3.1728 \(\int \frac {1}{(a+b x)^{9/4} (c+d x)^{5/4}} \, dx\)

Optimal. Leaf size=795 \[ \frac {48 (a+b x)^{3/4} d^2}{5 (b c-a d)^3 \sqrt [4]{c+d x}}-\frac {48 \sqrt {b} \sqrt {(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \sqrt {(a d+b (c+2 d x))^2} d^{3/2}}{5 (b c-a d)^4 \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )}+\frac {24 \sqrt {2} \sqrt [4]{b} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right ) d^{5/4}}{5 (b c-a d)^{3/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}-\frac {12 \sqrt {2} \sqrt [4]{b} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right ),\frac {1}{2}\right ) d^{5/4}}{5 (b c-a d)^{3/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}+\frac {24 d}{5 (b c-a d)^2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}-\frac {4}{5 (b c-a d) (a+b x)^{5/4} \sqrt [4]{c+d x}} \]

[Out]

-4/5/(-a*d+b*c)/(b*x+a)^(5/4)/(d*x+c)^(1/4)+24/5*d/(-a*d+b*c)^2/(b*x+a)^(1/4)/(d*x+c)^(1/4)+48/5*d^2*(b*x+a)^(
3/4)/(-a*d+b*c)^3/(d*x+c)^(1/4)-48/5*d^(3/2)*b^(1/2)*((b*x+a)*(d*x+c))^(1/2)*((2*b*d*x+a*d+b*c)^2)^(1/2)*((a*d
+b*(2*d*x+c))^2)^(1/2)/(-a*d+b*c)^4/(b*x+a)^(1/4)/(d*x+c)^(1/4)/(2*b*d*x+a*d+b*c)/(1+2*b^(1/2)*d^(1/2)*((b*x+a
)*(d*x+c))^(1/2)/(-a*d+b*c))+24/5*b^(1/4)*d^(5/4)*((b*x+a)*(d*x+c))^(1/4)*(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)))*EllipticE(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))*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)/(-a*d+b*c)
^(3/2)/(b*x+a)^(1/4)/(d*x+c)^(1/4)/(2*b*d*x+a*d+b*c)/((a*d+b*(2*d*x+c))^2)^(1/2)-12/5*b^(1/4)*d^(5/4)*((b*x+a)
*(d*x+c))^(1/4)*(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))*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)/(-a*d+b*c)^(3/2)/(b*x+a)^(1/4)/(d*x+c)^(1/4)/(2*b*d*x+a*d+b*c)/((a*d
+b*(2*d*x+c))^2)^(1/2)

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Rubi [A]  time = 0.86, antiderivative size = 795, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {51, 62, 623, 305, 220, 1196} \[ \frac {48 (a+b x)^{3/4} d^2}{5 (b c-a d)^3 \sqrt [4]{c+d x}}-\frac {48 \sqrt {b} \sqrt {(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \sqrt {(a d+b (c+2 d x))^2} d^{3/2}}{5 (b c-a d)^4 \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )}+\frac {24 \sqrt {2} \sqrt [4]{b} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right ) d^{5/4}}{5 (b c-a d)^{3/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}-\frac {12 \sqrt {2} \sqrt [4]{b} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right ) d^{5/4}}{5 (b c-a d)^{3/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}+\frac {24 d}{5 (b c-a d)^2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}-\frac {4}{5 (b c-a d) (a+b x)^{5/4} \sqrt [4]{c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(9/4)*(c + d*x)^(5/4)),x]

[Out]

-4/(5*(b*c - a*d)*(a + b*x)^(5/4)*(c + d*x)^(1/4)) + (24*d)/(5*(b*c - a*d)^2*(a + b*x)^(1/4)*(c + d*x)^(1/4))
+ (48*d^2*(a + b*x)^(3/4))/(5*(b*c - a*d)^3*(c + d*x)^(1/4)) - (48*Sqrt[b]*d^(3/2)*Sqrt[(a + b*x)*(c + d*x)]*S
qrt[(b*c + a*d + 2*b*d*x)^2]*Sqrt[(a*d + b*(c + 2*d*x))^2])/(5*(b*c - a*d)^4*(a + b*x)^(1/4)*(c + d*x)^(1/4)*(
b*c + a*d + 2*b*d*x)*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))) + (24*Sqrt[2]*b^(1/4)*d^
(5/4)*((a + b*x)*(c + d*x))^(1/4)*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)]*EllipticE[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])/(5*(b*c - a*d)^(3/2)*(a + b*x)^(1/4)*(c + d*x)^(1/4)*(b*c + a*d + 2*b*d*x)*Sqrt[(a*d + b*(c + 2*d*x))^2
]) - (12*Sqrt[2]*b^(1/4)*d^(5/4)*((a + b*x)*(c + d*x))^(1/4)*Sqrt[(b*c + a*d + 2*b*d*x)^2]*(1 + (2*Sqrt[b]*Sqr
t[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])/(5*(b*c - a*d)^(3/2)*(a + b*x)^(1/4)*(c + d*x)^(1/4)*(b*c + a*d + 2*b*d*x)*S
qrt[(a*d + b*(c + 2*d*x))^2])

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 62

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 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 305

