3.241 \(\int \frac{A+B x}{x^{7/2} (b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=216 \[ -\frac{35 c^3 \sqrt{x} (8 b B-9 A c)}{64 b^5 \sqrt{b x+c x^2}}-\frac{35 c^2 (8 b B-9 A c)}{192 b^4 \sqrt{x} \sqrt{b x+c x^2}}+\frac{35 c^3 (8 b B-9 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{64 b^{11/2}}+\frac{7 c (8 b B-9 A c)}{96 b^3 x^{3/2} \sqrt{b x+c x^2}}-\frac{8 b B-9 A c}{24 b^2 x^{5/2} \sqrt{b x+c x^2}}-\frac{A}{4 b x^{7/2} \sqrt{b x+c x^2}} \]

[Out]

-A/(4*b*x^(7/2)*Sqrt[b*x + c*x^2]) - (8*b*B - 9*A*c)/(24*b^2*x^(5/2)*Sqrt[b*x + c*x^2]) + (7*c*(8*b*B - 9*A*c)
)/(96*b^3*x^(3/2)*Sqrt[b*x + c*x^2]) - (35*c^2*(8*b*B - 9*A*c))/(192*b^4*Sqrt[x]*Sqrt[b*x + c*x^2]) - (35*c^3*
(8*b*B - 9*A*c)*Sqrt[x])/(64*b^5*Sqrt[b*x + c*x^2]) + (35*c^3*(8*b*B - 9*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[
b]*Sqrt[x])])/(64*b^(11/2))

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Rubi [A]  time = 0.185957, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {792, 672, 666, 660, 207} \[ -\frac{35 c^3 \sqrt{x} (8 b B-9 A c)}{64 b^5 \sqrt{b x+c x^2}}-\frac{35 c^2 (8 b B-9 A c)}{192 b^4 \sqrt{x} \sqrt{b x+c x^2}}+\frac{35 c^3 (8 b B-9 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{64 b^{11/2}}+\frac{7 c (8 b B-9 A c)}{96 b^3 x^{3/2} \sqrt{b x+c x^2}}-\frac{8 b B-9 A c}{24 b^2 x^{5/2} \sqrt{b x+c x^2}}-\frac{A}{4 b x^{7/2} \sqrt{b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(7/2)*(b*x + c*x^2)^(3/2)),x]

[Out]

-A/(4*b*x^(7/2)*Sqrt[b*x + c*x^2]) - (8*b*B - 9*A*c)/(24*b^2*x^(5/2)*Sqrt[b*x + c*x^2]) + (7*c*(8*b*B - 9*A*c)
)/(96*b^3*x^(3/2)*Sqrt[b*x + c*x^2]) - (35*c^2*(8*b*B - 9*A*c))/(192*b^4*Sqrt[x]*Sqrt[b*x + c*x^2]) - (35*c^3*
(8*b*B - 9*A*c)*Sqrt[x])/(64*b^5*Sqrt[b*x + c*x^2]) + (35*c^3*(8*b*B - 9*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[
b]*Sqrt[x])])/(64*b^(11/2))

