3.406 \(\int \frac{(a+b x)^{3/2} (A+B x)}{x^4} \, dx\)

Optimal. Leaf size=112 \[ \frac{b^2 (A b-6 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 a^{3/2}}+\frac{(a+b x)^{3/2} (A b-6 a B)}{12 a x^2}+\frac{b \sqrt{a+b x} (A b-6 a B)}{8 a x}-\frac{A (a+b x)^{5/2}}{3 a x^3} \]

[Out]

(b*(A*b - 6*a*B)*Sqrt[a + b*x])/(8*a*x) + ((A*b - 6*a*B)*(a + b*x)^(3/2))/(12*a*x^2) - (A*(a + b*x)^(5/2))/(3*
a*x^3) + (b^2*(A*b - 6*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(3/2))

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Rubi [A]  time = 0.0471317, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {78, 47, 63, 208} \[ \frac{b^2 (A b-6 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 a^{3/2}}+\frac{(a+b x)^{3/2} (A b-6 a B)}{12 a x^2}+\frac{b \sqrt{a+b x} (A b-6 a B)}{8 a x}-\frac{A (a+b x)^{5/2}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(3/2)*(A + B*x))/x^4,x]

[Out]

(b*(A*b - 6*a*B)*Sqrt[a + b*x])/(8*a*x) + ((A*b - 6*a*B)*(a + b*x)^(3/2))/(12*a*x^2) - (A*(a + b*x)^(5/2))/(3*
a*x^3) + (b^2*(A*b - 6*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(3/2))

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 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 208

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

Rubi steps

\begin{align*} \int \frac{(a+b x)^{3/2} (A+B x)}{x^4} \, dx &=-\frac{A (a+b x)^{5/2}}{3 a x^3}+\frac{\left (-\frac{A b}{2}+3 a B\right ) \int \frac{(a+b x)^{3/2}}{x^3} \, dx}{3 a}\\ &=\frac{(A b-6 a B) (a+b x)^{3/2}}{12 a x^2}-\frac{A (a+b x)^{5/2}}{3 a x^3}-\frac{(b (A b-6 a B)) \int \frac{\sqrt{a+b x}}{x^2} \, dx}{8 a}\\ &=\frac{b (A b-6 a B) \sqrt{a+b x}}{8 a x}+\frac{(A b-6 a B) (a+b x)^{3/2}}{12 a x^2}-\frac{A (a+b x)^{5/2}}{3 a x^3}-\frac{\left (b^2 (A b-6 a B)\right ) \int \frac{1}{x \sqrt{a+b x}} \, dx}{16 a}\\ &=\frac{b (A b-6 a B) \sqrt{a+b x}}{8 a x}+\frac{(A b-6 a B) (a+b x)^{3/2}}{12 a x^2}-\frac{A (a+b x)^{5/2}}{3 a x^3}-\frac{(b (A b-6 a B)) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )}{8 a}\\ &=\frac{b (A b-6 a B) \sqrt{a+b x}}{8 a x}+\frac{(A b-6 a B) (a+b x)^{3/2}}{12 a x^2}-\frac{A (a+b x)^{5/2}}{3 a x^3}+\frac{b^2 (A b-6 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 a^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0756756, size = 105, normalized size = 0.94 \[ \frac{-(a+b x) \left (4 a^2 (2 A+3 B x)+2 a b x (7 A+15 B x)+3 A b^2 x^2\right )-3 b^2 x^3 \sqrt{\frac{b x}{a}+1} (6 a B-A b) \tanh ^{-1}\left (\sqrt{\frac{b x}{a}+1}\right )}{24 a x^3 \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(3/2)*(A + B*x))/x^4,x]

[Out]

(-((a + b*x)*(3*A*b^2*x^2 + 4*a^2*(2*A + 3*B*x) + 2*a*b*x*(7*A + 15*B*x))) - 3*b^2*(-(A*b) + 6*a*B)*x^3*Sqrt[1
+ (b*x)/a]*ArcTanh[Sqrt[1 + (b*x)/a]])/(24*a*x^3*Sqrt[a + b*x])

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Maple [A]  time = 0.011, size = 96, normalized size = 0.9 \begin{align*} 2\,{b}^{2} \left ({\frac{1}{{b}^{3}{x}^{3}} \left ( -1/16\,{\frac{ \left ( Ab+10\,Ba \right ) \left ( bx+a \right ) ^{5/2}}{a}}+ \left ( -1/6\,Ab+Ba \right ) \left ( bx+a \right ) ^{3/2}+ \left ( -3/8\,B{a}^{2}+1/16\,Aba \right ) \sqrt{bx+a} \right ) }+1/16\,{\frac{Ab-6\,Ba}{{a}^{3/2}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*(B*x+A)/x^4,x)

[Out]

