\(\int \frac {\sqrt [4]{a+b x}}{(c+d x)^{3/4}} \, dx\) [1713]

Optimal result

Integrand size = 19, antiderivative size = 295 \[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{3/4}} \, dx=\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}-\frac {(b c-a d)^{3/2} ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{\sqrt {2} \sqrt [4]{b} d^{5/4} (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}} \]

[Out]

2*(b*x+a)^(1/4)*(d*x+c)^(1/4)/d-1/2*(-a*d+b*c)^(3/2)*((b*x+a)*(d*x+c))^(3/4)*(cos(2*arctan(b^(1/4)*d^(1/4)*((b
*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*
2^(1/2)/(-a*d+b*c)^(1/2)))*EllipticF(sin(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(
1/2))),1/2*2^(1/2))*(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/2)/(-a*d+b*c))*((2*b*d*x+a*d+b*c)^2)^(1/2)*((a*d
+b*(2*d*x+c))^2/(-a*d+b*c)^2/(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/2)/(-a*d+b*c))^2)^(1/2)/b^(1/4)/d^(5/4)
/(b*x+a)^(3/4)/(d*x+c)^(3/4)/(2*b*d*x+a*d+b*c)*2^(1/2)/((a*d+b*(2*d*x+c))^2)^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {52, 64, 637, 226} \[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{3/4}} \, dx=\frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}-\frac {(b c-a d)^{3/2} ((a+b x) (c+d x))^{3/4} \sqrt {(a d+b c+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right )}{\sqrt {2} \sqrt [4]{b} d^{5/4} (a+b x)^{3/4} (c+d x)^{3/4} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}} \]

[In]

Int[(a + b*x)^(1/4)/(c + d*x)^(3/4),x]

[Out]

(2*(a + b*x)^(1/4)*(c + d*x)^(1/4))/d - ((b*c - a*d)^(3/2)*((a + b*x)*(c + d*x))^(3/4)*Sqrt[(b*c + a*d + 2*b*d
*x)^2]*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))*Sqrt[(a*d + b*(c + 2*d*x))^2/((b*c - a*
d)^2*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))^2)]*EllipticF[2*ArcTan[(Sqrt[2]*b^(1/4)*d
^(1/4)*((a + b*x)*(c + d*x))^(1/4))/Sqrt[b*c - a*d]], 1/2])/(Sqrt[2]*b^(1/4)*d^(5/4)*(a + b*x)^(3/4)*(c + d*x)
^(3/4)*(b*c + a*d + 2*b*d*x)*Sqrt[(a*d + b*(c + 2*d*x))^2])

Rule 52

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

Rule 64

Int[((a_.) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Dist[(a + b*x)^m*((c + d*x)^m/((a + b*x)*
(c + d*x))^m), Int[(a*c + (b*c + a*d)*x + b*d*x^2)^m, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] &&
LtQ[-1, m, 0] && LeQ[3, Denominator[m], 4]

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 637

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{d = Denominator[p]}, Dist[d*(Sqrt[(b + 2*c*x)
^2]/(b + 2*c*x)), Subst[Int[x^(d*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4*c*x^d], x], x, (a + b*x + c*x^2)^(1/d)], x]
/; 3 <= d <= 4] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && RationalQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx}{2 d} \\ & = \frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}-\frac {\left ((b c-a d) ((a+b x) (c+d x))^{3/4}\right ) \int \frac {1}{\left (a c+(b c+a d) x+b d x^2\right )^{3/4}} \, dx}{2 d (a+b x)^{3/4} (c+d x)^{3/4}} \\ & = \frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}-\frac {\left (2 (b c-a d) ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-4 a b c d+(b c+a d)^2+4 b d x^4}} \, dx,x,\sqrt [4]{(a+b x) (c+d x)}\right )}{d (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x)} \\ & = \frac {2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{d}-\frac {(b c-a d)^{3/2} ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{\sqrt {2} \sqrt [4]{b} d^{5/4} (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.25 \[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{3/4}} \, dx=\frac {4 (a+b x)^{5/4} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {9}{4},\frac {d (a+b x)}{-b c+a d}\right )}{5 b (c+d x)^{3/4}} \]

[In]

Integrate[(a + b*x)^(1/4)/(c + d*x)^(3/4),x]

[Out]

(4*(a + b*x)^(5/4)*((b*(c + d*x))/(b*c - a*d))^(3/4)*Hypergeometric2F1[3/4, 5/4, 9/4, (d*(a + b*x))/(-(b*c) +
a*d)])/(5*b*(c + d*x)^(3/4))

Maple [F]

\[\int \frac {\left (b x +a \right )^{\frac {1}{4}}}{\left (d x +c \right )^{\frac {3}{4}}}d x\]

[In]

int((b*x+a)^(1/4)/(d*x+c)^(3/4),x)

[Out]

int((b*x+a)^(1/4)/(d*x+c)^(3/4),x)

Fricas [F]

\[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{3/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{{\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate((b*x+a)^(1/4)/(d*x+c)^(3/4),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{3/4}} \, dx=\int \frac {\sqrt [4]{a + b x}}{\left (c + d x\right )^{\frac {3}{4}}}\, dx \]

[In]

integrate((b*x+a)**(1/4)/(d*x+c)**(3/4),x)

[Out]

Integral((a + b*x)**(1/4)/(c + d*x)**(3/4), x)

Maxima [F]

\[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{3/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{{\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate((b*x+a)^(1/4)/(d*x+c)^(3/4),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(1/4)/(d*x + c)^(3/4), x)

Giac [F]

\[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{3/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{4}}}{{\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate((b*x+a)^(1/4)/(d*x+c)^(3/4),x, algorithm="giac")

[Out]

integrate((b*x + a)^(1/4)/(d*x + c)^(3/4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{3/4}} \, dx=\int \frac {{\left (a+b\,x\right )}^{1/4}}{{\left (c+d\,x\right )}^{3/4}} \,d x \]

[In]

int((a + b*x)^(1/4)/(c + d*x)^(3/4),x)

[Out]

int((a + b*x)^(1/4)/(c + d*x)^(3/4), x)