3.213 \(\int \frac{(A+B x) (b x+c x^2)^{3/2}}{x^{15/2}} \, dx\)

Optimal. Leaf size=216 \[ \frac{3 c^3 \sqrt{b x+c x^2} (2 b B-A c)}{128 b^3 x^{3/2}}-\frac{c^2 \sqrt{b x+c x^2} (2 b B-A c)}{64 b^2 x^{5/2}}-\frac{3 c^4 (2 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{128 b^{7/2}}-\frac{c \sqrt{b x+c x^2} (2 b B-A c)}{16 b x^{7/2}}-\frac{\left (b x+c x^2\right )^{3/2} (2 b B-A c)}{8 b x^{11/2}}-\frac{A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}} \]

[Out]

-(c*(2*b*B - A*c)*Sqrt[b*x + c*x^2])/(16*b*x^(7/2)) - (c^2*(2*b*B - A*c)*Sqrt[b*x + c*x^2])/(64*b^2*x^(5/2)) +
(3*c^3*(2*b*B - A*c)*Sqrt[b*x + c*x^2])/(128*b^3*x^(3/2)) - ((2*b*B - A*c)*(b*x + c*x^2)^(3/2))/(8*b*x^(11/2)
) - (A*(b*x + c*x^2)^(5/2))/(5*b*x^(15/2)) - (3*c^4*(2*b*B - A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])]
)/(128*b^(7/2))

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Rubi [A]  time = 0.191331, 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, 662, 672, 660, 207} \[ \frac{3 c^3 \sqrt{b x+c x^2} (2 b B-A c)}{128 b^3 x^{3/2}}-\frac{c^2 \sqrt{b x+c x^2} (2 b B-A c)}{64 b^2 x^{5/2}}-\frac{3 c^4 (2 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{128 b^{7/2}}-\frac{c \sqrt{b x+c x^2} (2 b B-A c)}{16 b x^{7/2}}-\frac{\left (b x+c x^2\right )^{3/2} (2 b B-A c)}{8 b x^{11/2}}-\frac{A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(3/2))/x^(15/2),x]

[Out]

-(c*(2*b*B - A*c)*Sqrt[b*x + c*x^2])/(16*b*x^(7/2)) - (c^2*(2*b*B - A*c)*Sqrt[b*x + c*x^2])/(64*b^2*x^(5/2)) +
(3*c^3*(2*b*B - A*c)*Sqrt[b*x + c*x^2])/(128*b^3*x^(3/2)) - ((2*b*B - A*c)*(b*x + c*x^2)^(3/2))/(8*b*x^(11/2)
) - (A*(b*x + c*x^2)^(5/2))/(5*b*x^(15/2)) - (3*c^4*(2*b*B - A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])]
)/(128*b^(7/2))

