3.2216 \(\int \frac{(a+b x)^{3/2} (A+B x)}{(d+e x)^{5/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.161261, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {78, 47, 50, 63, 217, 206} \[ -\frac{2 (a+b x)^{3/2} (-3 a B e-2 A b e+5 b B d)}{3 e^2 \sqrt{d+e x} (b d-a e)}+\frac{b \sqrt{a+b x} \sqrt{d+e x} (-3 a B e-2 A b e+5 b B d)}{e^3 (b d-a e)}-\frac{\sqrt{b} (-3 a B e-2 A b e+5 b B d) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{e^{7/2}}-\frac{2 (a+b x)^{5/2} (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(3/2)*(A + B*x))/(d + e*x)^(5/2),x]

[Out]

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

Mathematica [C]  time = 0.132795, size = 113, normalized size = 0.56 \[ \frac{2 (a+b x)^{5/2} \left (-\frac{\left (\frac{b (d+e x)}{b d-a e}\right )^{3/2} (-3 a B e-2 A b e+5 b B d) \, _2F_1\left (\frac{3}{2},\frac{5}{2};\frac{7}{2};\frac{e (a+b x)}{a e-b d}\right )}{b}-5 A e+5 B d\right )}{15 e (d+e x)^{3/2} (a e-b d)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(3/2)*(A + B*x))/(d + e*x)^(5/2),x]

[Out]

(2*(a + b*x)^(5/2)*(5*B*d - 5*A*e - ((5*b*B*d - 2*A*b*e - 3*a*B*e)*((b*(d + e*x))/(b*d - a*e))^(3/2)*Hypergeom
etric2F1[3/2, 5/2, 7/2, (e*(a + b*x))/(-(b*d) + a*e)])/b))/(15*e*(-(b*d) + a*e)*(d + e*x)^(3/2))

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Maple [B]  time = 0.022, size = 698, normalized size = 3.5 \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)^(5/2),x)

[Out]

1/6*(b*x+a)^(1/2)*(6*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^2*b^2*e^3
+9*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^2*a*b*e^3-15*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^2*b^2*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))*x*b^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))*x*a*b*d*e^2-30*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+6*B*x^2*b*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+6*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-16*A*x*b*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+9*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^3-12*B*x*a*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1
/2)+40*B*x*b*d*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-4*A*a*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-12*A*b*d*e*
((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-8*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*e)^(1/2)/((b*x+a)*(e*x+d))^(1/2)/e^3/(e*x+d)^(3/2)

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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+a)^(3/2)*(B*x+A)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 8.95025, size = 1200, normalized size = 5.94 \begin{align*} \left [-\frac{3 \,{\left (5 \, B b d^{3} -{\left (3 \, B a + 2 \, A b\right )} d^{2} e +{\left (5 \, B b d e^{2} -{\left (3 \, B a + 2 \, A b\right )} e^{3}\right )} x^{2} + 2 \,{\left (5 \, B b d^{2} e -{\left (3 \, B a + 2 \, A b\right )} d e^{2}\right )} x\right )} \sqrt{\frac{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^{2} x + b d e + a e^{2}\right )} \sqrt{b x + a} \sqrt{e x + d} \sqrt{\frac{b}{e}} + 8 \,{\left (b^{2} d e + a b e^{2}\right )} x\right ) - 4 \,{\left (3 \, B b e^{2} x^{2} + 15 \, B b d^{2} - 2 \, A a e^{2} - 2 \,{\left (2 \, B a + 3 \, A b\right )} d e + 2 \,{\left (10 \, B b d e -{\left (3 \, B a + 4 \, A b\right )} e^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{12 \,{\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}}, \frac{3 \,{\left (5 \, B b d^{3} -{\left (3 \, B a + 2 \, A b\right )} d^{2} e +{\left (5 \, B b d e^{2} -{\left (3 \, B a + 2 \, A b\right )} e^{3}\right )} x^{2} + 2 \,{\left (5 \, B b d^{2} e -{\left (3 \, B a + 2 \, A b\right )} d e^{2}\right )} x\right )} \sqrt{-\frac{b}{e}} \arctan \left (\frac{{\left (2 \, b e x + b d + a e\right )} \sqrt{b x + a} \sqrt{e x + d} \sqrt{-\frac{b}{e}}}{2 \,{\left (b^{2} e x^{2} + a b d +{\left (b^{2} d + a b e\right )} x\right )}}\right ) + 2 \,{\left (3 \, B b e^{2} x^{2} + 15 \, B b d^{2} - 2 \, A a e^{2} - 2 \,{\left (2 \, B a + 3 \, A b\right )} d e + 2 \,{\left (10 \, B b d e -{\left (3 \, B a + 4 \, A b\right )} e^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{e x + d}}{6 \,{\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\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)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(5*B*b*d^3 - (3*B*a + 2*A*b)*d^2*e + (5*B*b*d*e^2 - (3*B*a + 2*A*b)*e^3)*x^2 + 2*(5*B*b*d^2*e - (3*B
*a + 2*A*b)*d*e^2)*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^2*x + b*d*e + a*e
^2)*sqrt(b*x + a)*sqrt(e*x + d)*sqrt(b/e) + 8*(b^2*d*e + a*b*e^2)*x) - 4*(3*B*b*e^2*x^2 + 15*B*b*d^2 - 2*A*a*e
^2 - 2*(2*B*a + 3*A*b)*d*e + 2*(10*B*b*d*e - (3*B*a + 4*A*b)*e^2)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(e^5*x^2 + 2
*d*e^4*x + d^2*e^3), 1/6*(3*(5*B*b*d^3 - (3*B*a + 2*A*b)*d^2*e + (5*B*b*d*e^2 - (3*B*a + 2*A*b)*e^3)*x^2 + 2*(
5*B*b*d^2*e - (3*B*a + 2*A*b)*d*e^2)*x)*sqrt(-b/e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(b*x + a)*sqrt(e*x + d
)*sqrt(-b/e)/(b^2*e*x^2 + a*b*d + (b^2*d + a*b*e)*x)) + 2*(3*B*b*e^2*x^2 + 15*B*b*d^2 - 2*A*a*e^2 - 2*(2*B*a +
3*A*b)*d*e + 2*(10*B*b*d*e - (3*B*a + 4*A*b)*e^2)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(e^5*x^2 + 2*d*e^4*x + d^2*
e^3)]

