3.2215 \(\int \frac{(a+b x)^{3/2} (A+B x)}{(d+e x)^{3/2}} \, dx\)

Optimal. Leaf size=202 \[ \frac{(a+b x)^{3/2} \sqrt{d+e x} (-a B e-4 A b e+5 b B d)}{2 e^2 (b d-a e)}-\frac{3 \sqrt{a+b x} \sqrt{d+e x} (-a B e-4 A b e+5 b B d)}{4 e^3}+\frac{3 (b d-a e) (-a B e-4 A b e+5 b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{4 \sqrt{b} e^{7/2}}-\frac{2 (a+b x)^{5/2} (B d-A e)}{e \sqrt{d+e x} (b d-a e)} \]

[Out]

(-2*(B*d - A*e)*(a + b*x)^(5/2))/(e*(b*d - a*e)*Sqrt[d + e*x]) - (3*(5*b*B*d - 4*A*b*e - a*B*e)*Sqrt[a + b*x]*
Sqrt[d + e*x])/(4*e^3) + ((5*b*B*d - 4*A*b*e - a*B*e)*(a + b*x)^(3/2)*Sqrt[d + e*x])/(2*e^2*(b*d - a*e)) + (3*
(b*d - a*e)*(5*b*B*d - 4*A*b*e - a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(4*Sqrt[b]*e
^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.166587, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {78, 50, 63, 217, 206} \[ \frac{(a+b x)^{3/2} \sqrt{d+e x} (-a B e-4 A b e+5 b B d)}{2 e^2 (b d-a e)}-\frac{3 \sqrt{a+b x} \sqrt{d+e x} (-a B e-4 A b e+5 b B d)}{4 e^3}+\frac{3 (b d-a e) (-a B e-4 A b e+5 b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{4 \sqrt{b} e^{7/2}}-\frac{2 (a+b x)^{5/2} (B d-A e)}{e \sqrt{d+e x} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(B*d - A*e)*(a + b*x)^(5/2))/(e*(b*d - a*e)*Sqrt[d + e*x]) - (3*(5*b*B*d - 4*A*b*e - a*B*e)*Sqrt[a + b*x]*
Sqrt[d + e*x])/(4*e^3) + ((5*b*B*d - 4*A*b*e - a*B*e)*(a + b*x)^(3/2)*Sqrt[d + e*x])/(2*e^2*(b*d - a*e)) + (3*
(b*d - a*e)*(5*b*B*d - 4*A*b*e - a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(4*Sqrt[b]*e
^(7/2))

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

Mathematica [A]  time = 0.605539, size = 168, normalized size = 0.83 \[ \frac{\sqrt{e} \sqrt{a+b x} \left (a e (-8 A e+13 B d+5 B e x)+4 A b e (3 d+e x)+b B \left (-15 d^2-5 d e x+2 e^2 x^2\right )\right )+\frac{3 (b d-a e)^{3/2} \sqrt{\frac{b (d+e x)}{b d-a e}} (-a B e-4 A b e+5 b B d) \sinh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b d-a e}}\right )}{b}}{4 e^{7/2} \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e]*Sqrt[a + b*x]*(4*A*b*e*(3*d + e*x) + a*e*(13*B*d - 8*A*e + 5*B*e*x) + b*B*(-15*d^2 - 5*d*e*x + 2*e^2*
x^2)) + (3*(b*d - a*e)^(3/2)*(5*b*B*d - 4*A*b*e - a*B*e)*Sqrt[(b*(d + e*x))/(b*d - a*e)]*ArcSinh[(Sqrt[e]*Sqrt
[a + b*x])/Sqrt[b*d - a*e]])/b)/(4*e^(7/2)*Sqrt[d + e*x])

