3.2.2 \(\int \frac {x^{13/2}}{(b x+c x^2)^{3/2}} \, dx\) [102]

Optimal. Leaf size=164 \[ \frac {512 b^5 \sqrt {x}}{63 c^6 \sqrt {b x+c x^2}}+\frac {256 b^4 x^{3/2}}{63 c^5 \sqrt {b x+c x^2}}-\frac {64 b^3 x^{5/2}}{63 c^4 \sqrt {b x+c x^2}}+\frac {32 b^2 x^{7/2}}{63 c^3 \sqrt {b x+c x^2}}-\frac {20 b x^{9/2}}{63 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{11/2}}{9 c \sqrt {b x+c x^2}} \]

[Out]

256/63*b^4*x^(3/2)/c^5/(c*x^2+b*x)^(1/2)-64/63*b^3*x^(5/2)/c^4/(c*x^2+b*x)^(1/2)+32/63*b^2*x^(7/2)/c^3/(c*x^2+
b*x)^(1/2)-20/63*b*x^(9/2)/c^2/(c*x^2+b*x)^(1/2)+2/9*x^(11/2)/c/(c*x^2+b*x)^(1/2)+512/63*b^5*x^(1/2)/c^6/(c*x^
2+b*x)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {670, 662} \begin {gather*} \frac {512 b^5 \sqrt {x}}{63 c^6 \sqrt {b x+c x^2}}+\frac {256 b^4 x^{3/2}}{63 c^5 \sqrt {b x+c x^2}}-\frac {64 b^3 x^{5/2}}{63 c^4 \sqrt {b x+c x^2}}+\frac {32 b^2 x^{7/2}}{63 c^3 \sqrt {b x+c x^2}}-\frac {20 b x^{9/2}}{63 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{11/2}}{9 c \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(13/2)/(b*x + c*x^2)^(3/2),x]

[Out]

(512*b^5*Sqrt[x])/(63*c^6*Sqrt[b*x + c*x^2]) + (256*b^4*x^(3/2))/(63*c^5*Sqrt[b*x + c*x^2]) - (64*b^3*x^(5/2))
/(63*c^4*Sqrt[b*x + c*x^2]) + (32*b^2*x^(7/2))/(63*c^3*Sqrt[b*x + c*x^2]) - (20*b*x^(9/2))/(63*c^2*Sqrt[b*x +
c*x^2]) + (2*x^(11/2))/(9*c*Sqrt[b*x + c*x^2])

Rule 662

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

Rule 670

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

Rubi steps

\begin {align*} \int \frac {x^{13/2}}{\left (b x+c x^2\right )^{3/2}} \, dx &=\frac {2 x^{11/2}}{9 c \sqrt {b x+c x^2}}-\frac {(10 b) \int \frac {x^{11/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{9 c}\\ &=-\frac {20 b x^{9/2}}{63 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{11/2}}{9 c \sqrt {b x+c x^2}}+\frac {\left (80 b^2\right ) \int \frac {x^{9/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{63 c^2}\\ &=\frac {32 b^2 x^{7/2}}{63 c^3 \sqrt {b x+c x^2}}-\frac {20 b x^{9/2}}{63 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{11/2}}{9 c \sqrt {b x+c x^2}}-\frac {\left (32 b^3\right ) \int \frac {x^{7/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{21 c^3}\\ &=-\frac {64 b^3 x^{5/2}}{63 c^4 \sqrt {b x+c x^2}}+\frac {32 b^2 x^{7/2}}{63 c^3 \sqrt {b x+c x^2}}-\frac {20 b x^{9/2}}{63 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{11/2}}{9 c \sqrt {b x+c x^2}}+\frac {\left (128 b^4\right ) \int \frac {x^{5/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{63 c^4}\\ &=\frac {256 b^4 x^{3/2}}{63 c^5 \sqrt {b x+c x^2}}-\frac {64 b^3 x^{5/2}}{63 c^4 \sqrt {b x+c x^2}}+\frac {32 b^2 x^{7/2}}{63 c^3 \sqrt {b x+c x^2}}-\frac {20 b x^{9/2}}{63 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{11/2}}{9 c \sqrt {b x+c x^2}}-\frac {\left (256 b^5\right ) \int \frac {x^{3/2}}{\left (b x+c x^2\right )^{3/2}} \, dx}{63 c^5}\\ &=\frac {512 b^5 \sqrt {x}}{63 c^6 \sqrt {b x+c x^2}}+\frac {256 b^4 x^{3/2}}{63 c^5 \sqrt {b x+c x^2}}-\frac {64 b^3 x^{5/2}}{63 c^4 \sqrt {b x+c x^2}}+\frac {32 b^2 x^{7/2}}{63 c^3 \sqrt {b x+c x^2}}-\frac {20 b x^{9/2}}{63 c^2 \sqrt {b x+c x^2}}+\frac {2 x^{11/2}}{9 c \sqrt {b x+c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 75, normalized size = 0.46 \begin {gather*} \frac {2 \sqrt {x} \left (256 b^5+128 b^4 c x-32 b^3 c^2 x^2+16 b^2 c^3 x^3-10 b c^4 x^4+7 c^5 x^5\right )}{63 c^6 \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(13/2)/(b*x + c*x^2)^(3/2),x]