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Sympy [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((b*x+a)**(3/2)*(B*x+A)/(e*x+d)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.66163, size = 475, normalized size = 2.35 \begin{align*} \frac{{\left (5 \, B b d{\left | b \right |} - 3 \, B a{\left | b \right |} e - 2 \, A b{\left | b \right |} e\right )} e^{\left (-\frac{7}{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 )}{\sqrt{b}} + \frac{{\left ({\left (b x + a\right )}{\left (\frac{3 \,{\left (B b^{5} d{\left | b \right |} e^{4} - B a b^{4}{\left | b \right |} e^{5}\right )}{\left (b x + a\right )}}{b^{4} d e^{5} - a b^{3} e^{6}} + \frac{4 \,{\left (5 \, B b^{6} d^{2}{\left | b \right |} e^{3} - 8 \, B a b^{5} d{\left | b \right |} e^{4} - 2 \, A b^{6} d{\left | b \right |} e^{4} + 3 \, B a^{2} b^{4}{\left | b \right |} e^{5} + 2 \, A a b^{5}{\left | b \right |} e^{5}\right )}}{b^{4} d e^{5} - a b^{3} e^{6}}\right )} + \frac{3 \,{\left (5 \, B b^{7} d^{3}{\left | b \right |} e^{2} - 13 \, B a b^{6} d^{2}{\left | b \right |} e^{3} - 2 \, A b^{7} d^{2}{\left | b \right |} e^{3} + 11 \, B a^{2} b^{5} d{\left | b \right |} e^{4} + 4 \, A a b^{6} d{\left | b \right |} e^{4} - 3 \, B a^{3} b^{4}{\left | b \right |} e^{5} - 2 \, A a^{2} b^{5}{\left | b \right |} e^{5}\right )}}{b^{4} d e^{5} - a b^{3} e^{6}}\right )} \sqrt{b x + a}}{3 \,{\left (b^{2} d +{\left (b x + a\right )} b e - a b e\right )}^{\frac{3}{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)^(5/2),x, algorithm="giac")

[Out]

(5*B*b*d*abs(b) - 3*B*a*abs(b)*e - 2*A*b*abs(b)*e)*e^(-7/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*
d + (b*x + a)*b*e - a*b*e)))/sqrt(b) + 1/3*((b*x + a)*(3*(B*b^5*d*abs(b)*e^4 - B*a*b^4*abs(b)*e^5)*(b*x + a)/(
b^4*d*e^5 - a*b^3*e^6) + 4*(5*B*b^6*d^2*abs(b)*e^3 - 8*B*a*b^5*d*abs(b)*e^4 - 2*A*b^6*d*abs(b)*e^4 + 3*B*a^2*b
^4*abs(b)*e^5 + 2*A*a*b^5*abs(b)*e^5)/(b^4*d*e^5 - a*b^3*e^6)) + 3*(5*B*b^7*d^3*abs(b)*e^2 - 13*B*a*b^6*d^2*ab
s(b)*e^3 - 2*A*b^7*d^2*abs(b)*e^3 + 11*B*a^2*b^5*d*abs(b)*e^4 + 4*A*a*b^6*d*abs(b)*e^4 - 3*B*a^3*b^4*abs(b)*e^
5 - 2*A*a^2*b^5*abs(b)*e^5)/(b^4*d*e^5 - a*b^3*e^6))*sqrt(b*x + a)/(b^2*d + (b*x + a)*b*e - a*b*e)^(3/2)