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 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 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 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}{x^{7/2} \left (b x+c x^2\right )^{3/2}} \, dx &=-\frac{A}{4 b x^{7/2} \sqrt{b x+c x^2}}+\frac{\left (\frac{1}{2} (b B-2 A c)-\frac{7}{2} (-b B+A c)\right ) \int \frac{1}{x^{5/2} \left (b x+c x^2\right )^{3/2}} \, dx}{4 b}\\ &=-\frac{A}{4 b x^{7/2} \sqrt{b x+c x^2}}-\frac{8 b B-9 A c}{24 b^2 x^{5/2} \sqrt{b x+c x^2}}-\frac{(7 c (8 b B-9 A c)) \int \frac{1}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx}{48 b^2}\\ &=-\frac{A}{4 b x^{7/2} \sqrt{b x+c x^2}}-\frac{8 b B-9 A c}{24 b^2 x^{5/2} \sqrt{b x+c x^2}}+\frac{7 c (8 b B-9 A c)}{96 b^3 x^{3/2} \sqrt{b x+c x^2}}+\frac{\left (35 c^2 (8 b B-9 A c)\right ) \int \frac{1}{\sqrt{x} \left (b x+c x^2\right )^{3/2}} \, dx}{192 b^3}\\ &=-\frac{A}{4 b x^{7/2} \sqrt{b x+c x^2}}-\frac{8 b B-9 A c}{24 b^2 x^{5/2} \sqrt{b x+c x^2}}+\frac{7 c (8 b B-9 A c)}{96 b^3 x^{3/2} \sqrt{b x+c x^2}}-\frac{35 c^2 (8 b B-9 A c)}{192 b^4 \sqrt{x} \sqrt{b x+c x^2}}-\frac{\left (35 c^3 (8 b B-9 A c)\right ) \int \frac{\sqrt{x}}{\left (b x+c x^2\right )^{3/2}} \, dx}{128 b^4}\\ &=-\frac{A}{4 b x^{7/2} \sqrt{b x+c x^2}}-\frac{8 b B-9 A c}{24 b^2 x^{5/2} \sqrt{b x+c x^2}}+\frac{7 c (8 b B-9 A c)}{96 b^3 x^{3/2} \sqrt{b x+c x^2}}-\frac{35 c^2 (8 b B-9 A c)}{192 b^4 \sqrt{x} \sqrt{b x+c x^2}}-\frac{35 c^3 (8 b B-9 A c) \sqrt{x}}{64 b^5 \sqrt{b x+c x^2}}-\frac{\left (35 c^3 (8 b B-9 A c)\right ) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx}{128 b^5}\\ &=-\frac{A}{4 b x^{7/2} \sqrt{b x+c x^2}}-\frac{8 b B-9 A c}{24 b^2 x^{5/2} \sqrt{b x+c x^2}}+\frac{7 c (8 b B-9 A c)}{96 b^3 x^{3/2} \sqrt{b x+c x^2}}-\frac{35 c^2 (8 b B-9 A c)}{192 b^4 \sqrt{x} \sqrt{b x+c x^2}}-\frac{35 c^3 (8 b B-9 A c) \sqrt{x}}{64 b^5 \sqrt{b x+c x^2}}-\frac{\left (35 c^3 (8 b B-9 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{-b+x^2} \, dx,x,\frac{\sqrt{b x+c x^2}}{\sqrt{x}}\right )}{64 b^5}\\ &=-\frac{A}{4 b x^{7/2} \sqrt{b x+c x^2}}-\frac{8 b B-9 A c}{24 b^2 x^{5/2} \sqrt{b x+c x^2}}+\frac{7 c (8 b B-9 A c)}{96 b^3 x^{3/2} \sqrt{b x+c x^2}}-\frac{35 c^2 (8 b B-9 A c)}{192 b^4 \sqrt{x} \sqrt{b x+c x^2}}-\frac{35 c^3 (8 b B-9 A c) \sqrt{x}}{64 b^5 \sqrt{b x+c x^2}}+\frac{35 c^3 (8 b B-9 A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{64 b^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0250122, size = 62, normalized size = 0.29 \[ \frac{c^3 x^4 (9 A c-8 b B) \, _2F_1\left (-\frac{1}{2},4;\frac{1}{2};\frac{c x}{b}+1\right )-A b^4}{4 b^5 x^{7/2} \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(7/2)*(b*x + c*x^2)^(3/2)),x]

[Out]

(-(A*b^4) + c^3*(-8*b*B + 9*A*c)*x^4*Hypergeometric2F1[-1/2, 4, 1/2, 1 + (c*x)/b])/(4*b^5*x^(7/2)*Sqrt[x*(b +
c*x)])

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Maple [A]  time = 0.023, size = 174, normalized size = 0.8 \begin{align*} -{\frac{1}{192\,cx+192\,b}\sqrt{x \left ( cx+b \right ) } \left ( 945\,A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) \sqrt{cx+b}{x}^{4}{c}^{4}+64\,B{b}^{9/2}x-112\,B{b}^{7/2}{x}^{2}c+280\,B{b}^{5/2}{x}^{3}{c}^{2}+840\,B{b}^{3/2}{x}^{4}{c}^{3}-840\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ) \sqrt{cx+b}{x}^{4}b{c}^{3}+48\,A{b}^{9/2}-72\,A{b}^{7/2}xc+126\,A{b}^{5/2}{x}^{2}{c}^{2}-315\,A{b}^{3/2}{x}^{3}{c}^{3}-945\,A\sqrt{b}{x}^{4}{c}^{4} \right ){x}^{-{\frac{9}{2}}}{b}^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(7/2)/(c*x^2+b*x)^(3/2),x)

[Out]

-1/192/x^(9/2)*(x*(c*x+b))^(1/2)*(945*A*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x^4*c^4+64*B*b^(9/2)*x-11
2*B*b^(7/2)*x^2*c+280*B*b^(5/2)*x^3*c^2+840*B*b^(3/2)*x^4*c^3-840*B*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/
2)*x^4*b*c^3+48*A*b^(9/2)-72*A*b^(7/2)*x*c+126*A*b^(5/2)*x^2*c^2-315*A*b^(3/2)*x^3*c^3-945*A*b^(1/2)*x^4*c^4)/
(c*x+b)/b^(11/2)

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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{3}{2}} x^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)/((c*x^2 + b*x)^(3/2)*x^(7/2)), x)

