3.248 \(\int \frac{A+B x}{\sqrt{x} (b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=174 \[ -\frac{5 c \sqrt{x} (4 b B-7 A c)}{4 b^4 \sqrt{b x+c x^2}}-\frac{5 (4 b B-7 A c)}{12 b^3 \sqrt{x} \sqrt{b x+c x^2}}+\frac{\sqrt{x} (4 b B-7 A c)}{6 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{5 c (4 b B-7 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{4 b^{9/2}}-\frac{A}{2 b \sqrt{x} \left (b x+c x^2\right )^{3/2}} \]

[Out]

-A/(2*b*Sqrt[x]*(b*x + c*x^2)^(3/2)) + ((4*b*B - 7*A*c)*Sqrt[x])/(6*b^2*(b*x + c*x^2)^(3/2)) - (5*(4*b*B - 7*A
*c))/(12*b^3*Sqrt[x]*Sqrt[b*x + c*x^2]) - (5*c*(4*b*B - 7*A*c)*Sqrt[x])/(4*b^4*Sqrt[b*x + c*x^2]) + (5*c*(4*b*
B - 7*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(4*b^(9/2))

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Rubi [A]  time = 0.146951, antiderivative size = 174, 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 = {792, 666, 672, 660, 207} \[ -\frac{5 c \sqrt{x} (4 b B-7 A c)}{4 b^4 \sqrt{b x+c x^2}}-\frac{5 (4 b B-7 A c)}{12 b^3 \sqrt{x} \sqrt{b x+c x^2}}+\frac{\sqrt{x} (4 b B-7 A c)}{6 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{5 c (4 b B-7 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{4 b^{9/2}}-\frac{A}{2 b \sqrt{x} \left (b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(2*b*Sqrt[x]*(b*x + c*x^2)^(3/2)) + ((4*b*B - 7*A*c)*Sqrt[x])/(6*b^2*(b*x + c*x^2)^(3/2)) - (5*(4*b*B - 7*A
*c))/(12*b^3*Sqrt[x]*Sqrt[b*x + c*x^2]) - (5*c*(4*b*B - 7*A*c)*Sqrt[x])/(4*b^4*Sqrt[b*x + c*x^2]) + (5*c*(4*b*
B - 7*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(4*b^(9/2))

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rule 666

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((2*c*d - b*e)*(d +
e*x)^m*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*c*d - b*e)*(m + 2*p + 2))/((p + 1)*
(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && LtQ[0, m, 1] && IntegerQ[2*p]

