3.17.60 \(\int \frac {(a+b x)^{3/2}}{(c+d x)^{5/4}} \, dx\) [1660]

Optimal. Leaf size=220 \[ -\frac {4 (a+b x)^{3/2}}{d \sqrt [4]{c+d x}}+\frac {24 b \sqrt {a+b x} (c+d x)^{3/4}}{5 d^2}-\frac {48 \sqrt [4]{b} (b c-a d)^{7/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{5 d^3 \sqrt {a+b x}}+\frac {48 \sqrt [4]{b} (b c-a d)^{7/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{5 d^3 \sqrt {a+b x}} \]

[Out]

-4*(b*x+a)^(3/2)/d/(d*x+c)^(1/4)+24/5*b*(d*x+c)^(3/4)*(b*x+a)^(1/2)/d^2-48/5*b^(1/4)*(-a*d+b*c)^(7/4)*Elliptic
E(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)/d^3/(b*x+a)^(1/2)+48/5*b^(1/4)*(-a*d
+b*c)^(7/4)*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)/d^3/(b*x+a)^(1/2
)

________________________________________________________________________________________

Rubi [A]
time = 0.16, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {49, 52, 65, 313, 230, 227, 1214, 1213, 435} \begin {gather*} \frac {48 \sqrt [4]{b} (b c-a d)^{7/4} \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 )}{5 d^3 \sqrt {a+b x}}-\frac {48 \sqrt [4]{b} (b c-a d)^{7/4} \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 d^3 \sqrt {a+b x}}+\frac {24 b \sqrt {a+b x} (c+d x)^{3/4}}{5 d^2}-\frac {4 (a+b x)^{3/2}}{d \sqrt [4]{c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(a + b*x)^(3/2))/(d*(c + d*x)^(1/4)) + (24*b*Sqrt[a + b*x]*(c + d*x)^(3/4))/(5*d^2) - (48*b^(1/4)*(b*c - a
*d)^(7/4)*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*d^3*Sqrt[a + b*x]) + (48*b^(1/4)*(b*c - a*d)^(7/4)*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])/(5*d^3*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 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]

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 {(a+b x)^{3/2}}{(c+d x)^{5/4}} \, dx &=-\frac {4 (a+b x)^{3/2}}{d \sqrt [4]{c+d x}}+\frac {(6 b) \int \frac {\sqrt {a+b x}}{\sqrt [4]{c+d x}} \, dx}{d}\\ &=-\frac {4 (a+b x)^{3/2}}{d \sqrt [4]{c+d x}}+\frac {24 b \sqrt {a+b x} (c+d x)^{3/4}}{5 d^2}-\frac {(12 b (b c-a d)) \int \frac {1}{\sqrt {a+b x} \sqrt [4]{c+d x}} \, dx}{5 d^2}\\ &=-\frac {4 (a+b x)^{3/2}}{d \sqrt [4]{c+d x}}+\frac {24 b \sqrt {a+b x} (c+d x)^{3/4}}{5 d^2}-\frac {(48 b (b c-a d)) \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 d^3}\\ &=-\frac {4 (a+b x)^{3/2}}{d \sqrt [4]{c+d x}}+\frac {24 b \sqrt {a+b x} (c+d x)^{3/4}}{5 d^2}+\frac {\left (48 \sqrt {b} (b c-a d)^{3/2}\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 d^3}-\frac {\left (48 \sqrt {b} (b c-a d)^{3/2}\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 d^3}\\ &=-\frac {4 (a+b x)^{3/2}}{d \sqrt [4]{c+d x}}+\frac {24 b \sqrt {a+b x} (c+d x)^{3/4}}{5 d^2}+\frac {\left (48 \sqrt {b} (b c-a d)^{3/2} \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 d^3 \sqrt {a+b x}}-\frac {\left (48 \sqrt {b} (b c-a d)^{3/2} \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 d^3 \sqrt {a+b x}}\\ &=-\frac {4 (a+b x)^{3/2}}{d \sqrt [4]{c+d x}}+\frac {24 b \sqrt {a+b x} (c+d x)^{3/4}}{5 d^2}+\frac {48 \sqrt [4]{b} (b c-a d)^{7/4} \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 )}{5 d^3 \sqrt {a+b x}}-\frac {\left (48 \sqrt {b} (b c-a d)^{3/2} \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 d^3 \sqrt {a+b x}}\\ &=-\frac {4 (a+b x)^{3/2}}{d \sqrt [4]{c+d x}}+\frac {24 b \sqrt {a+b x} (c+d x)^{3/4}}{5 d^2}-\frac {48 \sqrt [4]{b} (b c-a d)^{7/4} \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 d^3 \sqrt {a+b x}}+\frac {48 \sqrt [4]{b} (b c-a d)^{7/4} \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 )}{5 d^3 \sqrt {a+b x}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.05, size = 73, normalized size = 0.33 \begin {gather*} \frac {2 (a+b x)^{5/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4} \, _2F_1\left (\frac {5}{4},\frac {5}{2};\frac {7}{2};\frac {d (a+b x)}{-b c+a d}\right )}{5 b (c+d x)^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(5/2)*((b*(c + d*x))/(b*c - a*d))^(5/4)*Hypergeometric2F1[5/4, 5/2, 7/2, (d*(a + b*x))/(-(b*c) +
a*d)])/(5*b*(c + d*x)^(5/4))

________________________________________________________________________________________

Mathics [F(-1)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[(a + b*x)^(3/2)/(c + d*x)^(5/4),x]')

[Out]

Timed out

________________________________________________________________________________________

Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{\frac {3}{2}}}{\left (d x +c \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)^(3/2)/(d*x+c)^(5/4),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.33, 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)^(3/2)/(d*x+c)^(5/4),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right )^{\frac {3}{2}}}{\left (c + d x\right )^{\frac {5}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)/(d*x+c)**(5/4),x)

[Out]

Integral((a + b*x)**(3/2)/(c + d*x)**(5/4), x)

________________________________________________________________________________________

Giac [F] N/A
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)^(3/2)/(d*x+c)^(5/4),x)

[Out]

Could not integrate

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{3/2}}{{\left (c+d\,x\right )}^{5/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________