### 3.93 $$\int \frac{(b x+c x^2)^{3/2}}{x^{15/2}} \, dx$$

Optimal. Leaf size=167 $-\frac{3 c^4 \sqrt{b x+c x^2}}{128 b^3 x^{3/2}}+\frac{c^3 \sqrt{b x+c x^2}}{64 b^2 x^{5/2}}+\frac{3 c^5 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{128 b^{7/2}}-\frac{c^2 \sqrt{b x+c x^2}}{80 b x^{7/2}}-\frac{3 c \sqrt{b x+c x^2}}{40 x^{9/2}}-\frac{\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}$

[Out]

(-3*c*Sqrt[b*x + c*x^2])/(40*x^(9/2)) - (c^2*Sqrt[b*x + c*x^2])/(80*b*x^(7/2)) + (c^3*Sqrt[b*x + c*x^2])/(64*b
^2*x^(5/2)) - (3*c^4*Sqrt[b*x + c*x^2])/(128*b^3*x^(3/2)) - (b*x + c*x^2)^(3/2)/(5*x^(13/2)) + (3*c^5*ArcTanh[
Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(128*b^(7/2))

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Rubi [A]  time = 0.0833269, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.21, Rules used = {662, 672, 660, 207} $-\frac{3 c^4 \sqrt{b x+c x^2}}{128 b^3 x^{3/2}}+\frac{c^3 \sqrt{b x+c x^2}}{64 b^2 x^{5/2}}+\frac{3 c^5 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{128 b^{7/2}}-\frac{c^2 \sqrt{b x+c x^2}}{80 b x^{7/2}}-\frac{3 c \sqrt{b x+c x^2}}{40 x^{9/2}}-\frac{\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*x + c*x^2)^(3/2)/x^(15/2),x]

[Out]

(-3*c*Sqrt[b*x + c*x^2])/(40*x^(9/2)) - (c^2*Sqrt[b*x + c*x^2])/(80*b*x^(7/2)) + (c^3*Sqrt[b*x + c*x^2])/(64*b
^2*x^(5/2)) - (3*c^4*Sqrt[b*x + c*x^2])/(128*b^3*x^(3/2)) - (b*x + c*x^2)^(3/2)/(5*x^(13/2)) + (3*c^5*ArcTanh[
Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(128*b^(7/2))

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{\left (b x+c x^2\right )^{3/2}}{x^{15/2}} \, dx &=-\frac{\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}+\frac{1}{10} (3 c) \int \frac{\sqrt{b x+c x^2}}{x^{11/2}} \, dx\\ &=-\frac{3 c \sqrt{b x+c x^2}}{40 x^{9/2}}-\frac{\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}+\frac{1}{80} \left (3 c^2\right ) \int \frac{1}{x^{7/2} \sqrt{b x+c x^2}} \, dx\\ &=-\frac{3 c \sqrt{b x+c x^2}}{40 x^{9/2}}-\frac{c^2 \sqrt{b x+c x^2}}{80 b x^{7/2}}-\frac{\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}-\frac{c^3 \int \frac{1}{x^{5/2} \sqrt{b x+c x^2}} \, dx}{32 b}\\ &=-\frac{3 c \sqrt{b x+c x^2}}{40 x^{9/2}}-\frac{c^2 \sqrt{b x+c x^2}}{80 b x^{7/2}}+\frac{c^3 \sqrt{b x+c x^2}}{64 b^2 x^{5/2}}-\frac{\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}+\frac{\left (3 c^4\right ) \int \frac{1}{x^{3/2} \sqrt{b x+c x^2}} \, dx}{128 b^2}\\ &=-\frac{3 c \sqrt{b x+c x^2}}{40 x^{9/2}}-\frac{c^2 \sqrt{b x+c x^2}}{80 b x^{7/2}}+\frac{c^3 \sqrt{b x+c x^2}}{64 b^2 x^{5/2}}-\frac{3 c^4 \sqrt{b x+c x^2}}{128 b^3 x^{3/2}}-\frac{\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}-\frac{\left (3 c^5\right ) \int \frac{1}{\sqrt{x} \sqrt{b x+c x^2}} \, dx}{256 b^3}\\ &=-\frac{3 c \sqrt{b x+c x^2}}{40 x^{9/2}}-\frac{c^2 \sqrt{b x+c x^2}}{80 b x^{7/2}}+\frac{c^3 \sqrt{b x+c x^2}}{64 b^2 x^{5/2}}-\frac{3 c^4 \sqrt{b x+c x^2}}{128 b^3 x^{3/2}}-\frac{\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}-\frac{\left (3 c^5\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{3 c \sqrt{b x+c x^2}}{40 x^{9/2}}-\frac{c^2 \sqrt{b x+c x^2}}{80 b x^{7/2}}+\frac{c^3 \sqrt{b x+c x^2}}{64 b^2 x^{5/2}}-\frac{3 c^4 \sqrt{b x+c x^2}}{128 b^3 x^{3/2}}-\frac{\left (b x+c x^2\right )^{3/2}}{5 x^{13/2}}+\frac{3 c^5 \tanh ^{-1}\left (\frac{\sqrt{b x+c x^2}}{\sqrt{b} \sqrt{x}}\right )}{128 b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0180711, size = 42, normalized size = 0.25 $\frac{2 c^5 (x (b+c x))^{5/2} \, _2F_1\left (\frac{5}{2},6;\frac{7}{2};\frac{c x}{b}+1\right )}{5 b^6 x^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.218, size = 126, normalized size = 0.8 \begin{align*}{\frac{1}{640}\sqrt{x \left ( cx+b \right ) } \left ( 15\,{\it Artanh} \left ({\frac{\sqrt{cx+b}}{\sqrt{b}}} \right ){x}^{5}{c}^{5}-15\,{x}^{4}{c}^{4}\sqrt{b}\sqrt{cx+b}+10\,{x}^{3}{b}^{3/2}{c}^{3}\sqrt{cx+b}-8\,{x}^{2}{b}^{5/2}{c}^{2}\sqrt{cx+b}-176\,x{b}^{7/2}c\sqrt{cx+b}-128\,{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((c*x^2+b*x)^(3/2)/x^(15/2),x)

