3.1671 \(\int \frac{(a+b x)^{7/2}}{(c+d x)^{9/4}} \, dx\)

Optimal. Leaf size=286 \[ -\frac{448 b^{5/4} (b c-a d)^{11/4} \sqrt{-\frac{d (a+b x)}{b c-a d}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{15 d^5 \sqrt{a+b x}}+\frac{112 b^2 (a+b x)^{3/2} (c+d x)^{3/4}}{9 d^3}-\frac{224 b^2 \sqrt{a+b x} (c+d x)^{3/4} (b c-a d)}{15 d^4}+\frac{448 b^{5/4} (b c-a d)^{11/4} \sqrt{-\frac{d (a+b x)}{b c-a d}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{15 d^5 \sqrt{a+b x}}-\frac{56 b (a+b x)^{5/2}}{5 d^2 \sqrt [4]{c+d x}}-\frac{4 (a+b x)^{7/2}}{5 d (c+d x)^{5/4}} \]

[Out]

(-4*(a + b*x)^(7/2))/(5*d*(c + d*x)^(5/4)) - (56*b*(a + b*x)^(5/2))/(5*d^2*(c + d*x)^(1/4)) - (224*b^2*(b*c -
a*d)*Sqrt[a + b*x]*(c + d*x)^(3/4))/(15*d^4) + (112*b^2*(a + b*x)^(3/2)*(c + d*x)^(3/4))/(9*d^3) + (448*b^(5/4
)*(b*c - a*d)^(11/4)*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])/(15*d^5*Sqrt[a + b*x]) - (448*b^(5/4)*(b*c - a*d)^(11/4)*Sqrt[-((d*(a + b*x))/(b*c - a*d))]*Elli
pticF[ArcSin[(b^(1/4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)], -1])/(15*d^5*Sqrt[a + b*x])

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Rubi [A]  time = 0.296989, antiderivative size = 286, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.474, Rules used = {47, 50, 63, 307, 224, 221, 1200, 1199, 424} \[ \frac{112 b^2 (a+b x)^{3/2} (c+d x)^{3/4}}{9 d^3}-\frac{224 b^2 \sqrt{a+b x} (c+d x)^{3/4} (b c-a d)}{15 d^4}-\frac{448 b^{5/4} (b c-a d)^{11/4} \sqrt{-\frac{d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{15 d^5 \sqrt{a+b x}}+\frac{448 b^{5/4} (b c-a d)^{11/4} \sqrt{-\frac{d (a+b x)}{b c-a d}} E\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{15 d^5 \sqrt{a+b x}}-\frac{56 b (a+b x)^{5/2}}{5 d^2 \sqrt [4]{c+d x}}-\frac{4 (a+b x)^{7/2}}{5 d (c+d x)^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(a + b*x)^(7/2))/(5*d*(c + d*x)^(5/4)) - (56*b*(a + b*x)^(5/2))/(5*d^2*(c + d*x)^(1/4)) - (224*b^2*(b*c -
a*d)*Sqrt[a + b*x]*(c + d*x)^(3/4))/(15*d^4) + (112*b^2*(a + b*x)^(3/2)*(c + d*x)^(3/4))/(9*d^3) + (448*b^(5/4
)*(b*c - a*d)^(11/4)*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])/(15*d^5*Sqrt[a + b*x]) - (448*b^(5/4)*(b*c - a*d)^(11/4)*Sqrt[-((d*(a + b*x))/(b*c - a*d))]*Elli
pticF[ArcSin[(b^(1/4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)], -1])/(15*d^5*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 50

