\(\int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx\) [1654]

Optimal result

Integrand size = 19, antiderivative size = 144 \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=-\frac {8 (b c-a d) \sqrt {a+b x} \sqrt [4]{c+d x}}{7 d^2}+\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}+\frac {16 (b c-a d)^{9/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{7 \sqrt [4]{b} d^3 \sqrt {a+b x}} \]

[Out]

4/7*(b*x+a)^(3/2)*(d*x+c)^(1/4)/d-8/7*(-a*d+b*c)*(d*x+c)^(1/4)*(b*x+a)^(1/2)/d^2+16/7*(-a*d+b*c)^(9/4)*Ellipti
cF(b^(1/4)*(d*x+c)^(1/4)/(-a*d+b*c)^(1/4),I)*(-d*(b*x+a)/(-a*d+b*c))^(1/2)/b^(1/4)/d^3/(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {52, 65, 230, 227} \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=\frac {16 (b c-a d)^{9/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{7 \sqrt [4]{b} d^3 \sqrt {a+b x}}-\frac {8 \sqrt {a+b x} \sqrt [4]{c+d x} (b c-a d)}{7 d^2}+\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d} \]

[In]

Int[(a + b*x)^(3/2)/(c + d*x)^(3/4),x]

[Out]

(-8*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(1/4))/(7*d^2) + (4*(a + b*x)^(3/2)*(c + d*x)^(1/4))/(7*d) + (16*(b*c
- a*d)^(9/4)*Sqrt[-((d*(a + b*x))/(b*c - a*d))]*EllipticF[ArcSin[(b^(1/4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)],
-1])/(7*b^(1/4)*d^3*Sqrt[a + b*x])

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 65

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 227

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[Rt[-b, 4]*(x/Rt[a, 4])], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 230

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + b*(x^4/a)]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + b*(x^4/
a)], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {(6 (b c-a d)) \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}} \, dx}{7 d} \\ & = -\frac {8 (b c-a d) \sqrt {a+b x} \sqrt [4]{c+d x}}{7 d^2}+\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}+\frac {\left (4 (b c-a d)^2\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx}{7 d^2} \\ & = -\frac {8 (b c-a d) \sqrt {a+b x} \sqrt [4]{c+d x}}{7 d^2}+\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}+\frac {\left (16 (b c-a d)^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a-\frac {b c}{d}+\frac {b x^4}{d}}} \, dx,x,\sqrt [4]{c+d x}\right )}{7 d^3} \\ & = -\frac {8 (b c-a d) \sqrt {a+b x} \sqrt [4]{c+d x}}{7 d^2}+\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}+\frac {\left (16 (b c-a d)^2 \sqrt {\frac {d (a+b x)}{-b c+a d}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {b x^4}{\left (a-\frac {b c}{d}\right ) d}}} \, dx,x,\sqrt [4]{c+d x}\right )}{7 d^3 \sqrt {a+b x}} \\ & = -\frac {8 (b c-a d) \sqrt {a+b x} \sqrt [4]{c+d x}}{7 d^2}+\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}+\frac {16 (b c-a d)^{9/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{7 \sqrt [4]{b} d^3 \sqrt {a+b x}} \\ \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 = 0.51 \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=\frac {2 (a+b x)^{5/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{2},\frac {d (a+b x)}{-b c+a d}\right )}{5 b (c+d x)^{3/4}} \]

[In]

Integrate[(a + b*x)^(3/2)/(c + d*x)^(3/4),x]

[Out]

(2*(a + b*x)^(5/2)*((b*(c + d*x))/(b*c - a*d))^(3/4)*Hypergeometric2F1[3/4, 5/2, 7/2, (d*(a + b*x))/(-(b*c) +
a*d)])/(5*b*(c + d*x)^(3/4))

Maple [F]

\[\int \frac {\left (b x +a \right )^{\frac {3}{2}}}{\left (d x +c \right )^{\frac {3}{4}}}d x\]

[In]

int((b*x+a)^(3/2)/(d*x+c)^(3/4),x)

[Out]

int((b*x+a)^(3/2)/(d*x+c)^(3/4),x)

Fricas [F]

\[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {3}{2}}}{{\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate((b*x+a)^(3/2)/(d*x+c)^(3/4),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=\int \frac {\left (a + b x\right )^{\frac {3}{2}}}{\left (c + d x\right )^{\frac {3}{4}}}\, dx \]

[In]

integrate((b*x+a)**(3/2)/(d*x+c)**(3/4),x)

[Out]

Integral((a + b*x)**(3/2)/(c + d*x)**(3/4), x)

Maxima [F]

\[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {3}{2}}}{{\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate((b*x+a)^(3/2)/(d*x+c)^(3/4),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(3/2)/(d*x + c)^(3/4), x)

Giac [F]

\[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {3}{2}}}{{\left (d x + c\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate((b*x+a)^(3/2)/(d*x+c)^(3/4),x, algorithm="giac")

[Out]

integrate((b*x + a)^(3/2)/(d*x + c)^(3/4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx=\int \frac {{\left (a+b\,x\right )}^{3/2}}{{\left (c+d\,x\right )}^{3/4}} \,d x \]

[In]

int((a + b*x)^(3/2)/(c + d*x)^(3/4),x)

[Out]

int((a + b*x)^(3/2)/(c + d*x)^(3/4), x)