3.1664 \(\int \frac {1}{(a+b x)^{5/2} (c+d x)^{5/4}} \, dx\)

Optimal. Leaf size=261 \[ \frac {7 \sqrt [4]{b} d \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{\sqrt {a+b x} (b c-a d)^{9/4}}+\frac {7 d^2 \sqrt {a+b x}}{\sqrt [4]{c+d x} (b c-a d)^3}+\frac {7 d}{3 \sqrt {a+b x} \sqrt [4]{c+d x} (b c-a d)^2}-\frac {2}{3 (a+b x)^{3/2} \sqrt [4]{c+d x} (b c-a d)}-\frac {7 \sqrt [4]{b} d \sqrt {-\frac {d (a+b x)}{b c-a d}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{\sqrt {a+b x} (b c-a d)^{9/4}} \]

[Out]

-2/3/(-a*d+b*c)/(b*x+a)^(3/2)/(d*x+c)^(1/4)+7/3*d/(-a*d+b*c)^2/(d*x+c)^(1/4)/(b*x+a)^(1/2)+7*d^2*(b*x+a)^(1/2)
/(-a*d+b*c)^3/(d*x+c)^(1/4)-7*b^(1/4)*d*EllipticE(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)^(9/4)/(b*x+a)^(1/2)+7*b^(1/4)*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)^(9/4)/(b*x+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.26, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {51, 63, 307, 224, 221, 1200, 1199, 424} \[ \frac {7 d^2 \sqrt {a+b x}}{\sqrt [4]{c+d x} (b c-a d)^3}+\frac {7 d}{3 \sqrt {a+b x} \sqrt [4]{c+d x} (b c-a d)^2}-\frac {2}{3 (a+b x)^{3/2} \sqrt [4]{c+d x} (b c-a d)}+\frac {7 \sqrt [4]{b} d \sqrt {-\frac {d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{\sqrt {a+b x} (b c-a d)^{9/4}}-\frac {7 \sqrt [4]{b} d \sqrt {-\frac {d (a+b x)}{b c-a d}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{\sqrt {a+b x} (b c-a d)^{9/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 63

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 221

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 224

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]

Rule 307

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

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 1200

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (c*x^4)/a]/Sqrt[a + c*x^4], In
t[(d + e*x^2)/Sqrt[1 + (c*x^4)/a], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] &&
!GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {1}{(a+b x)^{5/2} (c+d x)^{5/4}} \, dx &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}-\frac {(7 d) \int \frac {1}{(a+b x)^{3/2} (c+d x)^{5/4}} \, dx}{6 (b c-a d)}\\ &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}+\frac {7 d}{3 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}+\frac {\left (7 d^2\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/4}} \, dx}{4 (b c-a d)^2}\\ &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}+\frac {7 d}{3 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}+\frac {7 d^2 \sqrt {a+b x}}{(b c-a d)^3 \sqrt [4]{c+d x}}-\frac {\left (7 b d^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt [4]{c+d x}} \, dx}{4 (b c-a d)^3}\\ &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}+\frac {7 d}{3 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}+\frac {7 d^2 \sqrt {a+b x}}{(b c-a d)^3 \sqrt [4]{c+d x}}-\frac {(7 b d) \operatorname {Subst}\left (\int \frac {x^2}{\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} \sqrt [4]{c+d x}}+\frac {7 d}{3 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}+\frac {7 d^2 \sqrt {a+b x}}{(b c-a d)^3 \sqrt [4]{c+d x}}+\frac {\left (7 \sqrt {b} d\right ) \operatorname {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)^{5/2}}-\frac {\left (7 \sqrt {b} d\right ) \operatorname {Subst}\left (\int \frac {1+\frac {\sqrt {b} x^2}{\sqrt {b c-a d}}}{\sqrt {a-\frac {b c}{d}+\frac {b x^4}{d}}} \, dx,x,\sqrt [4]{c+d x}\right )}{(b c-a d)^{5/2}}\\ &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}+\frac {7 d}{3 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}+\frac {7 d^2 \sqrt {a+b x}}{(b c-a d)^3 \sqrt [4]{c+d x}}+\frac {\left (7 \sqrt {b} d \sqrt {\frac {d (a+b x)}{-b c+a d}}\right ) \operatorname {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)^{5/2} \sqrt {a+b x}}-\frac {\left (7 \sqrt {b} d \sqrt {\frac {d (a+b x)}{-b c+a d}}\right ) \operatorname {Subst}\left (\int \frac {1+\frac {\sqrt {b} x^2}{\sqrt {b c-a d}}}{\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)^{5/2} \sqrt {a+b x}}\\ &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}+\frac {7 d}{3 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}+\frac {7 d^2 \sqrt {a+b x}}{(b c-a d)^3 \sqrt [4]{c+d x}}+\frac {7 \sqrt [4]{b} 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)^{9/4} \sqrt {a+b x}}-\frac {\left (7 \sqrt {b} d \sqrt {\frac {d (a+b x)}{-b c+a d}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {\sqrt {b} x^2}{\sqrt {b c-a d}}}}{\sqrt {1-\frac {\sqrt {b} x^2}{\sqrt {b c-a d}}}} \, dx,x,\sqrt [4]{c+d x}\right )}{(b c-a d)^{5/2} \sqrt {a+b x}}\\ &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}+\frac {7 d}{3 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}+\frac {7 d^2 \sqrt {a+b x}}{(b c-a d)^3 \sqrt [4]{c+d x}}-\frac {7 \sqrt [4]{b} d \sqrt {-\frac {d (a+b x)}{b c-a d}} E\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)^{9/4} \sqrt {a+b x}}+\frac {7 \sqrt [4]{b} 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)^{9/4} \sqrt {a+b x}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.04, size = 73, normalized size = 0.28 \[ -\frac {2 \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4} \, _2F_1\left (-\frac {3}{2},\frac {5}{4};-\frac {1}{2};\frac {d (a+b x)}{a d-b c}\right )}{3 b (a+b x)^{3/2} (c+d x)^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*((b*(c + d*x))/(b*c - a*d))^(5/4)*Hypergeometric2F1[-3/2, 5/4, -1/2, (d*(a + b*x))/(-(b*c) + a*d)])/(3*b*(
a + b*x)^(3/2)*(c + d*x)^(5/4))

________________________________________________________________________________________

fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{4}}}{b^{3} d^{2} x^{5} + a^{3} c^{2} + {\left (2 \, b^{3} c d + 3 \, a b^{2} d^{2}\right )} x^{4} + {\left (b^{3} c^{2} + 6 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} x^{3} + {\left (3 \, a b^{2} c^{2} + 6 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2} + {\left (3 \, a^{2} b c^{2} + 2 \, a^{3} c d\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x + a)*(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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {5}{2}} {\left (d x + c\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {5}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {5}{2}} {\left (d x + c\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{5/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {5}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________