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Fricas [A]  time = 1.8804, size = 907, normalized size = 4.2 \begin{align*} \left [-\frac{105 \,{\left ({\left (8 \, B b c^{4} - 9 \, A c^{5}\right )} x^{6} +{\left (8 \, B b^{2} c^{3} - 9 \, A b c^{4}\right )} x^{5}\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 (48 \, A b^{5} + 105 \,{\left (8 \, B b^{2} c^{3} - 9 \, A b c^{4}\right )} x^{4} + 35 \,{\left (8 \, B b^{3} c^{2} - 9 \, A b^{2} c^{3}\right )} x^{3} - 14 \,{\left (8 \, B b^{4} c - 9 \, A b^{3} c^{2}\right )} x^{2} + 8 \,{\left (8 \, B b^{5} - 9 \, A b^{4} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{384 \,{\left (b^{6} c x^{6} + b^{7} x^{5}\right )}}, -\frac{105 \,{\left ({\left (8 \, B b c^{4} - 9 \, A c^{5}\right )} x^{6} +{\left (8 \, B b^{2} c^{3} - 9 \, A b c^{4}\right )} x^{5}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) +{\left (48 \, A b^{5} + 105 \,{\left (8 \, B b^{2} c^{3} - 9 \, A b c^{4}\right )} x^{4} + 35 \,{\left (8 \, B b^{3} c^{2} - 9 \, A b^{2} c^{3}\right )} x^{3} - 14 \,{\left (8 \, B b^{4} c - 9 \, A b^{3} c^{2}\right )} x^{2} + 8 \,{\left (8 \, B b^{5} - 9 \, A b^{4} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{192 \,{\left (b^{6} c x^{6} + b^{7} x^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

[-1/384*(105*((8*B*b*c^4 - 9*A*c^5)*x^6 + (8*B*b^2*c^3 - 9*A*b*c^4)*x^5)*sqrt(b)*log(-(c*x^2 + 2*b*x - 2*sqrt(
c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(48*A*b^5 + 105*(8*B*b^2*c^3 - 9*A*b*c^4)*x^4 + 35*(8*B*b^3*c^2 - 9*A*b
^2*c^3)*x^3 - 14*(8*B*b^4*c - 9*A*b^3*c^2)*x^2 + 8*(8*B*b^5 - 9*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^6*c*
x^6 + b^7*x^5), -1/192*(105*((8*B*b*c^4 - 9*A*c^5)*x^6 + (8*B*b^2*c^3 - 9*A*b*c^4)*x^5)*sqrt(-b)*arctan(sqrt(-
b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (48*A*b^5 + 105*(8*B*b^2*c^3 - 9*A*b*c^4)*x^4 + 35*(8*B*b^3*c^2 - 9*A*b^2*c^3)
*x^3 - 14*(8*B*b^4*c - 9*A*b^3*c^2)*x^2 + 8*(8*B*b^5 - 9*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^6*c*x^6 + b
^7*x^5)]

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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)/x**(7/2)/(c*x**2+b*x)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.33644, size = 266, normalized size = 1.23 \begin{align*} -\frac{35 \,{\left (8 \, B b c^{3} - 9 \, A c^{4}\right )} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{64 \, \sqrt{-b} b^{5}} - \frac{2 \,{\left (B b c^{3} - A c^{4}\right )}}{\sqrt{c x + b} b^{5}} - \frac{456 \,{\left (c x + b\right )}^{\frac{7}{2}} B b c^{3} - 1544 \,{\left (c x + b\right )}^{\frac{5}{2}} B b^{2} c^{3} + 1784 \,{\left (c x + b\right )}^{\frac{3}{2}} B b^{3} c^{3} - 696 \, \sqrt{c x + b} B b^{4} c^{3} - 561 \,{\left (c x + b\right )}^{\frac{7}{2}} A c^{4} + 1929 \,{\left (c x + b\right )}^{\frac{5}{2}} A b c^{4} - 2295 \,{\left (c x + b\right )}^{\frac{3}{2}} A b^{2} c^{4} + 975 \, \sqrt{c x + b} A b^{3} c^{4}}{192 \, b^{5} c^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

-35/64*(8*B*b*c^3 - 9*A*c^4)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^5) - 2*(B*b*c^3 - A*c^4)/(sqrt(c*x + b
)*b^5) - 1/192*(456*(c*x + b)^(7/2)*B*b*c^3 - 1544*(c*x + b)^(5/2)*B*b^2*c^3 + 1784*(c*x + b)^(3/2)*B*b^3*c^3
- 696*sqrt(c*x + b)*B*b^4*c^3 - 561*(c*x + b)^(7/2)*A*c^4 + 1929*(c*x + b)^(5/2)*A*b*c^4 - 2295*(c*x + b)^(3/2
)*A*b^2*c^4 + 975*sqrt(c*x + b)*A*b^3*c^4)/(b^5*c^4*x^4)