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 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(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] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && 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) \left (b x+c x^2\right )^{3/2}}{x^{15/2}} \, dx &=-\frac{A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}}+\frac{\left (-\frac{15}{2} (-b B+A c)+\frac{5}{2} (-b B+2 A c)\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx}{5 b}\\ &=-\frac{(2 b B-A c) \left (b x+c x^2\right )^{3/2}}{8 b x^{11/2}}-\frac{A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}}+\frac{(3 c (2 b B-A c)) \int \frac{\sqrt{b x+c x^2}}{x^{9/2}} \, dx}{16 b}\\ &=-\frac{c (2 b B-A c) \sqrt{b x+c x^2}}{16 b x^{7/2}}-\frac{(2 b B-A c) \left (b x+c x^2\right )^{3/2}}{8 b x^{11/2}}-\frac{A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}}+\frac{\left (c^2 (2 b B-A c)\right ) \int \frac{1}{x^{5/2} \sqrt{b x+c x^2}} \, dx}{32 b}\\ &=-\frac{c (2 b B-A c) \sqrt{b x+c x^2}}{16 b x^{7/2}}-\frac{c^2 (2 b B-A c) \sqrt{b x+c x^2}}{64 b^2 x^{5/2}}-\frac{(2 b B-A c) \left (b x+c x^2\right )^{3/2}}{8 b x^{11/2}}-\frac{A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}}-\frac{\left (3 c^3 (2 b B-A c)\right ) \int \frac{1}{x^{3/2} \sqrt{b x+c x^2}} \, dx}{128 b^2}\\ &=-\frac{c (2 b B-A c) \sqrt{b x+c x^2}}{16 b x^{7/2}}-\frac{c^2 (2 b B-A c) \sqrt{b x+c x^2}}{64 b^2 x^{5/2}}+\frac{3 c^3 (2 b B-A c) \sqrt{b x+c x^2}}{128 b^3 x^{3/2}}-\frac{(2 b B-A c) \left (b x+c x^2\right )^{3/2}}{8 b x^{11/2}}-\frac{A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}}+\frac{\left (3 c^4 (2 b B-A c)\right ) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx}{256 b^3}\\ &=-\frac{c (2 b B-A c) \sqrt{b x+c x^2}}{16 b x^{7/2}}-\frac{c^2 (2 b B-A c) \sqrt{b x+c x^2}}{64 b^2 x^{5/2}}+\frac{3 c^3 (2 b B-A c) \sqrt{b x+c x^2}}{128 b^3 x^{3/2}}-\frac{(2 b B-A c) \left (b x+c x^2\right )^{3/2}}{8 b x^{11/2}}-\frac{A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}}+\frac{\left (3 c^4 (2 b B-A c)\right ) \operatorname{Subst}\left (\int \frac{1}{-b+x^2} \, dx,x,\frac{\sqrt{b x+c x^2}}{\sqrt{x}}\right )}{128 b^3}\\ &=-\frac{c (2 b B-A c) \sqrt{b x+c x^2}}{16 b x^{7/2}}-\frac{c^2 (2 b B-A c) \sqrt{b x+c x^2}}{64 b^2 x^{5/2}}+\frac{3 c^3 (2 b B-A c) \sqrt{b x+c x^2}}{128 b^3 x^{3/2}}-\frac{(2 b B-A c) \left (b x+c x^2\right )^{3/2}}{8 b x^{11/2}}-\frac{A \left (b x+c x^2\right )^{5/2}}{5 b x^{15/2}}-\frac{3 c^4 (2 b B-A c) \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{128 b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0344737, size = 61, normalized size = 0.28 \[ -\frac{(x (b+c x))^{5/2} \left (A b^5+c^4 x^5 (2 b B-A c) \, _2F_1\left (\frac{5}{2},5;\frac{7}{2};\frac{c x}{b}+1\right )\right )}{5 b^6 x^{15/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(3/2))/x^(15/2),x]

[Out]

-((x*(b + c*x))^(5/2)*(A*b^5 + c^4*(2*b*B - A*c)*x^5*Hypergeometric2F1[5/2, 5, 7/2, 1 + (c*x)/b]))/(5*b^6*x^(1
5/2))

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Maple [A]  time = 0.021, size = 223, normalized size = 1. \begin{align*}{\frac{1}{640}\sqrt{x \left ( cx+b \right ) } \left ( 15\,A{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{5}{c}^{5}-30\,B{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{5}b{c}^{4}-15\,A{x}^{4}{c}^{4}\sqrt{b}\sqrt{cx+b}+30\,B{x}^{4}{b}^{3/2}{c}^{3}\sqrt{cx+b}+10\,A{x}^{3}{b}^{3/2}{c}^{3}\sqrt{cx+b}-20\,B{x}^{3}{b}^{5/2}{c}^{2}\sqrt{cx+b}-8\,A{x}^{2}{b}^{5/2}{c}^{2}\sqrt{cx+b}-240\,B{x}^{2}{b}^{7/2}c\sqrt{cx+b}-176\,Ax{b}^{7/2}c\sqrt{cx+b}-160\,Bx{b}^{9/2}\sqrt{cx+b}-128\,A{b}^{9/2}\sqrt{cx+b} \right ){b}^{-{\frac{7}{2}}}{x}^{-{\frac{11}{2}}}{\frac{1}{\sqrt{cx+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(3/2)/x^(15/2),x)

[Out]

1/640*(x*(c*x+b))^(1/2)/b^(7/2)*(15*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^5*c^5-30*B*arctanh((c*x+b)^(1/2)/b^(1/2
))*x^5*b*c^4-15*A*x^4*c^4*b^(1/2)*(c*x+b)^(1/2)+30*B*x^4*b^(3/2)*c^3*(c*x+b)^(1/2)+10*A*x^3*b^(3/2)*c^3*(c*x+b
)^(1/2)-20*B*x^3*b^(5/2)*c^2*(c*x+b)^(1/2)-8*A*x^2*b^(5/2)*c^2*(c*x+b)^(1/2)-240*B*x^2*b^(7/2)*c*(c*x+b)^(1/2)
-176*A*x*b^(7/2)*c*(c*x+b)^(1/2)-160*B*x*b^(9/2)*(c*x+b)^(1/2)-128*A*b^(9/2)*(c*x+b)^(1/2))/x^(11/2)/(c*x+b)^(
1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x\right )}^{\frac{3}{2}}{\left (B x + A\right )}}{x^{\frac{15}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^(15/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(3/2)*(B*x + A)/x^(15/2), x)