[Out]

(2*Sqrt[x]*(256*b^5 + 128*b^4*c*x - 32*b^3*c^2*x^2 + 16*b^2*c^3*x^3 - 10*b*c^4*x^4 + 7*c^5*x^5))/(63*c^6*Sqrt[
x*(b + c*x)])

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Maple [A]
time = 0.43, size = 77, normalized size = 0.47

method	result	size
gosper	\(\frac {2 \left (c x +b \right ) \left (7 c^{5} x^{5}-10 b \,x^{4} c^{4}+16 b^{2} c^{3} x^{3}-32 b^{3} x^{2} c^{2}+128 b^{4} c x +256 b^{5}\right ) x^{\frac {3}{2}}}{63 c^{6} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}\)	\(77\)
default	\(\frac {2 \sqrt {x \left (c x +b \right )}\, \left (7 c^{5} x^{5}-10 b \,x^{4} c^{4}+16 b^{2} c^{3} x^{3}-32 b^{3} x^{2} c^{2}+128 b^{4} c x +256 b^{5}\right )}{63 \sqrt {x}\, \left (c x +b \right ) c^{6}}\)	\(77\)
risch	\(\frac {2 \left (7 c^{4} x^{4}-17 b \,c^{3} x^{3}+33 b^{2} c^{2} x^{2}-65 b^{3} c x +193 b^{4}\right ) \left (c x +b \right ) \sqrt {x}}{63 c^{6} \sqrt {x \left (c x +b \right )}}+\frac {2 b^{5} \sqrt {x}}{c^{6} \sqrt {x \left (c x +b \right )}}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(13/2)/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/63/x^(1/2)*(x*(c*x+b))^(1/2)*(7*c^5*x^5-10*b*c^4*x^4+16*b^2*c^3*x^3-32*b^3*c^2*x^2+128*b^4*c*x+256*b^5)/(c*x
+b)/c^6

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

2/315*((35*c^6*x^5 - 5*b*c^5*x^4 + 8*b^2*c^4*x^3 - 16*b^3*c^3*x^2 + 64*b^4*c^2*x + 128*b^5*c)*x^5 - 2*(5*b*c^5
*x^5 - 2*b^2*c^4*x^4 + 5*b^3*c^3*x^3 - 28*b^4*c^2*x^2 - 104*b^5*c*x - 64*b^6)*x^4 + 6*(3*b^2*c^4*x^5 - 2*b^3*c
^3*x^4 + 11*b^4*c^2*x^3 + 40*b^5*c*x^2 + 24*b^6*x)*x^3 - 42*(b^3*c^3*x^5 - 2*b^4*c^2*x^4 - 7*b^5*c*x^3 - 4*b^6
*x^2)*x^2 + 210*(b^4*c^2*x^5 + 2*b^5*c*x^4 + b^6*x^3)*x)/((c^7*x^5 + b*c^6*x^4)*sqrt(c*x + b)) - integrate(2*(
b^5*c*x + b^6)*x/((c^7*x^3 + 2*b*c^6*x^2 + b^2*c^5*x)*sqrt(c*x + b)), x)

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Fricas [A]
time = 1.38, size = 84, normalized size = 0.51 \begin {gather*} \frac {2 \, {\left (7 \, c^{5} x^{5} - 10 \, b c^{4} x^{4} + 16 \, b^{2} c^{3} x^{3} - 32 \, b^{3} c^{2} x^{2} + 128 \, b^{4} c x + 256 \, b^{5}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{63 \, {\left (c^{7} x^{2} + b c^{6} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

2/63*(7*c^5*x^5 - 10*b*c^4*x^4 + 16*b^2*c^3*x^3 - 32*b^3*c^2*x^2 + 128*b^4*c*x + 256*b^5)*sqrt(c*x^2 + b*x)*sq
rt(x)/(c^7*x^2 + b*c^6*x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(13/2)/(c*x**2+b*x)**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3655 deep

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Giac [A]
time = 1.38, size = 100, normalized size = 0.61 \begin {gather*} -\frac {512 \, b^{\frac {9}{2}}}{63 \, c^{6}} + \frac {2 \, b^{5}}{\sqrt {c x + b} c^{6}} + \frac {2 \, {\left (7 \, {\left (c x + b\right )}^{\frac {9}{2}} c^{48} - 45 \, {\left (c x + b\right )}^{\frac {7}{2}} b c^{48} + 126 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} c^{48} - 210 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3} c^{48} + 315 \, \sqrt {c x + b} b^{4} c^{48}\right )}}{63 \, c^{54}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

-512/63*b^(9/2)/c^6 + 2*b^5/(sqrt(c*x + b)*c^6) + 2/63*(7*(c*x + b)^(9/2)*c^48 - 45*(c*x + b)^(7/2)*b*c^48 + 1
26*(c*x + b)^(5/2)*b^2*c^48 - 210*(c*x + b)^(3/2)*b^3*c^48 + 315*sqrt(c*x + b)*b^4*c^48)/c^54

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{13/2}}{{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(13/2)/(b*x + c*x^2)^(3/2),x)

[Out]

int(x^(13/2)/(b*x + c*x^2)^(3/2), x)

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