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 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{(a+b x)^{7/2}}{(c+d x)^{9/4}} \, dx &=-\frac{4 (a+b x)^{7/2}}{5 d (c+d x)^{5/4}}+\frac{(14 b) \int \frac{(a+b x)^{5/2}}{(c+d x)^{5/4}} \, dx}{5 d}\\ &=-\frac{4 (a+b x)^{7/2}}{5 d (c+d x)^{5/4}}-\frac{56 b (a+b x)^{5/2}}{5 d^2 \sqrt [4]{c+d x}}+\frac{\left (28 b^2\right ) \int \frac{(a+b x)^{3/2}}{\sqrt [4]{c+d x}} \, dx}{d^2}\\ &=-\frac{4 (a+b x)^{7/2}}{5 d (c+d x)^{5/4}}-\frac{56 b (a+b x)^{5/2}}{5 d^2 \sqrt [4]{c+d x}}+\frac{112 b^2 (a+b x)^{3/2} (c+d x)^{3/4}}{9 d^3}-\frac{\left (56 b^2 (b c-a d)\right ) \int \frac{\sqrt{a+b x}}{\sqrt [4]{c+d x}} \, dx}{3 d^3}\\ &=-\frac{4 (a+b x)^{7/2}}{5 d (c+d x)^{5/4}}-\frac{56 b (a+b x)^{5/2}}{5 d^2 \sqrt [4]{c+d x}}-\frac{224 b^2 (b c-a d) \sqrt{a+b x} (c+d x)^{3/4}}{15 d^4}+\frac{112 b^2 (a+b x)^{3/2} (c+d x)^{3/4}}{9 d^3}+\frac{\left (112 b^2 (b c-a d)^2\right ) \int \frac{1}{\sqrt{a+b x} \sqrt [4]{c+d x}} \, dx}{15 d^4}\\ &=-\frac{4 (a+b x)^{7/2}}{5 d (c+d x)^{5/4}}-\frac{56 b (a+b x)^{5/2}}{5 d^2 \sqrt [4]{c+d x}}-\frac{224 b^2 (b c-a d) \sqrt{a+b x} (c+d x)^{3/4}}{15 d^4}+\frac{112 b^2 (a+b x)^{3/2} (c+d x)^{3/4}}{9 d^3}+\frac{\left (448 b^2 (b c-a d)^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 )}{15 d^5}\\ &=-\frac{4 (a+b x)^{7/2}}{5 d (c+d x)^{5/4}}-\frac{56 b (a+b x)^{5/2}}{5 d^2 \sqrt [4]{c+d x}}-\frac{224 b^2 (b c-a d) \sqrt{a+b x} (c+d x)^{3/4}}{15 d^4}+\frac{112 b^2 (a+b x)^{3/2} (c+d x)^{3/4}}{9 d^3}-\frac{\left (448 b^{3/2} (b c-a d)^{5/2}\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 )}{15 d^5}+\frac{\left (448 b^{3/2} (b c-a d)^{5/2}\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 )}{15 d^5}\\ &=-\frac{4 (a+b x)^{7/2}}{5 d (c+d x)^{5/4}}-\frac{56 b (a+b x)^{5/2}}{5 d^2 \sqrt [4]{c+d x}}-\frac{224 b^2 (b c-a d) \sqrt{a+b x} (c+d x)^{3/4}}{15 d^4}+\frac{112 b^2 (a+b x)^{3/2} (c+d x)^{3/4}}{9 d^3}-\frac{\left (448 b^{3/2} (b c-a d)^{5/2} \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 )}{15 d^5 \sqrt{a+b x}}+\frac{\left (448 b^{3/2} (b c-a d)^{5/2} \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 )}{15 d^5 \sqrt{a+b x}}\\ &=-\frac{4 (a+b x)^{7/2}}{5 d (c+d x)^{5/4}}-\frac{56 b (a+b x)^{5/2}}{5 d^2 \sqrt [4]{c+d x}}-\frac{224 b^2 (b c-a d) \sqrt{a+b x} (c+d x)^{3/4}}{15 d^4}+\frac{112 b^2 (a+b x)^{3/2} (c+d x)^{3/4}}{9 d^3}-\frac{448 b^{5/4} (b c-a d)^{11/4} \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 )}{15 d^5 \sqrt{a+b x}}+\frac{\left (448 b^{3/2} (b c-a d)^{5/2} \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 )}{15 d^5 \sqrt{a+b x}}\\ &=-\frac{4 (a+b x)^{7/2}}{5 d (c+d x)^{5/4}}-\frac{56 b (a+b x)^{5/2}}{5 d^2 \sqrt [4]{c+d x}}-\frac{224 b^2 (b c-a d) \sqrt{a+b x} (c+d x)^{3/4}}{15 d^4}+\frac{112 b^2 (a+b x)^{3/2} (c+d x)^{3/4}}{9 d^3}+\frac{448 b^{5/4} (b c-a d)^{11/4} \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 )}{15 d^5 \sqrt{a+b x}}-\frac{448 b^{5/4} (b c-a d)^{11/4} \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 )}{15 d^5 \sqrt{a+b x}}\\ \end{align*}

Mathematica [C]  time = 0.0877278, size = 73, normalized size = 0.26 \[ \frac{2 (a+b x)^{9/2} \left (\frac{b (c+d x)}{b c-a d}\right )^{9/4} \, _2F_1\left (\frac{9}{4},\frac{9}{2};\frac{11}{2};\frac{d (a+b x)}{a d-b c}\right )}{9 b (c+d x)^{9/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(9/2)*((b*(c + d*x))/(b*c - a*d))^(9/4)*Hypergeometric2F1[9/4, 9/2, 11/2, (d*(a + b*x))/(-(b*c) +
a*d)])/(9*b*(c + d*x)^(9/4))

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Maple [F]  time = 0.092, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{{\frac{7}{2}}} \left ( dx+c \right ) ^{-{\frac{9}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(7/2)/(d*x+c)^(9/4),x)

[Out]

int((b*x+a)^(7/2)/(d*x+c)^(9/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{7}{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((b*x+a)^(7/2)/(d*x+c)^(9/4),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(7/2)/(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{{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \sqrt{b x + a}{\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/2)/(d*x+c)^(9/4),x, algorithm="fricas")

[Out]

integral((b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*sqrt(b*x + a)*(d*x + c)^(3/4)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^
2*d*x + c^3), 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((b*x+a)**(7/2)/(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{{\left (b x + a\right )}^{\frac{7}{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((b*x+a)^(7/2)/(d*x+c)^(9/4),x, algorithm="giac")

[Out]

integrate((b*x + a)^(7/2)/(d*x + c)^(9/4), x)