3.17.45 \(\int \frac {(c+d x)^{5/4}}{(a+b x)^{5/2}} \, dx\) [1645]

Optimal. Leaf size=135 \[ -\frac {5 d \sqrt [4]{c+d x}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/4}}{3 b (a+b x)^{3/2}}+\frac {5 d \sqrt [4]{b c-a d} \sqrt {-\frac {d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{3 b^{9/4} \sqrt {a+b x}} \]

[Out]

-2/3*(d*x+c)^(5/4)/b/(b*x+a)^(3/2)-5/3*d*(d*x+c)^(1/4)/b^2/(b*x+a)^(1/2)+5/3*d*(-a*d+b*c)^(1/4)*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)/b^(9/4)/(b*x+a)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {49, 65, 230, 227} \begin {gather*} \frac {5 d \sqrt [4]{b c-a 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 )}{3 b^{9/4} \sqrt {a+b x}}-\frac {5 d \sqrt [4]{c+d x}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/4}}{3 b (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*d*(c + d*x)^(1/4))/(3*b^2*Sqrt[a + b*x]) - (2*(c + d*x)^(5/4))/(3*b*(a + b*x)^(3/2)) + (5*d*(b*c - a*d)^(1
/4)*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])/(3*
b^(9/4)*Sqrt[a + b*x])

Rule 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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*} \int \frac {(c+d x)^{5/4}}{(a+b x)^{5/2}} \, dx &=-\frac {2 (c+d x)^{5/4}}{3 b (a+b x)^{3/2}}+\frac {(5 d) \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{3/2}} \, dx}{6 b}\\ &=-\frac {5 d \sqrt [4]{c+d x}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/4}}{3 b (a+b x)^{3/2}}+\frac {\left (5 d^2\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx}{12 b^2}\\ &=-\frac {5 d \sqrt [4]{c+d x}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/4}}{3 b (a+b x)^{3/2}}+\frac {(5 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 )}{3 b^2}\\ &=-\frac {5 d \sqrt [4]{c+d x}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/4}}{3 b (a+b x)^{3/2}}+\frac {\left (5 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 )}{3 b^2 \sqrt {a+b x}}\\ &=-\frac {5 d \sqrt [4]{c+d x}}{3 b^2 \sqrt {a+b x}}-\frac {2 (c+d x)^{5/4}}{3 b (a+b x)^{3/2}}+\frac {5 d \sqrt [4]{b c-a 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 )}{3 b^{9/4} \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 = 10.07, size = 73, normalized size = 0.54 \begin {gather*} -\frac {2 (c+d x)^{5/4} \, _2F_1\left (-\frac {3}{2},-\frac {5}{4};-\frac {1}{2};\frac {d (a+b x)}{-b c+a d}\right )}{3 b (a+b x)^{3/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (d x +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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