3.418 \(\int \frac{(a+b x)^{5/2} (A+B x)}{x^4} \, dx\)

Optimal. Leaf size=139 \[ \frac{5 b^2 \sqrt{a+b x} (6 a B+A b)}{8 a}-\frac{5 b^2 (6 a B+A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 \sqrt{a}}-\frac{(a+b x)^{5/2} (6 a B+A b)}{12 a x^2}-\frac{5 b (a+b x)^{3/2} (6 a B+A b)}{24 a x}-\frac{A (a+b x)^{7/2}}{3 a x^3} \]

[Out]

(5*b^2*(A*b + 6*a*B)*Sqrt[a + b*x])/(8*a) - (5*b*(A*b + 6*a*B)*(a + b*x)^(3/2))/(24*a*x) - ((A*b + 6*a*B)*(a +
b*x)^(5/2))/(12*a*x^2) - (A*(a + b*x)^(7/2))/(3*a*x^3) - (5*b^2*(A*b + 6*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])
/(8*Sqrt[a])

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Rubi [A]  time = 0.0581955, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {78, 47, 50, 63, 208} \[ \frac{5 b^2 \sqrt{a+b x} (6 a B+A b)}{8 a}-\frac{5 b^2 (6 a B+A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 \sqrt{a}}-\frac{(a+b x)^{5/2} (6 a B+A b)}{12 a x^2}-\frac{5 b (a+b x)^{3/2} (6 a B+A b)}{24 a x}-\frac{A (a+b x)^{7/2}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/x^4,x]

[Out]

(5*b^2*(A*b + 6*a*B)*Sqrt[a + b*x])/(8*a) - (5*b*(A*b + 6*a*B)*(a + b*x)^(3/2))/(24*a*x) - ((A*b + 6*a*B)*(a +
b*x)^(5/2))/(12*a*x^2) - (A*(a + b*x)^(7/2))/(3*a*x^3) - (5*b^2*(A*b + 6*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])
/(8*Sqrt[a])

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 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)^{5/2} (A+B x)}{x^4} \, dx &=-\frac{A (a+b x)^{7/2}}{3 a x^3}+\frac{\left (\frac{A b}{2}+3 a B\right ) \int \frac{(a+b x)^{5/2}}{x^3} \, dx}{3 a}\\ &=-\frac{(A b+6 a B) (a+b x)^{5/2}}{12 a x^2}-\frac{A (a+b x)^{7/2}}{3 a x^3}+\frac{(5 b (A b+6 a B)) \int \frac{(a+b x)^{3/2}}{x^2} \, dx}{24 a}\\ &=-\frac{5 b (A b+6 a B) (a+b x)^{3/2}}{24 a x}-\frac{(A b+6 a B) (a+b x)^{5/2}}{12 a x^2}-\frac{A (a+b x)^{7/2}}{3 a x^3}+\frac{\left (5 b^2 (A b+6 a B)\right ) \int \frac{\sqrt{a+b x}}{x} \, dx}{16 a}\\ &=\frac{5 b^2 (A b+6 a B) \sqrt{a+b x}}{8 a}-\frac{5 b (A b+6 a B) (a+b x)^{3/2}}{24 a x}-\frac{(A b+6 a B) (a+b x)^{5/2}}{12 a x^2}-\frac{A (a+b x)^{7/2}}{3 a x^3}+\frac{1}{16} \left (5 b^2 (A b+6 a B)\right ) \int \frac{1}{x \sqrt{a+b x}} \, dx\\ &=\frac{5 b^2 (A b+6 a B) \sqrt{a+b x}}{8 a}-\frac{5 b (A b+6 a B) (a+b x)^{3/2}}{24 a x}-\frac{(A b+6 a B) (a+b x)^{5/2}}{12 a x^2}-\frac{A (a+b x)^{7/2}}{3 a x^3}+\frac{1}{8} (5 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 )\\ &=\frac{5 b^2 (A b+6 a B) \sqrt{a+b x}}{8 a}-\frac{5 b (A b+6 a B) (a+b x)^{3/2}}{24 a x}-\frac{(A b+6 a B) (a+b x)^{5/2}}{12 a x^2}-\frac{A (a+b x)^{7/2}}{3 a x^3}-\frac{5 b^2 (A b+6 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{8 \sqrt{a}}\\ \end{align*}

Mathematica [C]  time = 0.0235202, size = 57, normalized size = 0.41 \[ -\frac{(a+b x)^{7/2} \left (7 a^3 A+b^2 x^3 (6 a B+A b) \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};\frac{b x}{a}+1\right )\right )}{21 a^4 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/x^4,x]

[Out]

-((a + b*x)^(7/2)*(7*a^3*A + b^2*(A*b + 6*a*B)*x^3*Hypergeometric2F1[3, 7/2, 9/2, 1 + (b*x)/a]))/(21*a^4*x^3)

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Maple [A]  time = 0.012, size = 108, normalized size = 0.8 \begin{align*} 2\,{b}^{2} \left ( B\sqrt{bx+a}+{\frac{1}{{b}^{3}{x}^{3}} \left ( \left ( -{\frac{11\,Ab}{16}}-{\frac{9\,Ba}{8}} \right ) \left ( bx+a \right ) ^{5/2}+ \left ( 5/6\,Aba+2\,B{a}^{2} \right ) \left ( bx+a \right ) ^{3/2}+ \left ( -{\frac{7\,B{a}^{3}}{8}}-{\frac{5\,Ab{a}^{2}}{16}} \right ) \sqrt{bx+a} \right ) }-{\frac{5\,Ab+30\,Ba}{16\,\sqrt{a}}{\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)^(5/2)*(B*x+A)/x^4,x)

