3.1727 \(\int \frac{1}{(a+b x)^{5/4} (c+d x)^{5/4}} \, dx\)

Optimal. Leaf size=750 \[ \frac{2 \sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right ),\frac{1}{2}\right )}{\sqrt [4]{a+b x} \sqrt [4]{c+d x} \sqrt{b c-a d} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}-\frac{8 d (a+b x)^{3/4}}{\sqrt [4]{c+d x} (b c-a d)^2}+\frac{8 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \sqrt{(a d+b (c+2 d x))^2}}{\sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c-a d)^3 (a d+b c+2 b d x) \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )}-\frac{4}{\sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c-a d)}-\frac{4 \sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{\sqrt [4]{a+b x} \sqrt [4]{c+d x} \sqrt{b c-a d} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}} \]

[Out]

-4/((b*c - a*d)*(a + b*x)^(1/4)*(c + d*x)^(1/4)) - (8*d*(a + b*x)^(3/4))/((b*c - a*d)^2*(c + d*x)^(1/4)) + (8*
Sqrt[b]*Sqrt[d]*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])/((b*c -
a*d)^3*(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))) - (4*Sqrt[2]*b^(1/4)*d^(1/4)*((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])/(Sqrt[b*c - a*d]*(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]) + (2*Sqrt[2]*b^(1/4)*d^(1/4)*((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)]*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[b*c - a*d]*(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])

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Rubi [A]  time = 0.739789, antiderivative size = 750, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{8 d (a+b x)^{3/4}}{\sqrt [4]{c+d x} (b c-a d)^2}+\frac{8 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \sqrt{(a d+b (c+2 d x))^2}}{\sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c-a d)^3 (a d+b c+2 b d x) \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )}-\frac{4}{\sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c-a d)}+\frac{2 \sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{\sqrt [4]{a+b x} \sqrt [4]{c+d x} \sqrt{b c-a d} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}-\frac{4 \sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{\sqrt [4]{a+b x} \sqrt [4]{c+d x} \sqrt{b c-a d} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-4/((b*c - a*d)*(a + b*x)^(1/4)*(c + d*x)^(1/4)) - (8*d*(a + b*x)^(3/4))/((b*c - a*d)^2*(c + d*x)^(1/4)) + (8*
Sqrt[b]*Sqrt[d]*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])/((b*c -
a*d)^3*(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))) - (4*Sqrt[2]*b^(1/4)*d^(1/4)*((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])/(Sqrt[b*c - a*d]*(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]) + (2*Sqrt[2]*b^(1/4)*d^(1/4)*((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)]*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[b*c - a*d]*(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])

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 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 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 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 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)^{5/4} (c+d x)^{5/4}} \, dx &=-\frac{4}{(b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}-\frac{(2 d) \int \frac{1}{\sqrt [4]{a+b x} (c+d x)^{5/4}} \, dx}{b c-a d}\\ &=-\frac{4}{(b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}-\frac{8 d (a+b x)^{3/4}}{(b c-a d)^2 \sqrt [4]{c+d x}}+\frac{(4 b d) \int \frac{1}{\sqrt [4]{a+b x} \sqrt [4]{c+d x}} \, dx}{(b c-a d)^2}\\ &=-\frac{4}{(b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}-\frac{8 d (a+b x)^{3/4}}{(b c-a d)^2 \sqrt [4]{c+d x}}+\frac{\left (4 b d \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}{(b c-a d)^2 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}\\ &=-\frac{4}{(b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}-\frac{8 d (a+b x)^{3/4}}{(b c-a d)^2 \sqrt [4]{c+d x}}+\frac{\left (16 b d \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 )}{(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}{(b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}-\frac{8 d (a+b x)^{3/4}}{(b c-a d)^2 \sqrt [4]{c+d x}}+\frac{\left (8 \sqrt{b} \sqrt{d} \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 )}{(b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}-\frac{\left (8 \sqrt{b} \sqrt{d} \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 )}{(b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}\\ &=-\frac{4}{(b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x}}-\frac{8 d (a+b x)^{3/4}}{(b c-a d)^2 \sqrt [4]{c+d x}}+\frac{8 \sqrt{b} \sqrt{d} \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}}{(b c-a d)^3 \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{4 \sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \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 )}{\sqrt{b c-a d} \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{2 \sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \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 )}{\sqrt{b c-a d} \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*}

Mathematica [C]  time = 0.0410293, size = 71, normalized size = 0.09 \[ -\frac{4 \left (\frac{b (c+d x)}{b c-a d}\right )^{5/4} \, _2F_1\left (-\frac{1}{4},\frac{5}{4};\frac{3}{4};\frac{d (a+b x)}{a d-b c}\right )}{b \sqrt [4]{a+b x} (c+d x)^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*((b*(c + d*x))/(b*c - a*d))^(5/4)*Hypergeometric2F1[-1/4, 5/4, 3/4, (d*(a + b*x))/(-(b*c) + a*d)])/(b*(a +
b*x)^(1/4)*(c + d*x)^(5/4))

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Maple [F]  time = 0.036, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{-{\frac{5}{4}}} \left ( dx+c \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x + a\right )}^{\frac{5}{4}}{\left (d x + c\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{\frac{3}{4}}{\left (d x + c\right )}^{\frac{3}{4}}}{b^{2} d^{2} x^{4} + a^{2} c^{2} + 2 \,{\left (b^{2} c d + a b d^{2}\right )} x^{3} +{\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} x^{2} + 2 \,{\left (a b c^{2} + a^{2} c d\right )} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x\right )^{\frac{5}{4}} \left (c + d x\right )^{\frac{5}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x + a\right )}^{\frac{5}{4}}{\left (d x + c\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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