Rule 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*(m + 2*p + 2))/((m + p + 1)*(2*c*d - b*e)), I
nt[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ
[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{A+B x}{\sqrt{x} \left (b x+c x^2\right )^{5/2}} \, dx &=-\frac{A}{2 b \sqrt{x} \left (b x+c x^2\right )^{3/2}}+\frac{\left (\frac{1}{2} (b B-A c)-\frac{3}{2} (-b B+2 A c)\right ) \int \frac{\sqrt{x}}{\left (b x+c x^2\right )^{5/2}} \, dx}{2 b}\\ &=-\frac{A}{2 b \sqrt{x} \left (b x+c x^2\right )^{3/2}}+\frac{(4 b B-7 A c) \sqrt{x}}{6 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{(5 (4 b B-7 A c)) \int \frac{1}{\sqrt{x} \left (b x+c x^2\right )^{3/2}} \, dx}{12 b^2}\\ &=-\frac{A}{2 b \sqrt{x} \left (b x+c x^2\right )^{3/2}}+\frac{(4 b B-7 A c) \sqrt{x}}{6 b^2 \left (b x+c x^2\right )^{3/2}}-\frac{5 (4 b B-7 A c)}{12 b^3 \sqrt{x} \sqrt{b x+c x^2}}-\frac{(5 c (4 b B-7 A c)) \int \frac{\sqrt{x}}{\left (b x+c x^2\right )^{3/2}} \, dx}{8 b^3}\\ &=-\frac{A}{2 b \sqrt{x} \left (b x+c x^2\right )^{3/2}}+\frac{(4 b B-7 A c) \sqrt{x}}{6 b^2 \left (b x+c x^2\right )^{3/2}}-\frac{5 (4 b B-7 A c)}{12 b^3 \sqrt{x} \sqrt{b x+c x^2}}-\frac{5 c (4 b B-7 A c) \sqrt{x}}{4 b^4 \sqrt{b x+c x^2}}-\frac{(5 c (4 b B-7 A c)) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx}{8 b^4}\\ &=-\frac{A}{2 b \sqrt{x} \left (b x+c x^2\right )^{3/2}}+\frac{(4 b B-7 A c) \sqrt{x}}{6 b^2 \left (b x+c x^2\right )^{3/2}}-\frac{5 (4 b B-7 A c)}{12 b^3 \sqrt{x} \sqrt{b x+c x^2}}-\frac{5 c (4 b B-7 A c) \sqrt{x}}{4 b^4 \sqrt{b x+c x^2}}-\frac{(5 c (4 b B-7 A c)) \operatorname{Subst}\left (\int \frac{1}{-b+x^2} \, dx,x,\frac{\sqrt{b x+c x^2}}{\sqrt{x}}\right )}{4 b^4}\\ &=-\frac{A}{2 b \sqrt{x} \left (b x+c x^2\right )^{3/2}}+\frac{(4 b B-7 A c) \sqrt{x}}{6 b^2 \left (b x+c x^2\right )^{3/2}}-\frac{5 (4 b B-7 A c)}{12 b^3 \sqrt{x} \sqrt{b x+c x^2}}-\frac{5 c (4 b B-7 A c) \sqrt{x}}{4 b^4 \sqrt{b x+c x^2}}+\frac{5 c (4 b B-7 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{4 b^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0270372, size = 60, normalized size = 0.34 \[ \frac{c x^2 (7 A c-4 b B) \, _2F_1\left (-\frac{3}{2},2;-\frac{1}{2};\frac{c x}{b}+1\right )-3 A b^2}{6 b^3 \sqrt{x} (x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.025, size = 208, normalized size = 1.2 \begin{align*} -{\frac{1}{12\, \left ( cx+b \right ) ^{2}}\sqrt{x \left ( cx+b \right ) } \left ( 105\,A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) \sqrt{cx+b}{x}^{3}{c}^{3}-60\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) \sqrt{cx+b}{x}^{3}b{c}^{2}+105\,A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{2}b{c}^{2}\sqrt{cx+b}-105\,A\sqrt{b}{x}^{3}{c}^{3}-60\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{2}{b}^{2}c\sqrt{cx+b}+60\,B{b}^{3/2}{x}^{3}{c}^{2}-140\,A{b}^{3/2}{x}^{2}{c}^{2}+80\,B{b}^{5/2}{x}^{2}c-21\,A{b}^{5/2}xc+12\,B{b}^{7/2}x+6\,A{b}^{7/2} \right ){x}^{-{\frac{5}{2}}}{b}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(c*x^2+b*x)^(5/2)/x^(1/2),x)

[Out]

-1/12*(x*(c*x+b))^(1/2)*(105*A*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x^3*c^3-60*B*arctanh((c*x+b)^(1/2)
/b^(1/2))*(c*x+b)^(1/2)*x^3*b*c^2+105*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*b*c^2*(c*x+b)^(1/2)-105*A*b^(1/2)*x
^3*c^3-60*B*arctanh((c*x+b)^(1/2)/b^(1/2))*x^2*b^2*c*(c*x+b)^(1/2)+60*B*b^(3/2)*x^3*c^2-140*A*b^(3/2)*x^2*c^2+
80*B*b^(5/2)*x^2*c-21*A*b^(5/2)*x*c+12*B*b^(7/2)*x+6*A*b^(7/2))/x^(5/2)/(c*x+b)^2/b^(9/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x + A}{{\left (c x^{2} + b x\right )}^{\frac{5}{2}} \sqrt{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x^2+b*x)^(5/2)/x^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)/((c*x^2 + b*x)^(5/2)*sqrt(x)), x)

