3.1564 \(\int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/2}} \, dx\)

Optimal. Leaf size=457 \[ -\frac {28 \sqrt {2-\sqrt {3}} d^2 \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\sin ^{-1}\left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),4 \sqrt {3}-7\right )}{135 \sqrt [4]{3} b^{4/3} \sqrt {a+b x} (b c-a d)^2 \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}+\frac {28 d^2 \sqrt [3]{c+d x}}{135 b \sqrt {a+b x} (b c-a d)^2}-\frac {4 d \sqrt [3]{c+d x}}{45 b (a+b x)^{3/2} (b c-a d)}-\frac {2 \sqrt [3]{c+d x}}{5 b (a+b x)^{5/2}} \]

[Out]

-2/5*(d*x+c)^(1/3)/b/(b*x+a)^(5/2)-4/45*d*(d*x+c)^(1/3)/b/(-a*d+b*c)/(b*x+a)^(3/2)+28/135*d^2*(d*x+c)^(1/3)/b/
(-a*d+b*c)^2/(b*x+a)^(1/2)-28/405*d^2*((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3))*EllipticF((-b^(1/3)*(d*x+c)^(1/
3)+(-a*d+b*c)^(1/3)*(1+3^(1/2)))/(-b^(1/3)*(d*x+c)^(1/3)+(-a*d+b*c)^(1/3)*(1-3^(1/2))),2*I-I*3^(1/2))*(((-a*d+
b*c)^(2/3)+b^(1/3)*(-a*d+b*c)^(1/3)*(d*x+c)^(1/3)+b^(2/3)*(d*x+c)^(2/3))/(-b^(1/3)*(d*x+c)^(1/3)+(-a*d+b*c)^(1
/3)*(1-3^(1/2)))^2)^(1/2)*(1/2*6^(1/2)-1/2*2^(1/2))*3^(3/4)/b^(4/3)/(-a*d+b*c)^2/(b*x+a)^(1/2)/(-(-a*d+b*c)^(1
/3)*((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3))/(-b^(1/3)*(d*x+c)^(1/3)+(-a*d+b*c)^(1/3)*(1-3^(1/2)))^2)^(1/2)

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Rubi [A]  time = 0.45, antiderivative size = 457, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {47, 51, 63, 219} \[ -\frac {28 \sqrt {2-\sqrt {3}} d^2 \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right )|-7+4 \sqrt {3}\right )}{135 \sqrt [4]{3} b^{4/3} \sqrt {a+b x} (b c-a d)^2 \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}+\frac {28 d^2 \sqrt [3]{c+d x}}{135 b \sqrt {a+b x} (b c-a d)^2}-\frac {4 d \sqrt [3]{c+d x}}{45 b (a+b x)^{3/2} (b c-a d)}-\frac {2 \sqrt [3]{c+d x}}{5 b (a+b x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(1/3)/(a + b*x)^(7/2),x]

[Out]

(-2*(c + d*x)^(1/3))/(5*b*(a + b*x)^(5/2)) - (4*d*(c + d*x)^(1/3))/(45*b*(b*c - a*d)*(a + b*x)^(3/2)) + (28*d^
2*(c + d*x)^(1/3))/(135*b*(b*c - a*d)^2*Sqrt[a + b*x]) - (28*Sqrt[2 - Sqrt[3]]*d^2*((b*c - a*d)^(1/3) - b^(1/3
)*(c + d*x)^(1/3))*Sqrt[((b*c - a*d)^(2/3) + b^(1/3)*(b*c - a*d)^(1/3)*(c + d*x)^(1/3) + b^(2/3)*(c + d*x)^(2/
3))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3])*(b*c - a*d)
^(1/3) - b^(1/3)*(c + d*x)^(1/3))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))], -7 + 4*Sqrt[3]
])/(135*3^(1/4)*b^(4/3)*(b*c - a*d)^2*Sqrt[a + b*x]*Sqrt[-(((b*c - a*d)^(1/3)*((b*c - a*d)^(1/3) - b^(1/3)*(c
+ d*x)^(1/3)))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2)])

