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

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

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Rubi [A]  time = 0.08, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {662, 672, 660, 207} \begin {gather*} -\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}} \end {gather*}

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 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])

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 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]

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*}

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Mathematica [C]  time = 0.02, size = 42, normalized size = 0.25 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.78, size = 104, normalized size = 0.62 \begin {gather*} \frac {3 c^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{128 b^{7/2}}+\frac {\sqrt {b x+c x^2} \left (-128 b^4-176 b^3 c x-8 b^2 c^2 x^2+10 b c^3 x^3-15 c^4 x^4\right )}{640 b^3 x^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-128*b^4 - 176*b^3*c*x - 8*b^2*c^2*x^2 + 10*b*c^3*x^3 - 15*c^4*x^4))/(640*b^3*x^(11/2)) +
(3*c^5*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*x^2]])/(128*b^(7/2))

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fricas [A]  time = 0.43, size = 218, normalized size = 1.31 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.27, size = 114, normalized size = 0.68 \begin {gather*} -\frac {\frac {15 \, c^{6} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} + \frac {15 \, {\left (c x + b\right )}^{\frac {9}{2}} c^{6} - 70 \, {\left (c x + b\right )}^{\frac {7}{2}} b c^{6} + 128 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} c^{6} + 70 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3} c^{6} - 15 \, \sqrt {c x + b} b^{4} c^{6}}{b^{3} c^{5} x^{5}}}{640 \, c} \end {gather*}

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*(15*c^6*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) + (15*(c*x + b)^(9/2)*c^6 - 70*(c*x + b)^(7/2)*b*
c^6 + 128*(c*x + b)^(5/2)*b^2*c^6 + 70*(c*x + b)^(3/2)*b^3*c^6 - 15*sqrt(c*x + b)*b^4*c^6)/(b^3*c^5*x^5))/c

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maple [A]  time = 0.06, size = 126, normalized size = 0.75 \begin {gather*} \frac {\sqrt {\left (c x +b \right ) x}\, \left (15 c^{5} x^{5} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-15 \sqrt {c x +b}\, \sqrt {b}\, c^{4} x^{4}+10 \sqrt {c x +b}\, b^{\frac {3}{2}} c^{3} x^{3}-8 \sqrt {c x +b}\, b^{\frac {5}{2}} c^{2} x^{2}-176 \sqrt {c x +b}\, b^{\frac {7}{2}} c x -128 \sqrt {c x +b}\, b^{\frac {9}{2}}\right )}{640 \sqrt {c x +b}\, b^{\frac {7}{2}} x^{\frac {11}{2}}} \end {gather*}

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*((c*x+b)*x)^(1/2)/b^(7/2)*(15*arctanh((c*x+b)^(1/2)/b^(1/2))*c^5*x^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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{x^{\frac {15}{2}}}\,{d x} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{x^{15/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(3/2)/x^(15/2),x)

[Out]

int((b*x + c*x^2)^(3/2)/x^(15/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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