3.1676 \(\int \frac{1}{(a+b x)^{3/2} (c+d x)^{9/4}} \, dx\)

Optimal. Leaf size=262 \[ -\frac{42 b^{5/4} \sqrt{-\frac{d (a+b x)}{b c-a d}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{5 \sqrt{a+b x} (b c-a d)^{9/4}}+\frac{42 b^{5/4} \sqrt{-\frac{d (a+b x)}{b c-a d}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{5 \sqrt{a+b x} (b c-a d)^{9/4}}-\frac{42 b d \sqrt{a+b x}}{5 \sqrt [4]{c+d x} (b c-a d)^3}-\frac{14 d \sqrt{a+b x}}{5 (c+d x)^{5/4} (b c-a d)^2}-\frac{2}{\sqrt{a+b x} (c+d x)^{5/4} (b c-a d)} \]

[Out]

-2/((b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(5/4)) - (14*d*Sqrt[a + b*x])/(5*(b*c - a*d)^2*(c + d*x)^(5/4)) - (42*
b*d*Sqrt[a + b*x])/(5*(b*c - a*d)^3*(c + d*x)^(1/4)) + (42*b^(5/4)*Sqrt[-((d*(a + b*x))/(b*c - a*d))]*Elliptic
E[ArcSin[(b^(1/4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)], -1])/(5*(b*c - a*d)^(9/4)*Sqrt[a + b*x]) - (42*b^(5/4)*
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*(b*c
- a*d)^(9/4)*Sqrt[a + b*x])

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Rubi [A]  time = 0.256844, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {51, 63, 307, 224, 221, 1200, 1199, 424} \[ -\frac{42 b^{5/4} \sqrt{-\frac{d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{5 \sqrt{a+b x} (b c-a d)^{9/4}}+\frac{42 b^{5/4} \sqrt{-\frac{d (a+b x)}{b c-a d}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{5 \sqrt{a+b x} (b c-a d)^{9/4}}-\frac{42 b d \sqrt{a+b x}}{5 \sqrt [4]{c+d x} (b c-a d)^3}-\frac{14 d \sqrt{a+b x}}{5 (c+d x)^{5/4} (b c-a d)^2}-\frac{2}{\sqrt{a+b x} (c+d x)^{5/4} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2/((b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(5/4)) - (14*d*Sqrt[a + b*x])/(5*(b*c - a*d)^2*(c + d*x)^(5/4)) - (42*
b*d*Sqrt[a + b*x])/(5*(b*c - a*d)^3*(c + d*x)^(1/4)) + (42*b^(5/4)*Sqrt[-((d*(a + b*x))/(b*c - a*d))]*Elliptic
E[ArcSin[(b^(1/4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)], -1])/(5*(b*c - a*d)^(9/4)*Sqrt[a + b*x]) - (42*b^(5/4)*
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*(b*c
- a*d)^(9/4)*Sqrt[a + b*x])

Rule 51

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 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 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 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 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 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]

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 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]