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 219

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 - Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 + Sqrt[3
])*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[-((s*(s + r*x))/((1 - S
qrt[3])*s + r*x)^2)]), x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/2}} \, dx &=-\frac {2 \sqrt [3]{c+d x}}{5 b (a+b x)^{5/2}}+\frac {(2 d) \int \frac {1}{(a+b x)^{5/2} (c+d x)^{2/3}} \, dx}{15 b}\\ &=-\frac {2 \sqrt [3]{c+d x}}{5 b (a+b x)^{5/2}}-\frac {4 d \sqrt [3]{c+d x}}{45 b (b c-a d) (a+b x)^{3/2}}-\frac {\left (14 d^2\right ) \int \frac {1}{(a+b x)^{3/2} (c+d x)^{2/3}} \, dx}{135 b (b c-a d)}\\ &=-\frac {2 \sqrt [3]{c+d x}}{5 b (a+b x)^{5/2}}-\frac {4 d \sqrt [3]{c+d x}}{45 b (b c-a d) (a+b x)^{3/2}}+\frac {28 d^2 \sqrt [3]{c+d x}}{135 b (b c-a d)^2 \sqrt {a+b x}}+\frac {\left (14 d^3\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{2/3}} \, dx}{405 b (b c-a d)^2}\\ &=-\frac {2 \sqrt [3]{c+d x}}{5 b (a+b x)^{5/2}}-\frac {4 d \sqrt [3]{c+d x}}{45 b (b c-a d) (a+b x)^{3/2}}+\frac {28 d^2 \sqrt [3]{c+d x}}{135 b (b c-a d)^2 \sqrt {a+b x}}+\frac {\left (14 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-\frac {b c}{d}+\frac {b x^3}{d}}} \, dx,x,\sqrt [3]{c+d x}\right )}{135 b (b c-a d)^2}\\ &=-\frac {2 \sqrt [3]{c+d x}}{5 b (a+b x)^{5/2}}-\frac {4 d \sqrt [3]{c+d x}}{45 b (b c-a d) (a+b x)^{3/2}}+\frac {28 d^2 \sqrt [3]{c+d x}}{135 b (b c-a d)^2 \sqrt {a+b x}}-\frac {28 \sqrt {2-\sqrt {3}} d^2 \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right )|-7+4 \sqrt {3}\right )}{135 \sqrt [4]{3} b^{4/3} (b c-a d)^2 \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 73, normalized size = 0.16 \[ -\frac {2 \sqrt [3]{c+d x} \, _2F_1\left (-\frac {5}{2},-\frac {1}{3};-\frac {3}{2};\frac {d (a+b x)}{a d-b c}\right )}{5 b (a+b x)^{5/2} \sqrt [3]{\frac {b (c+d x)}{b c-a d}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(1/3)/(a + b*x)^(7/2),x]

[Out]

(-2*(c + d*x)^(1/3)*Hypergeometric2F1[-5/2, -1/3, -3/2, (d*(a + b*x))/(-(b*c) + a*d)])/(5*b*(a + b*x)^(5/2)*((
b*(c + d*x))/(b*c - a*d))^(1/3))

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x + a} {\left (d x + c\right )}^{\frac {1}{3}}}{b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x + a)*(d*x + c)^(1/3)/(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(7/2),x, algorithm="giac")

[Out]

integrate((d*x + c)^(1/3)/(b*x + a)^(7/2), x)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{\frac {1}{3}}}{\left (b x +a \right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(1/3)/(b*x+a)^(7/2),x)

[Out]

int((d*x+c)^(1/3)/(b*x+a)^(7/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(7/2),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(1/3)/(b*x + a)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c+d\,x\right )}^{1/3}}{{\left (a+b\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(1/3)/(a + b*x)^(7/2),x)

[Out]

int((c + d*x)^(1/3)/(a + b*x)^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt [3]{c + d x}}{\left (a + b x\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(1/3)/(b*x+a)**(7/2),x)

[Out]

Integral((c + d*x)**(1/3)/(a + b*x)**(7/2), x)

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