________________________________________________________________________________________

Maple [B]  time = 0.02, size = 740, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x+a)^(1/2)*(12*A*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x*a*b*e^3-
12*A*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x*b^2*d*e^2+3*B*ln(1/2*(2*b*x
*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x*a^2*e^3-18*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+
d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*x*a*b*d*e^2+15*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(
1/2)+a*e+b*d)/(b*e)^(1/2))*x*b^2*d^2*e+4*B*x^2*b*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+12*A*ln(1/2*(2*b*x*e+
2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b*d*e^2-12*A*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))
^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^2*d^2*e+8*A*x*b*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+3*B*ln(1/2*
(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*d*e^2-18*B*ln(1/2*(2*b*x*e+2*((b*x+a)
*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b*d^2*e+15*B*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*
e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^2*d^3+10*B*x*a*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-10*B*x*b*d*e*((b*x+a)*
(e*x+d))^(1/2)*(b*e)^(1/2)-16*A*a*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+24*A*b*d*e*((b*x+a)*(e*x+d))^(1/2)*(
b*e)^(1/2)+26*B*a*d*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-30*B*b*d^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/((b*
x+a)*(e*x+d))^(1/2)/(b*e)^(1/2)/(e*x+d)^(1/2)/e^3

________________________________________________________________________________________

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+a)^(3/2)*(B*x+A)/(e*x+d)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 4.92089, size = 1281, normalized size = 6.34 \begin{align*} \left [\frac{3 \,{\left (5 \, B b^{2} d^{3} - 2 \,{\left (3 \, B a b + 2 \, A b^{2}\right )} d^{2} e +{\left (B a^{2} + 4 \, A a b\right )} d e^{2} +{\left (5 \, B b^{2} d^{2} e - 2 \,{\left (3 \, B a b + 2 \, A b^{2}\right )} d e^{2} +{\left (B a^{2} + 4 \, A a b\right )} e^{3}\right )} x\right )} \sqrt{b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \,{\left (2 \, b e x + b d + a e\right )} \sqrt{b e} \sqrt{b x + a} \sqrt{e x + d} + 8 \,{\left (b^{2} d e + a b e^{2}\right )} x\right ) + 4 \,{\left (2 \, B b^{2} e^{3} x^{2} - 15 \, B b^{2} d^{2} e - 8 \, A a b e^{3} +{\left (13 \, B a b + 12 \, A b^{2}\right )} d e^{2} -{\left (5 \, B b^{2} d e^{2} -{\left (5 \, B a b + 4 \, A b^{2}\right )} e^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{16 \,{\left (b e^{5} x + b d e^{4}\right )}}, -\frac{3 \,{\left (5 \, B b^{2} d^{3} - 2 \,{\left (3 \, B a b + 2 \, A b^{2}\right )} d^{2} e +{\left (B a^{2} + 4 \, A a b\right )} d e^{2} +{\left (5 \, B b^{2} d^{2} e - 2 \,{\left (3 \, B a b + 2 \, A b^{2}\right )} d e^{2} +{\left (B a^{2} + 4 \, A a b\right )} e^{3}\right )} x\right )} \sqrt{-b e} \arctan \left (\frac{{\left (2 \, b e x + b d + a e\right )} \sqrt{-b e} \sqrt{b x + a} \sqrt{e x + d}}{2 \,{\left (b^{2} e^{2} x^{2} + a b d e +{\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) - 2 \,{\left (2 \, B b^{2} e^{3} x^{2} - 15 \, B b^{2} d^{2} e - 8 \, A a b e^{3} +{\left (13 \, B a b + 12 \, A b^{2}\right )} d e^{2} -{\left (5 \, B b^{2} d e^{2} -{\left (5 \, B a b + 4 \, A b^{2}\right )} e^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{8 \,{\left (b e^{5} x + b d e^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*(5*B*b^2*d^3 - 2*(3*B*a*b + 2*A*b^2)*d^2*e + (B*a^2 + 4*A*a*b)*d*e^2 + (5*B*b^2*d^2*e - 2*(3*B*a*b +
2*A*b^2)*d*e^2 + (B*a^2 + 4*A*a*b)*e^3)*x)*sqrt(b*e)*log(8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*
b*e*x + b*d + a*e)*sqrt(b*e)*sqrt(b*x + a)*sqrt(e*x + d) + 8*(b^2*d*e + a*b*e^2)*x) + 4*(2*B*b^2*e^3*x^2 - 15*
B*b^2*d^2*e - 8*A*a*b*e^3 + (13*B*a*b + 12*A*b^2)*d*e^2 - (5*B*b^2*d*e^2 - (5*B*a*b + 4*A*b^2)*e^3)*x)*sqrt(b*
x + a)*sqrt(e*x + d))/(b*e^5*x + b*d*e^4), -1/8*(3*(5*B*b^2*d^3 - 2*(3*B*a*b + 2*A*b^2)*d^2*e + (B*a^2 + 4*A*a
*b)*d*e^2 + (5*B*b^2*d^2*e - 2*(3*B*a*b + 2*A*b^2)*d*e^2 + (B*a^2 + 4*A*a*b)*e^3)*x)*sqrt(-b*e)*arctan(1/2*(2*
b*e*x + b*d + a*e)*sqrt(-b*e)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*e^2*x^2 + a*b*d*e + (b^2*d*e + a*b*e^2)*x)) - 2
*(2*B*b^2*e^3*x^2 - 15*B*b^2*d^2*e - 8*A*a*b*e^3 + (13*B*a*b + 12*A*b^2)*d*e^2 - (5*B*b^2*d*e^2 - (5*B*a*b + 4
*A*b^2)*e^3)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b*e^5*x + b*d*e^4)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (a + b x\right )^{\frac{3}{2}}}{\left (d + e x\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)*(a + b*x)**(3/2)/(d + e*x)**(3/2), x)