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Fricas [A]  time = 1.99501, size = 932, normalized size = 5.36 \begin{align*} \left [-\frac{15 \,{\left ({\left (4 \, B b c^{3} - 7 \, A c^{4}\right )} x^{5} + 2 \,{\left (4 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{4} +{\left (4 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{3}\right )} \sqrt{b} \log \left (-\frac{c x^{2} + 2 \, b x - 2 \, \sqrt{c x^{2} + b x} \sqrt{b} \sqrt{x}}{x^{2}}\right ) + 2 \,{\left (6 \, A b^{4} + 15 \,{\left (4 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{3} + 20 \,{\left (4 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{2} + 3 \,{\left (4 \, B b^{4} - 7 \, A b^{3} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{24 \,{\left (b^{5} c^{2} x^{5} + 2 \, b^{6} c x^{4} + b^{7} x^{3}\right )}}, -\frac{15 \,{\left ({\left (4 \, B b c^{3} - 7 \, A c^{4}\right )} x^{5} + 2 \,{\left (4 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{4} +{\left (4 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{3}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) +{\left (6 \, A b^{4} + 15 \,{\left (4 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{3} + 20 \,{\left (4 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{2} + 3 \,{\left (4 \, B b^{4} - 7 \, A b^{3} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{12 \,{\left (b^{5} c^{2} x^{5} + 2 \, b^{6} c x^{4} + b^{7} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x^2+b*x)^(5/2)/x^(1/2),x, algorithm="fricas")

[Out]

[-1/24*(15*((4*B*b*c^3 - 7*A*c^4)*x^5 + 2*(4*B*b^2*c^2 - 7*A*b*c^3)*x^4 + (4*B*b^3*c - 7*A*b^2*c^2)*x^3)*sqrt(
b)*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(6*A*b^4 + 15*(4*B*b^2*c^2 - 7*A*b*c^3)
*x^3 + 20*(4*B*b^3*c - 7*A*b^2*c^2)*x^2 + 3*(4*B*b^4 - 7*A*b^3*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^5*c^2*x^5 +
2*b^6*c*x^4 + b^7*x^3), -1/12*(15*((4*B*b*c^3 - 7*A*c^4)*x^5 + 2*(4*B*b^2*c^2 - 7*A*b*c^3)*x^4 + (4*B*b^3*c -
7*A*b^2*c^2)*x^3)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (6*A*b^4 + 15*(4*B*b^2*c^2 - 7*A*b*c^
3)*x^3 + 20*(4*B*b^3*c - 7*A*b^2*c^2)*x^2 + 3*(4*B*b^4 - 7*A*b^3*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^5*c^2*x^5
+ 2*b^6*c*x^4 + b^7*x^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)/(c*x**2+b*x)**(5/2)/x**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.27845, size = 201, normalized size = 1.16 \begin{align*} -\frac{5 \,{\left (4 \, B b c - 7 \, A c^{2}\right )} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{4 \, \sqrt{-b} b^{4}} - \frac{2 \,{\left (6 \,{\left (c x + b\right )} B b c + B b^{2} c - 9 \,{\left (c x + b\right )} A c^{2} - A b c^{2}\right )}}{3 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{4}} - \frac{4 \,{\left (c x + b\right )}^{\frac{3}{2}} B b c - 4 \, \sqrt{c x + b} B b^{2} c - 11 \,{\left (c x + b\right )}^{\frac{3}{2}} A c^{2} + 13 \, \sqrt{c x + b} A b c^{2}}{4 \, b^{4} c^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x^2+b*x)^(5/2)/x^(1/2),x, algorithm="giac")

[Out]

-5/4*(4*B*b*c - 7*A*c^2)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^4) - 2/3*(6*(c*x + b)*B*b*c + B*b^2*c - 9*
(c*x + b)*A*c^2 - A*b*c^2)/((c*x + b)^(3/2)*b^4) - 1/4*(4*(c*x + b)^(3/2)*B*b*c - 4*sqrt(c*x + b)*B*b^2*c - 11
*(c*x + b)^(3/2)*A*c^2 + 13*sqrt(c*x + b)*A*b*c^2)/(b^4*c^2*x^2)