3.1673 \(\int \frac {(a+b x)^{3/2}}{(c+d x)^{9/4}} \, dx\)

Optimal. Leaf size=222 \[ -\frac {48 b^{5/4} (b c-a d)^{3/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{5 d^3 \sqrt {a+b x}}+\frac {48 b^{5/4} (b c-a d)^{3/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{5 d^3 \sqrt {a+b x}}-\frac {24 b \sqrt {a+b x}}{5 d^2 \sqrt [4]{c+d x}}-\frac {4 (a+b x)^{3/2}}{5 d (c+d x)^{5/4}} \]

[Out]

-4/5*(b*x+a)^(3/2)/d/(d*x+c)^(5/4)-24/5*b*(b*x+a)^(1/2)/d^2/(d*x+c)^(1/4)+48/5*b^(5/4)*(-a*d+b*c)^(3/4)*Ellipt
icE(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^(5/4)*(-a
*d+b*c)^(3/4)*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 = 0.23, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {47, 63, 307, 224, 221, 1200, 1199, 424} \[ -\frac {48 b^{5/4} (b c-a d)^{3/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{5 d^3 \sqrt {a+b x}}+\frac {48 b^{5/4} (b c-a d)^{3/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{5 d^3 \sqrt {a+b x}}-\frac {24 b \sqrt {a+b x}}{5 d^2 \sqrt [4]{c+d x}}-\frac {4 (a+b x)^{3/2}}{5 d (c+d x)^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(a + b*x)^(3/2))/(5*d*(c + d*x)^(5/4)) - (24*b*Sqrt[a + b*x])/(5*d^2*(c + d*x)^(1/4)) + (48*b^(5/4)*(b*c -
a*d)^(3/4)*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^(5/4)*(b*c - a*d)^(3/4)*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 47

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 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 {(a+b x)^{3/2}}{(c+d x)^{9/4}} \, dx &=-\frac {4 (a+b x)^{3/2}}{5 d (c+d x)^{5/4}}+\frac {(6 b) \int \frac {\sqrt {a+b x}}{(c+d x)^{5/4}} \, dx}{5 d}\\ &=-\frac {4 (a+b x)^{3/2}}{5 d (c+d x)^{5/4}}-\frac {24 b \sqrt {a+b x}}{5 d^2 \sqrt [4]{c+d x}}+\frac {\left (12 b^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt [4]{c+d x}} \, dx}{5 d^2}\\ &=-\frac {4 (a+b x)^{3/2}}{5 d (c+d x)^{5/4}}-\frac {24 b \sqrt {a+b x}}{5 d^2 \sqrt [4]{c+d x}}+\frac {\left (48 b^2\right ) \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 )}{5 d^3}\\ &=-\frac {4 (a+b x)^{3/2}}{5 d (c+d x)^{5/4}}-\frac {24 b \sqrt {a+b x}}{5 d^2 \sqrt [4]{c+d x}}-\frac {\left (48 b^{3/2} \sqrt {b c-a 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 )}{5 d^3}+\frac {\left (48 b^{3/2} \sqrt {b c-a 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 )}{5 d^3}\\ &=-\frac {4 (a+b x)^{3/2}}{5 d (c+d x)^{5/4}}-\frac {24 b \sqrt {a+b x}}{5 d^2 \sqrt [4]{c+d x}}-\frac {\left (48 b^{3/2} \sqrt {b c-a 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 )}{5 d^3 \sqrt {a+b x}}+\frac {\left (48 b^{3/2} \sqrt {b c-a 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 )}{5 d^3 \sqrt {a+b x}}\\ &=-\frac {4 (a+b x)^{3/2}}{5 d (c+d x)^{5/4}}-\frac {24 b \sqrt {a+b x}}{5 d^2 \sqrt [4]{c+d x}}-\frac {48 b^{5/4} (b c-a d)^{3/4} \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 b^{3/2} \sqrt {b c-a 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 )}{5 d^3 \sqrt {a+b x}}\\ &=-\frac {4 (a+b x)^{3/2}}{5 d (c+d x)^{5/4}}-\frac {24 b \sqrt {a+b x}}{5 d^2 \sqrt [4]{c+d x}}+\frac {48 b^{5/4} (b c-a d)^{3/4} \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 b^{5/4} (b c-a d)^{3/4} \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]  time = 0.06, size = 73, normalized size = 0.33 \[ \frac {2 (a+b x)^{5/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{9/4} \, _2F_1\left (\frac {9}{4},\frac {5}{2};\frac {7}{2};\frac {d (a+b x)}{a d-b c}\right )}{5 b (c+d x)^{9/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(5/2)*((b*(c + d*x))/(b*c - a*d))^(9/4)*Hypergeometric2F1[9/4, 5/2, 7/2, (d*(a + b*x))/(-(b*c) +
a*d)])/(5*b*(c + d*x)^(9/4))

________________________________________________________________________________________

fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {3}{4}}}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/(d*x+c)^(9/4),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{\frac {3}{2}}}{{\left (d x + c\right )}^{\frac {9}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/(d*x+c)^(9/4),x, algorithm="giac")

[Out]

integrate((b*x + a)^(3/2)/(d*x + c)^(9/4), x)

________________________________________________________________________________________

maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{\frac {3}{2}}}{\left (d x +c \right )^{\frac {9}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)/(d*x+c)^(9/4),x)

[Out]

int((b*x+a)^(3/2)/(d*x+c)^(9/4),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{\frac {3}{2}}}{{\left (d x + c\right )}^{\frac {9}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/(d*x+c)^(9/4),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(3/2)/(d*x + c)^(9/4), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,x\right )}^{3/2}}{{\left (c+d\,x\right )}^{9/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(3/2)/(c + d*x)^(9/4),x)

[Out]

int((a + b*x)^(3/2)/(c + d*x)^(9/4), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x\right )^{\frac {3}{2}}}{\left (c + d x\right )^{\frac {9}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)/(d*x+c)**(9/4),x)

[Out]

Integral((a + b*x)**(3/2)/(c + d*x)**(9/4), x)

________________________________________________________________________________________