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Fricas [A]  time = 1.65089, size = 767, normalized size = 3.55 \begin{align*} \left [-\frac{15 \,{\left (2 \, B b c^{4} - A c^{5}\right )} \sqrt{b} x^{6} \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 (128 \, A b^{5} - 15 \,{\left (2 \, B b^{2} c^{3} - A b c^{4}\right )} x^{4} + 10 \,{\left (2 \, B b^{3} c^{2} - A b^{2} c^{3}\right )} x^{3} + 8 \,{\left (30 \, B b^{4} c + A b^{3} c^{2}\right )} x^{2} + 16 \,{\left (10 \, B b^{5} + 11 \, A b^{4} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{1280 \, b^{4} x^{6}}, \frac{15 \,{\left (2 \, B b c^{4} - A c^{5}\right )} \sqrt{-b} x^{6} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) -{\left (128 \, A b^{5} - 15 \,{\left (2 \, B b^{2} c^{3} - A b c^{4}\right )} x^{4} + 10 \,{\left (2 \, B b^{3} c^{2} - A b^{2} c^{3}\right )} x^{3} + 8 \,{\left (30 \, B b^{4} c + A b^{3} c^{2}\right )} x^{2} + 16 \,{\left (10 \, B b^{5} + 11 \, A b^{4} c\right )} x\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{640 \, b^{4} x^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^(15/2),x, algorithm="fricas")

[Out]

[-1/1280*(15*(2*B*b*c^4 - A*c^5)*sqrt(b)*x^6*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) +
2*(128*A*b^5 - 15*(2*B*b^2*c^3 - A*b*c^4)*x^4 + 10*(2*B*b^3*c^2 - A*b^2*c^3)*x^3 + 8*(30*B*b^4*c + A*b^3*c^2)
*x^2 + 16*(10*B*b^5 + 11*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*x^6), 1/640*(15*(2*B*b*c^4 - A*c^5)*sqrt(
-b)*x^6*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) - (128*A*b^5 - 15*(2*B*b^2*c^3 - A*b*c^4)*x^4 + 10*(2*B*b^3
*c^2 - A*b^2*c^3)*x^3 + 8*(30*B*b^4*c + A*b^3*c^2)*x^2 + 16*(10*B*b^5 + 11*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(
x))/(b^4*x^6)]

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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)**(3/2)/x**(15/2),x)

[Out]

Timed out

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Giac [A]  time = 1.30993, size = 259, normalized size = 1.2 \begin{align*} \frac{\frac{15 \,{\left (2 \, B b c^{5} - A c^{6}\right )} \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{3}} + \frac{30 \,{\left (c x + b\right )}^{\frac{9}{2}} B b c^{5} - 140 \,{\left (c x + b\right )}^{\frac{7}{2}} B b^{2} c^{5} + 140 \,{\left (c x + b\right )}^{\frac{3}{2}} B b^{4} c^{5} - 30 \, \sqrt{c x + b} B b^{5} c^{5} - 15 \,{\left (c x + b\right )}^{\frac{9}{2}} A c^{6} + 70 \,{\left (c x + b\right )}^{\frac{7}{2}} A b c^{6} - 128 \,{\left (c x + b\right )}^{\frac{5}{2}} A b^{2} c^{6} - 70 \,{\left (c x + b\right )}^{\frac{3}{2}} A b^{3} c^{6} + 15 \, \sqrt{c x + b} A b^{4} c^{6}}{b^{3} c^{5} x^{5}}}{640 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(3/2)/x^(15/2),x, algorithm="giac")

[Out]

1/640*(15*(2*B*b*c^5 - A*c^6)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) + (30*(c*x + b)^(9/2)*B*b*c^5 - 14
0*(c*x + b)^(7/2)*B*b^2*c^5 + 140*(c*x + b)^(3/2)*B*b^4*c^5 - 30*sqrt(c*x + b)*B*b^5*c^5 - 15*(c*x + b)^(9/2)*
A*c^6 + 70*(c*x + b)^(7/2)*A*b*c^6 - 128*(c*x + b)^(5/2)*A*b^2*c^6 - 70*(c*x + b)^(3/2)*A*b^3*c^6 + 15*sqrt(c*
x + b)*A*b^4*c^6)/(b^3*c^5*x^5))/c