + b*x)/Abs(a) > 1), (-I*a**(5/2)*sqrt(a + b*x)/(16
*sqrt(b)*sqrt(-b*x/a)) + I*a**(3/2)*(a + b*x)**(3/2)/(48*sqrt(b)*sqrt(-b*x/a)) + 5*I*sqrt(a)*(a + b*x)**(5/2)/
(24*sqrt(b)*sqrt(-b*x/a)) + I*a**3*asin(sqrt(a + b*x)/sqrt(a))/(16*sqrt(b)) - I*(a + b*x)**(7/2)/(6*sqrt(a)*sq
rt(b)*sqrt(-b*x/a)), True))/b**2 + 2*A*Piecewise((5*a**(7/2)*sqrt(a + b*x)/(128*sqrt(b)*sqrt(b*x/a)) - 5*a**(5
/2)*(a + b*x)**(3/2)/(384*sqrt(b)*sqrt(b*x/a)) - a**(3/2)*(a + b*x)**(5/2)/(192*sqrt(b)*sqrt(b*x/a)) - 7*sqrt(
a)*(a + b*x)**(7/2)/(48*sqrt(b)*sqrt(b*x/a)) - 5*a**4*acosh(sqrt(a + b*x)/sqrt(a))/(128*sqrt(b)) + (a + b*x)**
(9/2)/(8*sqrt(a)*sqrt(b)*sqrt(b*x/a)), Abs(a + b*x)/Abs(a) > 1), (-5*I*a**(7/2)*sqrt(a + b*x)/(128*sqrt(b)*sqr
t(-b*x/a)) + 5*I*a**(5/2)*(a + b*x)**(3/2)/(384*sqrt(b)*sqrt(-b*x/a)) + I*a**(3/2)*(a + b*x)**(5/2)/(192*sqrt(
b)*sqrt(-b*x/a)) + 7*I*sqrt(a)*(a + b*x)**(7/2)/(48*sqrt(b)*sqrt(-b*x/a)) + 5*I*a**4*asin(sqrt(a + b*x)/sqrt(a
))/(128*sqrt(b)) - I*(a + b*x)**(9/2)/(8*sqrt(a)*sqrt(b)*sqrt(-b*x/a)), True))/b**2 + 2*B*a**2*Piecewise((a**(
5/2)*sqrt(a + b*x)/(16*sqrt(b)*sqrt(b*x/a)) - a**(3/2)*(a + b*x)**(3/2)/(48*sqrt(b)*sqrt(b*x/a)) - 5*sqrt(a)*(
a + b*x)**(5/2)/(24*sqrt(b)*sqrt(b*x/a)) - a**3*acosh(sqrt(a + b*x)/sqrt(a))/(16*sqrt(b)) + (a + b*x)**(7/2)/(
6*sqrt(a)*sqrt(b)*sqrt(b*x/a)), Abs(a + b*x)/Abs(a) > 1), (-I*a**(5/2)*sqrt(a + b*x)/(16*sqrt(b)*sqrt(-b*x/a))
+ I*a**(3/2)*(a + b*x)**(3/2)/(48*sqrt(b)*sqrt(-b*x/a)) + 5*I*sqrt(a)*(a + b*x)**(5/2)/(24*sqrt(b)*sqrt(-b*x/
a)) + I*a**3*asin(sqrt(a + b*x)/sqrt(a))/(16*sqrt(b)) - I*(a + b*x)**(7/2)/(6*sqrt(a)*sqrt(b)*sqrt(-b*x/a)), T
rue))/b**3 - 4*B*a*Piecewise((5*a**(7/2)*sqrt(a + b*x)/(128*sqrt(b)*sqrt(b*x/a)) - 5*a**(5/2)*(a + b*x)**(3/2)
/(384*sqrt(b)*sqrt(b*x/a)) - a**(3/2)*(a + b*x)**(5/2)/(192*sqrt(b)*sqrt(b*x/a)) - 7*sqrt(a)*(a + b*x)**(7/2)/
(48*sqrt(b)*sqrt(b*x/a)) - 5*a**4*acosh(sqrt(a + b*x)/sqrt(a))/(128*sqrt(b)) + (a + b*x)**(9/2)/(8*sqrt(a)*sqr
t(b)*sqrt(b*x/a)), Abs(a + b*x)/Abs(a) > 1), (-5*I*a**(7/2)*sqrt(a + b*x)/(128*sqrt(b)*sqrt(-b*x/a)) + 5*I*a**
(5/2)*(a + b*x)**(3/2)/(384*sqrt(b)*sqrt(-b*x/a)) + I*a**(3/2)*(a + b*x)**(5/2)/(192*sqrt(b)*sqrt(-b*x/a)) + 7
*I*sqrt(a)*(a + b*x)**(7/2)/(48*sqrt(b)*sqrt(-b*x/a)) + 5*I*a**4*asin(sqrt(a + b*x)/sqrt(a))/(128*sqrt(b)) - I
*(a + b*x)**(9/2)/(8*sqrt(a)*sqrt(b)*sqrt(-b*x/a)), True))/b**3 + 2*B*Piecewise((7*a**(9/2)*sqrt(a + b*x)/(256
*sqrt(b)*sqrt(b*x/a)) - 7*a**(7/2)*(a + b*x)**(3/2)/(768*sqrt(b)*sqrt(b*x/a)) - 7*a**(5/2)*(a + b*x)**(5/2)/(1
920*sqrt(b)*sqrt(b*x/a)) - a**(3/2)*(a + b*x)**(7/2)/(480*sqrt(b)*sqrt(b*x/a)) - 9*sqrt(a)*(a + b*x)**(9/2)/(8
0*sqrt(b)*sqrt(b*x/a)) - 7*a**5*acosh(sqrt(a + b*x)/sqrt(a))/(256*sqrt(b)) + (a + b*x)**(11/2)/(10*sqrt(a)*sqr
t(b)*sqrt(b*x/a)), Abs(a + b*x)/Abs(a) > 1), (-7*I*a**(9/2)*sqrt(a + b*x)/(256*sqrt(b)*sqrt(-b*x/a)) + 7*I*a**
(7/2)*(a + b*x)**(3/2)/(768*sqrt(b)*sqrt(-b*x/a)) + 7*I*a**(5/2)*(a + b*x)**(5/2)/(1920*sqrt(b)*sqrt(-b*x/a))
+ I*a**(3/2)*(a + b*x)**(7/2)/(480*sqrt(b)*sqrt(-b*x/a)) + 9*I*sqrt(a)*(a + b*x)**(9/2)/(80*sqrt(b)*sqrt(-b*x/
a)) + 7*I*a**5*asin(sqrt(a + b*x)/sqrt(a))/(256*sqrt(b)) - I*(a + b*x)**(11/2)/(10*sqrt(a)*sqrt(b)*sqrt(-b*x/a
)), True))/b**3

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out