3.17.77 \(\int \frac {1}{(a+b x)^{5/2} (c+d x)^{9/4}} \, dx\) [1677]

Optimal. Leaf size=303 \[ -\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{5/4}}+\frac {11 d}{3 (b c-a d)^2 \sqrt {a+b x} (c+d x)^{5/4}}+\frac {77 d^2 \sqrt {a+b x}}{15 (b c-a d)^3 (c+d x)^{5/4}}+\frac {77 b d^2 \sqrt {a+b x}}{5 (b c-a d)^4 \sqrt [4]{c+d x}}-\frac {77 b^{5/4} d \sqrt {-\frac {d (a+b x)}{b c-a d}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{5 (b c-a d)^{13/4} \sqrt {a+b x}}+\frac {77 b^{5/4} d \sqrt {-\frac {d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{5 (b c-a d)^{13/4} \sqrt {a+b x}} \]

[Out]

-2/3/(-a*d+b*c)/(b*x+a)^(3/2)/(d*x+c)^(5/4)+11/3*d/(-a*d+b*c)^2/(d*x+c)^(5/4)/(b*x+a)^(1/2)+77/15*d^2*(b*x+a)^
(1/2)/(-a*d+b*c)^3/(d*x+c)^(5/4)+77/5*b*d^2*(b*x+a)^(1/2)/(-a*d+b*c)^4/(d*x+c)^(1/4)-77/5*b^(5/4)*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)^(13/4)/(b*x+a)^(1/2)+77/5*b
^(5/4)*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)^(13/4)/(
b*x+a)^(1/2)

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Rubi [A]
time = 0.21, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {53, 65, 313, 230, 227, 1214, 1213, 435} \begin {gather*} \frac {77 b^{5/4} d \sqrt {-\frac {d (a+b x)}{b c-a d}} F\left (\left .\text {ArcSin}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{5 \sqrt {a+b x} (b c-a d)^{13/4}}-\frac {77 b^{5/4} d \sqrt {-\frac {d (a+b x)}{b c-a d}} E\left (\left .\text {ArcSin}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{5 \sqrt {a+b x} (b c-a d)^{13/4}}+\frac {77 b d^2 \sqrt {a+b x}}{5 \sqrt [4]{c+d x} (b c-a d)^4}+\frac {77 d^2 \sqrt {a+b x}}{15 (c+d x)^{5/4} (b c-a d)^3}+\frac {11 d}{3 \sqrt {a+b x} (c+d x)^{5/4} (b c-a d)^2}-\frac {2}{3 (a+b x)^{3/2} (c+d x)^{5/4} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(5/2)*(c + d*x)^(9/4)),x]

[Out]

-2/(3*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(5/4)) + (11*d)/(3*(b*c - a*d)^2*Sqrt[a + b*x]*(c + d*x)^(5/4)) +
(77*d^2*Sqrt[a + b*x])/(15*(b*c - a*d)^3*(c + d*x)^(5/4)) + (77*b*d^2*Sqrt[a + b*x])/(5*(b*c - a*d)^4*(c + d*x
)^(1/4)) - (77*b^(5/4)*d*Sqrt[-((d*(a + b*x))/(b*c - a*d))]*EllipticE[ArcSin[(b^(1/4)*(c + d*x)^(1/4))/(b*c -
a*d)^(1/4)], -1])/(5*(b*c - a*d)^(13/4)*Sqrt[a + b*x]) + (77*b^(5/4)*d*Sqrt[-((d*(a + b*x))/(b*c - a*d))]*Elli
pticF[ArcSin[(b^(1/4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)], -1])/(5*(b*c - a*d)^(13/4)*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]