[Out]

2*b^2*(B*(b*x+a)^(1/2)+((-11/16*A*b-9/8*B*a)*(b*x+a)^(5/2)+(5/6*A*b*a+2*B*a^2)*(b*x+a)^(3/2)+(-7/8*B*a^3-5/16*
A*b*a^2)*(b*x+a)^(1/2))/b^3/x^3-5/16*(A*b+6*B*a)/a^(1/2)*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)^(5/2)*(B*x+A)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.41879, size = 539, normalized size = 3.88 \begin{align*} \left [\frac{15 \,{\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 (48 \, B a b^{2} x^{3} - 8 \, A a^{3} - 3 \,{\left (18 \, B a^{2} b + 11 \, A a b^{2}\right )} x^{2} - 2 \,{\left (6 \, B a^{3} + 13 \, A a^{2} b\right )} x\right )} \sqrt{b x + a}}{48 \, a x^{3}}, \frac{15 \,{\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 (48 \, B a b^{2} x^{3} - 8 \, A a^{3} - 3 \,{\left (18 \, B a^{2} b + 11 \, A a b^{2}\right )} x^{2} - 2 \,{\left (6 \, B a^{3} + 13 \, A a^{2} b\right )} x\right )} \sqrt{b x + a}}{24 \, a x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^4,x, algorithm="fricas")

[Out]

[1/48*(15*(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*(48*B*a*b^2*x^3 - 8
*A*a^3 - 3*(18*B*a^2*b + 11*A*a*b^2)*x^2 - 2*(6*B*a^3 + 13*A*a^2*b)*x)*sqrt(b*x + a))/(a*x^3), 1/24*(15*(6*B*a
*b^2 + A*b^3)*sqrt(-a)*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (48*B*a*b^2*x^3 - 8*A*a^3 - 3*(18*B*a^2*b + 11*A
*a*b^2)*x^2 - 2*(6*B*a^3 + 13*A*a^2*b)*x)*sqrt(b*x + a))/(a*x^3)]

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Sympy [B]  time = 87.1026, size = 877, normalized size = 6.31 \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)**(5/2)*(B*x+A)/x**4,x)

[Out]

-66*A*a**5*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
**4*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**3*
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) - 30*A*a**3*b**3
*sqrt(a + b*x)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) - 5*A*a**3*b**3*sqrt(a**(-7))*log(-a**4*sqrt(a**(
-7)) + sqrt(a + b*x))/16 + 5*A*a**3*b**3*sqrt(a**(-7))*log(a**4*sqrt(a**(-7)) + sqrt(a + b*x))/16 + 18*A*a**2*
b**3*(a + b*x)**(3/2)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 9*A*a**2*b**3*sqrt(a**(-5))*log(-a**3*sq
rt(a**(-5)) + sqrt(a + b*x))/8 - 9*A*a**2*b**3*sqrt(a**(-5))*log(a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - 3*A*a
*b**3*sqrt(a**(-3))*log(-a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 + 3*A*a*b**3*sqrt(a**(-3))*log(a**2*sqrt(a**(-3
)) + sqrt(a + b*x))/2 + 2*A*b**3*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) - 3*A*b**2*sqrt(a + b*x)/x - 10*B*a**4*
b**2*sqrt(a + b*x)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 6*B*a**3*b**2*(a + b*x)**(3/2)/(-8*a**4 - 1
6*a**3*b*x + 8*a**2*(a + b*x)**2) + 3*B*a**3*b**2*sqrt(a**(-5))*log(-a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - 3
*B*a**3*b**2*sqrt(a**(-5))*log(a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - 3*B*a**2*b**2*sqrt(a**(-3))*log(-a**2*s
qrt(a**(-3)) + sqrt(a + b*x))/2 + 3*B*a**2*b**2*sqrt(a**(-3))*log(a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 + 6*B*
a*b**2*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) - 3*B*a*b*sqrt(a + b*x)/x + 2*B*b**2*sqrt(a + b*x)

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Giac [A]  time = 1.17461, size = 204, normalized size = 1.47 \begin{align*} \frac{48 \, \sqrt{b x + a} B b^{3} + \frac{15 \,{\left (6 \, B a b^{3} + A b^{4}\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{54 \,{\left (b x + a\right )}^{\frac{5}{2}} B a b^{3} - 96 \,{\left (b x + a\right )}^{\frac{3}{2}} B a^{2} b^{3} + 42 \, \sqrt{b x + a} B a^{3} b^{3} + 33 \,{\left (b x + a\right )}^{\frac{5}{2}} A b^{4} - 40 \,{\left (b x + a\right )}^{\frac{3}{2}} A a b^{4} + 15 \, \sqrt{b x + a} A a^{2} b^{4}}{b^{3} x^{3}}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^4,x, algorithm="giac")

[Out]

1/24*(48*sqrt(b*x + a)*B*b^3 + 15*(6*B*a*b^3 + A*b^4)*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) - (54*(b*x + a)^
(5/2)*B*a*b^3 - 96*(b*x + a)^(3/2)*B*a^2*b^3 + 42*sqrt(b*x + a)*B*a^3*b^3 + 33*(b*x + a)^(5/2)*A*b^4 - 40*(b*x
+ a)^(3/2)*A*a*b^4 + 15*sqrt(b*x + a)*A*a^2*b^4)/(b^3*x^3))/b