3.1646 \(\int \frac{(c+d x)^{5/4}}{(a+b x)^{7/2}} \, dx\)

Optimal. Leaf size=175 \[ -\frac{d^2 \sqrt{-\frac{d (a+b x)}{b c-a d}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{6 b^{9/4} \sqrt{a+b x} (b c-a d)^{3/4}}-\frac{d^2 \sqrt [4]{c+d x}}{6 b^2 \sqrt{a+b x} (b c-a d)}-\frac{d \sqrt [4]{c+d x}}{3 b^2 (a+b x)^{3/2}}-\frac{2 (c+d x)^{5/4}}{5 b (a+b x)^{5/2}} \]

[Out]

-(d*(c + d*x)^(1/4))/(3*b^2*(a + b*x)^(3/2)) - (d^2*(c + d*x)^(1/4))/(6*b^2*(b*c - a*d)*Sqrt[a + b*x]) - (2*(c
+ d*x)^(5/4))/(5*b*(a + b*x)^(5/2)) - (d^2*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])/(6*b^(9/4)*(b*c - a*d)^(3/4)*Sqrt[a + b*x])

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Rubi [A]  time = 0.103906, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {47, 51, 63, 224, 221} \[ -\frac{d^2 \sqrt [4]{c+d x}}{6 b^2 \sqrt{a+b x} (b c-a d)}-\frac{d^2 \sqrt{-\frac{d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{6 b^{9/4} \sqrt{a+b x} (b c-a d)^{3/4}}-\frac{d \sqrt [4]{c+d x}}{3 b^2 (a+b x)^{3/2}}-\frac{2 (c+d x)^{5/4}}{5 b (a+b x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(c + d*x)^(1/4))/(3*b^2*(a + b*x)^(3/2)) - (d^2*(c + d*x)^(1/4))/(6*b^2*(b*c - a*d)*Sqrt[a + b*x]) - (2*(c
+ d*x)^(5/4))/(5*b*(a + b*x)^(5/2)) - (d^2*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])/(6*b^(9/4)*(b*c - a*d)^(3/4)*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 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 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]

Rubi steps

\begin{align*} \int \frac{(c+d x)^{5/4}}{(a+b x)^{7/2}} \, dx &=-\frac{2 (c+d x)^{5/4}}{5 b (a+b x)^{5/2}}+\frac{d \int \frac{\sqrt [4]{c+d x}}{(a+b x)^{5/2}} \, dx}{2 b}\\ &=-\frac{d \sqrt [4]{c+d x}}{3 b^2 (a+b x)^{3/2}}-\frac{2 (c+d x)^{5/4}}{5 b (a+b x)^{5/2}}+\frac{d^2 \int \frac{1}{(a+b x)^{3/2} (c+d x)^{3/4}} \, dx}{12 b^2}\\ &=-\frac{d \sqrt [4]{c+d x}}{3 b^2 (a+b x)^{3/2}}-\frac{d^2 \sqrt [4]{c+d x}}{6 b^2 (b c-a d) \sqrt{a+b x}}-\frac{2 (c+d x)^{5/4}}{5 b (a+b x)^{5/2}}-\frac{d^3 \int \frac{1}{\sqrt{a+b x} (c+d x)^{3/4}} \, dx}{24 b^2 (b c-a d)}\\ &=-\frac{d \sqrt [4]{c+d x}}{3 b^2 (a+b x)^{3/2}}-\frac{d^2 \sqrt [4]{c+d x}}{6 b^2 (b c-a d) \sqrt{a+b x}}-\frac{2 (c+d x)^{5/4}}{5 b (a+b x)^{5/2}}-\frac{d^2 \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 )}{6 b^2 (b c-a d)}\\ &=-\frac{d \sqrt [4]{c+d x}}{3 b^2 (a+b x)^{3/2}}-\frac{d^2 \sqrt [4]{c+d x}}{6 b^2 (b c-a d) \sqrt{a+b x}}-\frac{2 (c+d x)^{5/4}}{5 b (a+b x)^{5/2}}-\frac{\left (d^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 )}{6 b^2 (b c-a d) \sqrt{a+b x}}\\ &=-\frac{d \sqrt [4]{c+d x}}{3 b^2 (a+b x)^{3/2}}-\frac{d^2 \sqrt [4]{c+d x}}{6 b^2 (b c-a d) \sqrt{a+b x}}-\frac{2 (c+d x)^{5/4}}{5 b (a+b x)^{5/2}}-\frac{d^2 \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 )}{6 b^{9/4} (b c-a d)^{3/4} \sqrt{a+b x}}\\ \end{align*}

Mathematica [C]  time = 0.046409, size = 73, normalized size = 0.42 \[ -\frac{2 (c+d x)^{5/4} \, _2F_1\left (-\frac{5}{2},-\frac{5}{4};-\frac{3}{2};\frac{d (a+b x)}{a d-b c}\right )}{5 b (a+b x)^{5/2} \left (\frac{b (c+d x)}{b c-a d}\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.04, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dx+c \right ) ^{{\frac{5}{4}}} \left ( bx+a \right ) ^{-{\frac{7}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/4)/(b*x+a)^(7/2),x)

[Out]

int((d*x+c)^(5/4)/(b*x+a)^(7/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{\frac{5}{4}}}{{\left (b x + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(7/2),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(5/4)/(b*x + a)^(7/2), 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{5}{4}}}{b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x + a)*(d*x + c)^(5/4)/(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4), 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((d*x+c)**(5/4)/(b*x+a)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{\frac{5}{4}}}{{\left (b x + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/4)/(b*x+a)^(7/2),x, algorithm="giac")

[Out]

integrate((d*x + c)^(5/4)/(b*x + a)^(7/2), x)