Rubi steps

\begin{align*} \int \frac{1}{(a+b x)^{3/2} (c+d x)^{9/4}} \, dx &=-\frac{2}{(b c-a d) \sqrt{a+b x} (c+d x)^{5/4}}-\frac{(7 d) \int \frac{1}{\sqrt{a+b x} (c+d x)^{9/4}} \, dx}{2 (b c-a d)}\\ &=-\frac{2}{(b c-a d) \sqrt{a+b x} (c+d x)^{5/4}}-\frac{14 d \sqrt{a+b x}}{5 (b c-a d)^2 (c+d x)^{5/4}}-\frac{(21 b d) \int \frac{1}{\sqrt{a+b x} (c+d x)^{5/4}} \, dx}{10 (b c-a d)^2}\\ &=-\frac{2}{(b c-a d) \sqrt{a+b x} (c+d x)^{5/4}}-\frac{14 d \sqrt{a+b x}}{5 (b c-a d)^2 (c+d x)^{5/4}}-\frac{42 b d \sqrt{a+b x}}{5 (b c-a d)^3 \sqrt [4]{c+d x}}+\frac{\left (21 b^2 d\right ) \int \frac{1}{\sqrt{a+b x} \sqrt [4]{c+d x}} \, dx}{10 (b c-a d)^3}\\ &=-\frac{2}{(b c-a d) \sqrt{a+b x} (c+d x)^{5/4}}-\frac{14 d \sqrt{a+b x}}{5 (b c-a d)^2 (c+d x)^{5/4}}-\frac{42 b d \sqrt{a+b x}}{5 (b c-a d)^3 \sqrt [4]{c+d x}}+\frac{\left (42 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 (b c-a d)^3}\\ &=-\frac{2}{(b c-a d) \sqrt{a+b x} (c+d x)^{5/4}}-\frac{14 d \sqrt{a+b x}}{5 (b c-a d)^2 (c+d x)^{5/4}}-\frac{42 b d \sqrt{a+b x}}{5 (b c-a d)^3 \sqrt [4]{c+d x}}-\frac{\left (42 b^{3/2}\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 (b c-a d)^{5/2}}+\frac{\left (42 b^{3/2}\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 (b c-a d)^{5/2}}\\ &=-\frac{2}{(b c-a d) \sqrt{a+b x} (c+d x)^{5/4}}-\frac{14 d \sqrt{a+b x}}{5 (b c-a d)^2 (c+d x)^{5/4}}-\frac{42 b d \sqrt{a+b x}}{5 (b c-a d)^3 \sqrt [4]{c+d x}}-\frac{\left (42 b^{3/2} \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 (b c-a d)^{5/2} \sqrt{a+b x}}+\frac{\left (42 b^{3/2} \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 (b c-a d)^{5/2} \sqrt{a+b x}}\\ &=-\frac{2}{(b c-a d) \sqrt{a+b x} (c+d x)^{5/4}}-\frac{14 d \sqrt{a+b x}}{5 (b c-a d)^2 (c+d x)^{5/4}}-\frac{42 b d \sqrt{a+b x}}{5 (b c-a d)^3 \sqrt [4]{c+d x}}-\frac{42 b^{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 )}{5 (b c-a d)^{9/4} \sqrt{a+b x}}+\frac{\left (42 b^{3/2} \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 (b c-a d)^{5/2} \sqrt{a+b x}}\\ &=-\frac{2}{(b c-a d) \sqrt{a+b x} (c+d x)^{5/4}}-\frac{14 d \sqrt{a+b x}}{5 (b c-a d)^2 (c+d x)^{5/4}}-\frac{42 b d \sqrt{a+b x}}{5 (b c-a d)^3 \sqrt [4]{c+d x}}+\frac{42 b^{5/4} \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 (b c-a d)^{9/4} \sqrt{a+b x}}-\frac{42 b^{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 )}{5 (b c-a d)^{9/4} \sqrt{a+b x}}\\ \end{align*}

Mathematica [C]  time = 0.046052, size = 71, normalized size = 0.27 \[ -\frac{2 \left (\frac{b (c+d x)}{b c-a d}\right )^{9/4} \, _2F_1\left (-\frac{1}{2},\frac{9}{4};\frac{1}{2};\frac{d (a+b x)}{a d-b c}\right )}{b \sqrt{a+b x} (c+d x)^{9/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.041, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{-{\frac{3}{2}}} \left ( dx+c \right ) ^{-{\frac{9}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x + a\right )}^{\frac{3}{2}}{\left (d x + c\right )}^{\frac{9}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(3/2)/(d*x+c)^(9/4),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x + a}{\left (d x + c\right )}^{\frac{3}{4}}}{b^{2} d^{3} x^{5} + a^{2} c^{3} +{\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{4} +{\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{3} +{\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{2} +{\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(3/2)/(d*x+c)^(9/4),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(3/2)/(d*x+c)**(9/4),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x + a\right )}^{\frac{3}{2}}{\left (d x + c\right )}^{\frac{9}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(3/2)/(d*x+c)^(9/4),x, algorithm="giac")

[Out]

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