2*b^2*((-1/16*(A*b+10*B*a)/a*(b*x+a)^(5/2)+(-1/6*A*b+B*a)*(b*x+a)^(3/2)+(-3/8*B*a^2+1/16*A*b*a)*(b*x+a)^(1/2))
/b^3/x^3+1/16*(A*b-6*B*a)/a^(3/2)*arctanh((b*x+a)^(1/2)/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+a)^(3/2)*(B*x+A)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.43122, size = 486, normalized size = 4.34 \begin{align*} \left [-\frac{3 \,{\left (6 \, B a b^{2} - A b^{3}\right )} \sqrt{a} x^{3} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (8 \, A a^{3} + 3 \,{\left (10 \, B a^{2} b + A a b^{2}\right )} x^{2} + 2 \,{\left (6 \, B a^{3} + 7 \, A a^{2} b\right )} x\right )} \sqrt{b x + a}}{48 \, a^{2} x^{3}}, \frac{3 \,{\left (6 \, B a b^{2} - A b^{3}\right )} \sqrt{-a} x^{3} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) -{\left (8 \, A a^{3} + 3 \,{\left (10 \, B a^{2} b + A a b^{2}\right )} x^{2} + 2 \,{\left (6 \, B a^{3} + 7 \, A a^{2} b\right )} x\right )} \sqrt{b x + a}}{24 \, a^{2} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^4,x, algorithm="fricas")

[Out]

[-1/48*(3*(6*B*a*b^2 - A*b^3)*sqrt(a)*x^3*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(8*A*a^3 + 3*(10*B*
a^2*b + A*a*b^2)*x^2 + 2*(6*B*a^3 + 7*A*a^2*b)*x)*sqrt(b*x + a))/(a^2*x^3), 1/24*(3*(6*B*a*b^2 - A*b^3)*sqrt(-
a)*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) - (8*A*a^3 + 3*(10*B*a^2*b + A*a*b^2)*x^2 + 2*(6*B*a^3 + 7*A*a^2*b)*x)
*sqrt(b*x + a))/(a^2*x^3)]

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Sympy [B]  time = 65.4678, size = 806, normalized size = 7.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)*(B*x+A)/x**4,x)

[Out]

-66*A*a**4*b**3*sqrt(a + b*x)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) + 80*A*a
**3*b**3*(a + b*x)**(3/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) - 30*A*a**2*
b**3*(a + b*x)**(5/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) - 20*A*a**2*b**3
*sqrt(a + b*x)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) - 5*A*a**2*b**3*sqrt(a**(-7))*log(-a**4*sqrt(a**(
-7)) + sqrt(a + b*x))/16 + 5*A*a**2*b**3*sqrt(a**(-7))*log(a**4*sqrt(a**(-7)) + sqrt(a + b*x))/16 + 12*A*a*b**
3*(a + b*x)**(3/2)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 3*A*a*b**3*sqrt(a**(-5))*log(-a**3*sqrt(a**
(-5)) + sqrt(a + b*x))/4 - 3*A*a*b**3*sqrt(a**(-5))*log(a**3*sqrt(a**(-5)) + sqrt(a + b*x))/4 - A*b**3*sqrt(a*
*(-3))*log(-a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 + A*b**3*sqrt(a**(-3))*log(a**2*sqrt(a**(-3)) + sqrt(a + b*x
))/2 - A*b**2*sqrt(a + b*x)/(a*x) - 10*B*a**3*b**2*sqrt(a + b*x)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2)
+ 6*B*a**2*b**2*(a + b*x)**(3/2)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 3*B*a**2*b**2*sqrt(a**(-5))*
log(-a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - 3*B*a**2*b**2*sqrt(a**(-5))*log(a**3*sqrt(a**(-5)) + sqrt(a + b*x
))/8 - B*a*b**2*sqrt(a**(-3))*log(-a**2*sqrt(a**(-3)) + sqrt(a + b*x)) + B*a*b**2*sqrt(a**(-3))*log(a**2*sqrt(
a**(-3)) + sqrt(a + b*x)) + 2*B*b**2*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) - 2*B*b*sqrt(a + b*x)/x

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Giac [A]  time = 1.2696, size = 196, normalized size = 1.75 \begin{align*} \frac{\frac{3 \,{\left (6 \, B a b^{3} - A b^{4}\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} - \frac{30 \,{\left (b x + a\right )}^{\frac{5}{2}} B a b^{3} - 48 \,{\left (b x + a\right )}^{\frac{3}{2}} B a^{2} b^{3} + 18 \, \sqrt{b x + a} B a^{3} b^{3} + 3 \,{\left (b x + a\right )}^{\frac{5}{2}} A b^{4} + 8 \,{\left (b x + a\right )}^{\frac{3}{2}} A a b^{4} - 3 \, \sqrt{b x + a} A a^{2} b^{4}}{a b^{3} x^{3}}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^4,x, algorithm="giac")

[Out]

1/24*(3*(6*B*a*b^3 - A*b^4)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a) - (30*(b*x + a)^(5/2)*B*a*b^3 - 48*(b*
x + a)^(3/2)*B*a^2*b^3 + 18*sqrt(b*x + a)*B*a^3*b^3 + 3*(b*x + a)^(5/2)*A*b^4 + 8*(b*x + a)^(3/2)*A*a*b^4 - 3*
sqrt(b*x + a)*A*a^2*b^4)/(a*b^3*x^3))/b