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

Rule 623

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]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {1}{(a+b x)^{9/4} (c+d x)^{5/4}} \, dx &=-\frac {4}{5 (b c-a d) (a+b x)^{5/4} \sqrt [4]{c+d x}}-\frac {(6 d) \int \frac {1}{(a+b x)^{5/4} (c+d x)^{5/4}} \, dx}{5 (b c-a d)}\\ &=-\frac {4}{5 (b c-a d) (a+b x)^{5/4} \sqrt [4]{c+d x}}+\frac {24 d}{5 (b c-a d)^2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}+\frac {\left (12 d^2\right ) \int \frac {1}{\sqrt [4]{a+b x} (c+d x)^{5/4}} \, dx}{5 (b c-a d)^2}\\ &=-\frac {4}{5 (b c-a d) (a+b x)^{5/4} \sqrt [4]{c+d x}}+\frac {24 d}{5 (b c-a d)^2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}+\frac {48 d^2 (a+b x)^{3/4}}{5 (b c-a d)^3 \sqrt [4]{c+d x}}-\frac {\left (24 b d^2\right ) \int \frac {1}{\sqrt [4]{a+b x} \sqrt [4]{c+d x}} \, dx}{5 (b c-a d)^3}\\ &=-\frac {4}{5 (b c-a d) (a+b x)^{5/4} \sqrt [4]{c+d x}}+\frac {24 d}{5 (b c-a d)^2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}+\frac {48 d^2 (a+b x)^{3/4}}{5 (b c-a d)^3 \sqrt [4]{c+d x}}-\frac {\left (24 b d^2 \sqrt [4]{(a+b x) (c+d x)}\right ) \int \frac {1}{\sqrt [4]{a c+(b c+a d) x+b d x^2}} \, dx}{5 (b c-a d)^3 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}\\ &=-\frac {4}{5 (b c-a d) (a+b x)^{5/4} \sqrt [4]{c+d x}}+\frac {24 d}{5 (b c-a d)^2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}+\frac {48 d^2 (a+b x)^{3/4}}{5 (b c-a d)^3 \sqrt [4]{c+d x}}-\frac {\left (96 b d^2 \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\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 )}{5 (b c-a d)^3 \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}\\ &=-\frac {4}{5 (b c-a d) (a+b x)^{5/4} \sqrt [4]{c+d x}}+\frac {24 d}{5 (b c-a d)^2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}+\frac {48 d^2 (a+b x)^{3/4}}{5 (b c-a d)^3 \sqrt [4]{c+d x}}-\frac {\left (48 \sqrt {b} d^{3/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2}\right ) \operatorname {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 )}{5 (b c-a d)^2 \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}+\frac {\left (48 \sqrt {b} d^{3/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {2 \sqrt {b} \sqrt {d} x^2}{b c-a d}}{\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 )}{5 (b c-a d)^2 \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}\\ &=-\frac {4}{5 (b c-a d) (a+b x)^{5/4} \sqrt [4]{c+d x}}+\frac {24 d}{5 (b c-a d)^2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}+\frac {48 d^2 (a+b x)^{3/4}}{5 (b c-a d)^3 \sqrt [4]{c+d x}}-\frac {48 \sqrt {b} d^{3/2} \sqrt {(a+b x) (c+d x)} \sqrt {(b c+a d+2 b d x)^2} \sqrt {(a d+b (c+2 d x))^2}}{5 (b c-a d)^4 \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )}+\frac {24 \sqrt {2} \sqrt [4]{b} d^{5/4} \sqrt [4]{(a+b x) (c+d x)} \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}} E\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 )}{5 (b c-a d)^{3/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}-\frac {12 \sqrt {2} \sqrt [4]{b} d^{5/4} \sqrt [4]{(a+b x) (c+d x)} \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 )}{5 (b c-a d)^{3/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 73, normalized size = 0.09 \[ -\frac {4 \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4} \, _2F_1\left (-\frac {5}{4},\frac {5}{4};-\frac {1}{4};\frac {d (a+b x)}{a d-b c}\right )}{5 b (a+b x)^{5/4} (c+d x)^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(9/4)*(c + d*x)^(5/4)),x]

[Out]

(-4*((b*(c + d*x))/(b*c - a*d))^(5/4)*Hypergeometric2F1[-5/4, 5/4, -1/4, (d*(a + b*x))/(-(b*c) + a*d)])/(5*b*(
a + b*x)^(5/4)*(c + d*x)^(5/4))

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{b^{3} d^{2} x^{5} + a^{3} c^{2} + {\left (2 \, b^{3} c d + 3 \, a b^{2} d^{2}\right )} x^{4} + {\left (b^{3} c^{2} + 6 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} x^{3} + {\left (3 \, a b^{2} c^{2} + 6 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2} + {\left (3 \, a^{2} b c^{2} + 2 \, a^{3} c d\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(9/4)/(d*x+c)^(5/4),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {9}{4}} {\left (d x + c\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(9/4)/(d*x+c)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(9/4)*(d*x + c)^(5/4)), x)

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b x +a \right )^{\frac {9}{4}} \left (d x +c \right )^{\frac {5}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(9/4)/(d*x+c)^(5/4),x)

[Out]

int(1/(b*x+a)^(9/4)/(d*x+c)^(5/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {9}{4}} {\left (d x + c\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(9/4)/(d*x+c)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(9/4)*(d*x + c)^(5/4)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (a+b\,x\right )}^{9/4}\,{\left (c+d\,x\right )}^{5/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(9/4)*(c + d*x)^(5/4)),x)

[Out]

int(1/((a + b*x)^(9/4)*(c + d*x)^(5/4)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x\right )^{\frac {9}{4}} \left (c + d x\right )^{\frac {5}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(9/4)/(d*x+c)**(5/4),x)

[Out]

Integral(1/((a + b*x)**(9/4)*(c + d*x)**(5/4)), x)

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