3.17.55 \(\int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}} \, dx\) [1655]

Optimal. Leaf size=111 \[ \frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d)^{5/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{3 \sqrt [4]{b} d^2 \sqrt {a+b x}} \]

[Out]

4/3*(d*x+c)^(1/4)*(b*x+a)^(1/2)/d-8/3*(-a*d+b*c)^(5/4)*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^(1/4)/d^2/(b*x+a)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {52, 65, 230, 227} \begin {gather*} \frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d)^{5/4} \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 \sqrt [4]{b} d^2 \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x]/(c + d*x)^(3/4),x]

[Out]

(4*Sqrt[a + b*x]*(c + d*x)^(1/4))/(3*d) - (8*(b*c - a*d)^(5/4)*Sqrt[-((d*(a + b*x))/(b*c - a*d))]*EllipticF[Ar
cSin[(b^(1/4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)], -1])/(3*b^(1/4)*d^2*Sqrt[a + b*x])

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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 {\sqrt {a+b x}}{(c+d x)^{3/4}} \, dx &=\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {(2 (b c-a d)) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx}{3 d}\\ &=\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {(8 (b c-a 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 d^2}\\ &=\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {\left (8 (b c-a 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 d^2 \sqrt {a+b x}}\\ &=\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d)^{5/4} \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 \sqrt [4]{b} d^2 \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 = 10.02, size = 73, normalized size = 0.66 \begin {gather*} \frac {2 (a+b x)^{3/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/4} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {5}{2};\frac {d (a+b x)}{-b c+a d}\right )}{3 b (c+d x)^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x]/(c + d*x)^(3/4),x]

[Out]

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

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {b x +a}}{\left (d x +c \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(1/2)/(d*x+c)^(3/4),x)

[Out]

int((b*x+a)^(1/2)/(d*x+c)^(3/4),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((b*x+a)^(1/2)/(d*x+c)^(3/4),x, algorithm="maxima")

[Out]

integrate(sqrt(b*x + a)/(d*x + c)^(3/4), 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((b*x+a)^(1/2)/(d*x+c)^(3/4),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b x}}{\left (c + d x\right )^{\frac {3}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(1/2)/(d*x+c)**(3/4),x)

[Out]

Integral(sqrt(a + b*x)/(c + d*x)**(3/4), 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((b*x+a)^(1/2)/(d*x+c)^(3/4),x, algorithm="giac")

[Out]

integrate(sqrt(b*x + a)/(d*x + c)^(3/4), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {a+b\,x}}{{\left (c+d\,x\right )}^{3/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(1/2)/(c + d*x)^(3/4),x)

[Out]

int((a + b*x)^(1/2)/(c + d*x)^(3/4), x)

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