3.1641 \(\int (a+b x)^{3/2} (c+d x)^{5/4} \, dx\)

Optimal. Leaf size=220 \[ \frac{16 (b c-a d)^{17/4} \sqrt{-\frac{d (a+b x)}{b c-a d}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{231 b^{9/4} d^3 \sqrt{a+b x}}-\frac{8 \sqrt{a+b x} \sqrt [4]{c+d x} (b c-a d)^3}{231 b^2 d^2}+\frac{4 (a+b x)^{3/2} \sqrt [4]{c+d x} (b c-a d)^2}{231 b^2 d}+\frac{4 (a+b x)^{5/2} \sqrt [4]{c+d x} (b c-a d)}{33 b^2}+\frac{4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b} \]

[Out]

(-8*(b*c - a*d)^3*Sqrt[a + b*x]*(c + d*x)^(1/4))/(231*b^2*d^2) + (4*(b*c - a*d)^2*(a + b*x)^(3/2)*(c + d*x)^(1
/4))/(231*b^2*d) + (4*(b*c - a*d)*(a + b*x)^(5/2)*(c + d*x)^(1/4))/(33*b^2) + (4*(a + b*x)^(5/2)*(c + d*x)^(5/
4))/(15*b) + (16*(b*c - a*d)^(17/4)*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])/(231*b^(9/4)*d^3*Sqrt[a + b*x])

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Rubi [A]  time = 0.168037, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {50, 63, 224, 221} \[ -\frac{8 \sqrt{a+b x} \sqrt [4]{c+d x} (b c-a d)^3}{231 b^2 d^2}+\frac{16 (b c-a d)^{17/4} \sqrt{-\frac{d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{231 b^{9/4} d^3 \sqrt{a+b x}}+\frac{4 (a+b x)^{3/2} \sqrt [4]{c+d x} (b c-a d)^2}{231 b^2 d}+\frac{4 (a+b x)^{5/2} \sqrt [4]{c+d x} (b c-a d)}{33 b^2}+\frac{4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(b*c - a*d)^3*Sqrt[a + b*x]*(c + d*x)^(1/4))/(231*b^2*d^2) + (4*(b*c - a*d)^2*(a + b*x)^(3/2)*(c + d*x)^(1
/4))/(231*b^2*d) + (4*(b*c - a*d)*(a + b*x)^(5/2)*(c + d*x)^(1/4))/(33*b^2) + (4*(a + b*x)^(5/2)*(c + d*x)^(5/
4))/(15*b) + (16*(b*c - a*d)^(17/4)*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])/(231*b^(9/4)*d^3*Sqrt[a + b*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 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 (a+b x)^{3/2} (c+d x)^{5/4} \, dx &=\frac{4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b}+\frac{(b c-a d) \int (a+b x)^{3/2} \sqrt [4]{c+d x} \, dx}{3 b}\\ &=\frac{4 (b c-a d) (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 b^2}+\frac{4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b}+\frac{(b c-a d)^2 \int \frac{(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx}{33 b^2}\\ &=\frac{4 (b c-a d)^2 (a+b x)^{3/2} \sqrt [4]{c+d x}}{231 b^2 d}+\frac{4 (b c-a d) (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 b^2}+\frac{4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b}-\frac{\left (2 (b c-a d)^3\right ) \int \frac{\sqrt{a+b x}}{(c+d x)^{3/4}} \, dx}{77 b^2 d}\\ &=-\frac{8 (b c-a d)^3 \sqrt{a+b x} \sqrt [4]{c+d x}}{231 b^2 d^2}+\frac{4 (b c-a d)^2 (a+b x)^{3/2} \sqrt [4]{c+d x}}{231 b^2 d}+\frac{4 (b c-a d) (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 b^2}+\frac{4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b}+\frac{\left (4 (b c-a d)^4\right ) \int \frac{1}{\sqrt{a+b x} (c+d x)^{3/4}} \, dx}{231 b^2 d^2}\\ &=-\frac{8 (b c-a d)^3 \sqrt{a+b x} \sqrt [4]{c+d x}}{231 b^2 d^2}+\frac{4 (b c-a d)^2 (a+b x)^{3/2} \sqrt [4]{c+d x}}{231 b^2 d}+\frac{4 (b c-a d) (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 b^2}+\frac{4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b}+\frac{\left (16 (b c-a d)^4\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 )}{231 b^2 d^3}\\ &=-\frac{8 (b c-a d)^3 \sqrt{a+b x} \sqrt [4]{c+d x}}{231 b^2 d^2}+\frac{4 (b c-a d)^2 (a+b x)^{3/2} \sqrt [4]{c+d x}}{231 b^2 d}+\frac{4 (b c-a d) (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 b^2}+\frac{4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b}+\frac{\left (16 (b c-a d)^4 \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 )}{231 b^2 d^3 \sqrt{a+b x}}\\ &=-\frac{8 (b c-a d)^3 \sqrt{a+b x} \sqrt [4]{c+d x}}{231 b^2 d^2}+\frac{4 (b c-a d)^2 (a+b x)^{3/2} \sqrt [4]{c+d x}}{231 b^2 d}+\frac{4 (b c-a d) (a+b x)^{5/2} \sqrt [4]{c+d x}}{33 b^2}+\frac{4 (a+b x)^{5/2} (c+d x)^{5/4}}{15 b}+\frac{16 (b c-a d)^{17/4} \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 )}{231 b^{9/4} d^3 \sqrt{a+b x}}\\ \end{align*}

Mathematica [C]  time = 0.0680344, size = 73, normalized size = 0.33 \[ \frac{2 (a+b x)^{5/2} (c+d x)^{5/4} \, _2F_1\left (-\frac{5}{4},\frac{5}{2};\frac{7}{2};\frac{d (a+b x)}{a d-b c}\right )}{5 b \left (\frac{b (c+d x)}{b c-a d}\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.022, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{{\frac{3}{2}}} \left ( dx+c \right ) ^{{\frac{5}{4}}}\, dx \end{align*}

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)

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

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)

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

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*d*x^2 + a*c + (b*c + a*d)*x)*sqrt(b*x + a)*(d*x + c)^(1/4), x)

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

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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