[Out]

1/640*(x*(c*x+b))^(1/2)/b^(7/2)*(15*arctanh((c*x+b)^(1/2)/b^(1/2))*x^5*c^5-15*x^4*c^4*b^(1/2)*(c*x+b)^(1/2)+10
*x^3*b^(3/2)*c^3*(c*x+b)^(1/2)-8*x^2*b^(5/2)*c^2*(c*x+b)^(1/2)-176*x*b^(7/2)*c*(c*x+b)^(1/2)-128*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}}}{x^{\frac{15}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.1178, size = 529, normalized size = 3.17 \begin{align*} \left [\frac{15 \, \sqrt{b} c^{5} 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 (15 \, b c^{4} x^{4} - 10 \, b^{2} c^{3} x^{3} + 8 \, b^{3} c^{2} x^{2} + 176 \, b^{4} c x + 128 \, b^{5}\right )} \sqrt{c x^{2} + b x} \sqrt{x}}{1280 \, b^{4} x^{6}}, -\frac{15 \, \sqrt{-b} c^{5} x^{6} \arctan \left (\frac{\sqrt{-b} \sqrt{x}}{\sqrt{c x^{2} + b x}}\right ) +{\left (15 \, b c^{4} x^{4} - 10 \, b^{2} c^{3} x^{3} + 8 \, b^{3} c^{2} x^{2} + 176 \, b^{4} c x + 128 \, b^{5}\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((c*x^2+b*x)^(3/2)/x^(15/2),x, algorithm="fricas")

[Out]

[1/1280*(15*sqrt(b)*c^5*x^6*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) - 2*(15*b*c^4*x^4
- 10*b^2*c^3*x^3 + 8*b^3*c^2*x^2 + 176*b^4*c*x + 128*b^5)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*x^6), -1/640*(15*sqr
t(-b)*c^5*x^6*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (15*b*c^4*x^4 - 10*b^2*c^3*x^3 + 8*b^3*c^2*x^2 + 17
6*b^4*c*x + 128*b^5)*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((c*x**2+b*x)**(3/2)/x**(15/2),x)

[Out]

Timed out

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Giac [A]  time = 1.35627, size = 130, normalized size = 0.78 \begin{align*} -\frac{1}{640} \, c^{5}{\left (\frac{15 \, \arctan \left (\frac{\sqrt{c x + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{3}} + \frac{15 \,{\left (c x + b\right )}^{\frac{9}{2}} - 70 \,{\left (c x + b\right )}^{\frac{7}{2}} b + 128 \,{\left (c x + b\right )}^{\frac{5}{2}} b^{2} + 70 \,{\left (c x + b\right )}^{\frac{3}{2}} b^{3} - 15 \, \sqrt{c x + b} b^{4}}{b^{3} c^{5} x^{5}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/640*c^5*(15*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) + (15*(c*x + b)^(9/2) - 70*(c*x + b)^(7/2)*b + 12
8*(c*x + b)^(5/2)*b^2 + 70*(c*x + b)^(3/2)*b^3 - 15*sqrt(c*x + b)*b^4)/(b^3*c^5*x^5))