________________________________________________________________________________________

Giac [A]  time = 2.55408, size = 367, normalized size = 1.82 \begin{align*} \frac{{\left ({\left (\frac{2 \,{\left (b x + a\right )} B b{\left | b \right |} e^{4}}{b^{8} d e^{6} - a b^{7} e^{7}} - \frac{5 \, B b^{2} d{\left | b \right |} e^{3} - B a b{\left | b \right |} e^{4} - 4 \, A b^{2}{\left | b \right |} e^{4}}{b^{8} d e^{6} - a b^{7} e^{7}}\right )}{\left (b x + a\right )} - \frac{3 \,{\left (5 \, B b^{3} d^{2}{\left | b \right |} e^{2} - 6 \, B a b^{2} d{\left | b \right |} e^{3} - 4 \, A b^{3} d{\left | b \right |} e^{3} + B a^{2} b{\left | b \right |} e^{4} + 4 \, A a b^{2}{\left | b \right |} e^{4}\right )}}{b^{8} d e^{6} - a b^{7} e^{7}}\right )} \sqrt{b x + a}}{1536 \, \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e}} - \frac{{\left (5 \, B b d{\left | b \right |} - B a{\left | b \right |} e - 4 \, A b{\left | b \right |} e\right )} e^{\left (-\frac{9}{2}\right )} \log \left ({\left | -\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} + \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \right |}\right )}{512 \, b^{\frac{13}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1536*((2*(b*x + a)*B*b*abs(b)*e^4/(b^8*d*e^6 - a*b^7*e^7) - (5*B*b^2*d*abs(b)*e^3 - B*a*b*abs(b)*e^4 - 4*A*b
^2*abs(b)*e^4)/(b^8*d*e^6 - a*b^7*e^7))*(b*x + a) - 3*(5*B*b^3*d^2*abs(b)*e^2 - 6*B*a*b^2*d*abs(b)*e^3 - 4*A*b
^3*d*abs(b)*e^3 + B*a^2*b*abs(b)*e^4 + 4*A*a*b^2*abs(b)*e^4)/(b^8*d*e^6 - a*b^7*e^7))*sqrt(b*x + a)/sqrt(b^2*d
+ (b*x + a)*b*e - a*b*e) - 1/512*(5*B*b*d*abs(b) - B*a*abs(b)*e - 4*A*b*abs(b)*e)*e^(-9/2)*log(abs(-sqrt(b*x
+ a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/b^(13/2)