Rule 313

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 435

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

Rule 1213

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 1214

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)^{9/4}} \, dx &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{5/4}}-\frac {(11 d) \int \frac {1}{(a+b x)^{3/2} (c+d x)^{9/4}} \, dx}{6 (b c-a d)}\\ &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{5/4}}+\frac {11 d}{3 (b c-a d)^2 \sqrt {a+b x} (c+d x)^{5/4}}+\frac {\left (77 d^2\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{9/4}} \, dx}{12 (b c-a d)^2}\\ &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{5/4}}+\frac {11 d}{3 (b c-a d)^2 \sqrt {a+b x} (c+d x)^{5/4}}+\frac {77 d^2 \sqrt {a+b x}}{15 (b c-a d)^3 (c+d x)^{5/4}}+\frac {\left (77 b d^2\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/4}} \, dx}{20 (b c-a d)^3}\\ &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{5/4}}+\frac {11 d}{3 (b c-a d)^2 \sqrt {a+b x} (c+d x)^{5/4}}+\frac {77 d^2 \sqrt {a+b x}}{15 (b c-a d)^3 (c+d x)^{5/4}}+\frac {77 b d^2 \sqrt {a+b x}}{5 (b c-a d)^4 \sqrt [4]{c+d x}}-\frac {\left (77 b^2 d^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt [4]{c+d x}} \, dx}{20 (b c-a d)^4}\\ &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{5/4}}+\frac {11 d}{3 (b c-a d)^2 \sqrt {a+b x} (c+d x)^{5/4}}+\frac {77 d^2 \sqrt {a+b x}}{15 (b c-a d)^3 (c+d x)^{5/4}}+\frac {77 b d^2 \sqrt {a+b x}}{5 (b c-a d)^4 \sqrt [4]{c+d x}}-\frac {\left (77 b^2 d\right ) \text {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 )}{5 (b c-a d)^4}\\ &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{5/4}}+\frac {11 d}{3 (b c-a d)^2 \sqrt {a+b x} (c+d x)^{5/4}}+\frac {77 d^2 \sqrt {a+b x}}{15 (b c-a d)^3 (c+d x)^{5/4}}+\frac {77 b d^2 \sqrt {a+b x}}{5 (b c-a d)^4 \sqrt [4]{c+d x}}+\frac {\left (77 b^{3/2} d\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 )}{5 (b c-a d)^{7/2}}-\frac {\left (77 b^{3/2} d\right ) \text {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 )}{5 (b c-a d)^{7/2}}\\ &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{5/4}}+\frac {11 d}{3 (b c-a d)^2 \sqrt {a+b x} (c+d x)^{5/4}}+\frac {77 d^2 \sqrt {a+b x}}{15 (b c-a d)^3 (c+d x)^{5/4}}+\frac {77 b d^2 \sqrt {a+b x}}{5 (b c-a d)^4 \sqrt [4]{c+d x}}+\frac {\left (77 b^{3/2} 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 )}{5 (b c-a d)^{7/2} \sqrt {a+b x}}-\frac {\left (77 b^{3/2} d \sqrt {\frac {d (a+b x)}{-b c+a d}}\right ) \text {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 )}{5 (b c-a d)^{7/2} \sqrt {a+b x}}\\ &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{5/4}}+\frac {11 d}{3 (b c-a d)^2 \sqrt {a+b x} (c+d x)^{5/4}}+\frac {77 d^2 \sqrt {a+b x}}{15 (b c-a d)^3 (c+d x)^{5/4}}+\frac {77 b d^2 \sqrt {a+b x}}{5 (b c-a d)^4 \sqrt [4]{c+d x}}+\frac {77 b^{5/4} 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 )}{5 (b c-a d)^{13/4} \sqrt {a+b x}}-\frac {\left (77 b^{3/2} d \sqrt {\frac {d (a+b x)}{-b c+a d}}\right ) \text {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 )}{5 (b c-a d)^{7/2} \sqrt {a+b x}}\\ &=-\frac {2}{3 (b c-a d) (a+b x)^{3/2} (c+d x)^{5/4}}+\frac {11 d}{3 (b c-a d)^2 \sqrt {a+b x} (c+d x)^{5/4}}+\frac {77 d^2 \sqrt {a+b x}}{15 (b c-a d)^3 (c+d x)^{5/4}}+\frac {77 b d^2 \sqrt {a+b x}}{5 (b c-a d)^4 \sqrt [4]{c+d x}}-\frac {77 b^{5/4} 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 )}{5 (b c-a d)^{13/4} \sqrt {a+b x}}+\frac {77 b^{5/4} 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 )}{5 (b c-a d)^{13/4} \sqrt {a+b x}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.04, size = 73, normalized size = 0.24 \begin {gather*} -\frac {2 \left (\frac {b (c+d x)}{b c-a d}\right )^{9/4} \, _2F_1\left (-\frac {3}{2},\frac {9}{4};-\frac {1}{2};\frac {d (a+b x)}{-b c+a d}\right )}{3 b (a+b x)^{3/2} (c+d x)^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(5/2)*(c + d*x)^(9/4)),x]

[Out]

(-2*((b*(c + d*x))/(b*c - a*d))^(9/4)*Hypergeometric2F1[-3/2, 9/4, -1/2, (d*(a + b*x))/(-(b*c) + a*d)])/(3*b*(
a + b*x)^(3/2)*(c + d*x)^(9/4))

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Maple [F]
time = 0.09, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b x +a \right )^{\frac {5}{2}} \left (d x +c \right )^{\frac {9}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(5/2)/(d*x+c)^(9/4),x)

[Out]

int(1/(b*x+a)^(5/2)/(d*x+c)^(9/4),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(5/2)/(d*x+c)^(9/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(5/2)*(d*x + c)^(9/4)), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(5/2)/(d*x+c)^(9/4),x, algorithm="fricas")

[Out]

integral(sqrt(b*x + a)*(d*x + c)^(3/4)/(b^3*d^3*x^6 + a^3*c^3 + 3*(b^3*c*d^2 + a*b^2*d^3)*x^5 + 3*(b^3*c^2*d +
3*a*b^2*c*d^2 + a^2*b*d^3)*x^4 + (b^3*c^3 + 9*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*x^3 + 3*(a*b^2*c^3 + 3*a
^2*b*c^2*d + a^3*c*d^2)*x^2 + 3*(a^2*b*c^3 + a^3*c^2*d)*x), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {9}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(5/2)/(d*x+c)**(9/4),x)

[Out]

Integral(1/((a + b*x)**(5/2)*(c + d*x)**(9/4)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(5/2)/(d*x+c)^(9/4),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(5/2)*(d*x + c)^(9/4)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{9/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(5/2)*(c + d*x)^(9/4)),x)

[Out]

int(1/((a + b*x)^(5/2)*(c + d*x)^(9/4)), x)

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