\(\int \frac {1}{(a+b x)^{5/2} \sqrt [5]{c+d x}} \, dx\) [1735]

Optimal result

Integrand size = 19, antiderivative size = 74 \[ \int \frac {1}{(a+b x)^{5/2} \sqrt [5]{c+d x}} \, dx=-\frac {2 \sqrt [5]{\frac {b (c+d x)}{b c-a d}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{5},-\frac {1}{2},-\frac {d (a+b x)}{b c-a d}\right )}{3 b (a+b x)^{3/2} \sqrt [5]{c+d x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {72, 71} \[ \int \frac {1}{(a+b x)^{5/2} \sqrt [5]{c+d x}} \, dx=-\frac {2 \sqrt [5]{\frac {b (c+d x)}{b c-a d}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{5},-\frac {1}{2},-\frac {d (a+b x)}{b c-a d}\right )}{3 b (a+b x)^{3/2} \sqrt [5]{c+d x}} \]

[In]

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

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

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

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [5]{\frac {b (c+d x)}{b c-a d}} \int \frac {1}{(a+b x)^{5/2} \sqrt [5]{\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}}} \, dx}{\sqrt [5]{c+d x}} \\ & = -\frac {2 \sqrt [5]{\frac {b (c+d x)}{b c-a d}} \, _2F_1\left (-\frac {3}{2},\frac {1}{5};-\frac {1}{2};-\frac {d (a+b x)}{b c-a d}\right )}{3 b (a+b x)^{3/2} \sqrt [5]{c+d x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(a+b x)^{5/2} \sqrt [5]{c+d x}} \, dx=-\frac {2 \sqrt [5]{\frac {b (c+d x)}{b c-a d}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{5},-\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b (a+b x)^{3/2} \sqrt [5]{c+d x}} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {1}{\left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {1}{5}}}d x\]

[In]

int(1/(b*x+a)^(5/2)/(d*x+c)^(1/5),x)

[Out]

int(1/(b*x+a)^(5/2)/(d*x+c)^(1/5),x)

Fricas [F]

\[ \int \frac {1}{(a+b x)^{5/2} \sqrt [5]{c+d x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{2}} {\left (d x + c\right )}^{\frac {1}{5}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(5/2)/(d*x+c)^(1/5),x, algorithm="fricas")

[Out]

integral(sqrt(b*x + a)*(d*x + c)^(4/5)/(b^3*d*x^4 + a^3*c + (b^3*c + 3*a*b^2*d)*x^3 + 3*(a*b^2*c + a^2*b*d)*x^
2 + (3*a^2*b*c + a^3*d)*x), x)

Sympy [F]

\[ \int \frac {1}{(a+b x)^{5/2} \sqrt [5]{c+d x}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {5}{2}} \sqrt [5]{c + d x}}\, dx \]

[In]

integrate(1/(b*x+a)**(5/2)/(d*x+c)**(1/5),x)

[Out]

Integral(1/((a + b*x)**(5/2)*(c + d*x)**(1/5)), x)

Maxima [F]

\[ \int \frac {1}{(a+b x)^{5/2} \sqrt [5]{c+d x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{2}} {\left (d x + c\right )}^{\frac {1}{5}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(5/2)/(d*x+c)^(1/5),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(5/2)*(d*x + c)^(1/5)), x)

Giac [F]

\[ \int \frac {1}{(a+b x)^{5/2} \sqrt [5]{c+d x}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{2}} {\left (d x + c\right )}^{\frac {1}{5}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(5/2)/(d*x+c)^(1/5),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(5/2)*(d*x + c)^(1/5)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^{5/2} \sqrt [5]{c+d x}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{1/5}} \,d x \]

[In]

int(1/((a + b*x)^(5/2)*(c + d*x)^(1/5)),x)

[Out]

int(1/((a + b*x)^(5/2)*(c + d*x)^(1/5)), x)