\(\int \frac {1}{(a+b x)^{5/2} (c+d x)^{7/4}} \, dx\) [1670]

Optimal result

Integrand size = 19, antiderivative size = 178 \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{7/4}} \, dx=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/4}}+\frac {3 d}{(b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/4}}+\frac {5 d^2 \sqrt {a+b x}}{(b c-a d)^3 (c+d x)^{3/4}}+\frac {5 b^{3/4} d \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{(b c-a d)^{11/4} \sqrt {a+b x}} \]

[Out]

-2/3/(-a*d+b*c)/(b*x+a)^(3/2)/(d*x+c)^(3/4)+3*d/(-a*d+b*c)^2/(d*x+c)^(3/4)/(b*x+a)^(1/2)+5*d^2*(b*x+a)^(1/2)/(
-a*d+b*c)^3/(d*x+c)^(3/4)+5*b^(3/4)*d*EllipticF(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)/(-a*d+b*c)^(11/4)/(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {53, 65, 230, 227} \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{7/4}} \, dx=\frac {5 b^{3/4} d \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{\sqrt {a+b x} (b c-a d)^{11/4}}+\frac {5 d^2 \sqrt {a+b x}}{(c+d x)^{3/4} (b c-a d)^3}+\frac {3 d}{\sqrt {a+b x} (c+d x)^{3/4} (b c-a d)^2}-\frac {2}{3 (a+b x)^{3/2} (c+d x)^{3/4} (b c-a d)} \]

[In]

Int[1/((a + b*x)^(5/2)*(c + d*x)^(7/4)),x]

[Out]

-2/(3*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/4)) + (3*d)/((b*c - a*d)^2*Sqrt[a + b*x]*(c + d*x)^(3/4)) + (5*
d^2*Sqrt[a + b*x])/((b*c - a*d)^3*(c + d*x)^(3/4)) + (5*b^(3/4)*d*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])/((b*c - a*d)^(11/4)*Sqrt[a + b*x])

Rule 53

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 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 {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/4}}-\frac {(3 d) \int \frac {1}{(a+b x)^{3/2} (c+d x)^{7/4}} \, dx}{2 (b c-a d)} \\ & = -\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/4}}+\frac {3 d}{(b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/4}}+\frac {\left (15 d^2\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{7/4}} \, dx}{4 (b c-a d)^2} \\ & = -\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/4}}+\frac {3 d}{(b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/4}}+\frac {5 d^2 \sqrt {a+b x}}{(b c-a d)^3 (c+d x)^{3/4}}+\frac {\left (5 b d^2\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx}{4 (b c-a d)^3} \\ & = -\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/4}}+\frac {3 d}{(b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/4}}+\frac {5 d^2 \sqrt {a+b x}}{(b c-a d)^3 (c+d x)^{3/4}}+\frac {(5 b d) \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 )}{(b c-a d)^3} \\ & = -\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/4}}+\frac {3 d}{(b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/4}}+\frac {5 d^2 \sqrt {a+b x}}{(b c-a d)^3 (c+d x)^{3/4}}+\frac {\left (5 b d \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 )}{(b c-a d)^3 \sqrt {a+b x}} \\ & = -\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/4}}+\frac {3 d}{(b c-a d)^2 \sqrt {a+b x} (c+d x)^{3/4}}+\frac {5 d^2 \sqrt {a+b x}}{(b c-a d)^3 (c+d x)^{3/4}}+\frac {5 b^{3/4} d \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 )}{(b c-a d)^{11/4} \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 = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.41 \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{7/4}} \, dx=-\frac {2 \left (\frac {b (c+d x)}{b c-a d}\right )^{7/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {7}{4},-\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b (a+b x)^{3/2} (c+d x)^{7/4}} \]

[In]

Integrate[1/((a + b*x)^(5/2)*(c + d*x)^(7/4)),x]

[Out]

(-2*((b*(c + d*x))/(b*c - a*d))^(7/4)*Hypergeometric2F1[-3/2, 7/4, -1/2, (d*(a + b*x))/(-(b*c) + a*d)])/(3*b*(
a + b*x)^(3/2)*(c + d*x)^(7/4))

Maple [F]

\[\int \frac {1}{\left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {7}{4}}}d x\]

[In]

int(1/(b*x+a)^(5/2)/(d*x+c)^(7/4),x)

[Out]

int(1/(b*x+a)^(5/2)/(d*x+c)^(7/4),x)

Fricas [F]

\[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{7/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{2}} {\left (d x + c\right )}^{\frac {7}{4}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(5/2)/(d*x+c)^(7/4),x, algorithm="fricas")

[Out]

integral(sqrt(b*x + a)*(d*x + c)^(1/4)/(b^3*d^2*x^5 + a^3*c^2 + (2*b^3*c*d + 3*a*b^2*d^2)*x^4 + (b^3*c^2 + 6*a
*b^2*c*d + 3*a^2*b*d^2)*x^3 + (3*a*b^2*c^2 + 6*a^2*b*c*d + a^3*d^2)*x^2 + (3*a^2*b*c^2 + 2*a^3*c*d)*x), x)

Sympy [F]

\[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{7/4}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {7}{4}}}\, dx \]

[In]

integrate(1/(b*x+a)**(5/2)/(d*x+c)**(7/4),x)

[Out]

Integral(1/((a + b*x)**(5/2)*(c + d*x)**(7/4)), x)

Maxima [F]

\[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{7/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{2}} {\left (d x + c\right )}^{\frac {7}{4}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(5/2)/(d*x+c)^(7/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(5/2)*(d*x + c)^(7/4)), x)

Giac [F]

\[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{7/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {5}{2}} {\left (d x + c\right )}^{\frac {7}{4}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(5/2)/(d*x+c)^(7/4),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(5/2)*(d*x + c)^(7/4)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^{5/2} (c+d x)^{7/4}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{7/4}} \,d x \]

[In]

int(1/((a + b*x)^(5/2)*(c + d*x)^(7/4)),x)

[Out]

int(1/((a + b*x)^(5/2)*(c + d*x)^(7/4)), x)