3.1701 \(\int \frac{(a+b x)^{3/4}}{\sqrt [4]{c+d x}} \, dx\)

Optimal. Leaf size=705 \[ -\frac{(b c-a d)^{5/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right ),\frac{1}{2}\right )}{2 \sqrt{2} b^{3/4} d^{7/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}+\frac{(b c-a d)^{5/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{\sqrt{2} b^{3/4} d^{7/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}-\frac{\sqrt{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \sqrt{(a d+b (c+2 d x))^2}}{\sqrt{b} d^{3/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )}+\frac{2 (a+b x)^{3/4} (c+d x)^{3/4}}{3 d} \]

[Out]

(2*(a + b*x)^(3/4)*(c + d*x)^(3/4))/(3*d) - (Sqrt[(a + b*x)*(c + d*x)]*Sqrt[(b*c + a*d + 2*b*d*x)^2]*Sqrt[(a*d
+ b*(c + 2*d*x))^2])/(Sqrt[b]*d^(3/2)*(a + b*x)^(1/4)*(c + d*x)^(1/4)*(b*c + a*d + 2*b*d*x)*(1 + (2*Sqrt[b]*S
qrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))) + ((b*c - a*d)^(5/2)*((a + b*x)*(c + d*x))^(1/4)*Sqrt[(b*c + a
*d + 2*b*d*x)^2]*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))*Sqrt[(a*d + b*(c + 2*d*x))^2/
((b*c - a*d)^2*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))^2)]*EllipticE[2*ArcTan[(Sqrt[2]
*b^(1/4)*d^(1/4)*((a + b*x)*(c + d*x))^(1/4))/Sqrt[b*c - a*d]], 1/2])/(Sqrt[2]*b^(3/4)*d^(7/4)*(a + b*x)^(1/4)
*(c + d*x)^(1/4)*(b*c + a*d + 2*b*d*x)*Sqrt[(a*d + b*(c + 2*d*x))^2]) - ((b*c - a*d)^(5/2)*((a + b*x)*(c + d*x
))^(1/4)*Sqrt[(b*c + a*d + 2*b*d*x)^2]*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))*Sqrt[(a
*d + b*(c + 2*d*x))^2/((b*c - a*d)^2*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))^2)]*Ellip
ticF[2*ArcTan[(Sqrt[2]*b^(1/4)*d^(1/4)*((a + b*x)*(c + d*x))^(1/4))/Sqrt[b*c - a*d]], 1/2])/(2*Sqrt[2]*b^(3/4)
*d^(7/4)*(a + b*x)^(1/4)*(c + d*x)^(1/4)*(b*c + a*d + 2*b*d*x)*Sqrt[(a*d + b*(c + 2*d*x))^2])

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Rubi [A]  time = 0.616858, antiderivative size = 705, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {50, 62, 623, 305, 220, 1196} \[ -\frac{(b c-a d)^{5/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{2 \sqrt{2} b^{3/4} d^{7/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}+\frac{(b c-a d)^{5/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{\sqrt{2} b^{3/4} d^{7/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}-\frac{\sqrt{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \sqrt{(a d+b (c+2 d x))^2}}{\sqrt{b} d^{3/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )}+\frac{2 (a+b x)^{3/4} (c+d x)^{3/4}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(3/4)*(c + d*x)^(3/4))/(3*d) - (Sqrt[(a + b*x)*(c + d*x)]*Sqrt[(b*c + a*d + 2*b*d*x)^2]*Sqrt[(a*d
+ b*(c + 2*d*x))^2])/(Sqrt[b]*d^(3/2)*(a + b*x)^(1/4)*(c + d*x)^(1/4)*(b*c + a*d + 2*b*d*x)*(1 + (2*Sqrt[b]*S
qrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))) + ((b*c - a*d)^(5/2)*((a + b*x)*(c + d*x))^(1/4)*Sqrt[(b*c + a
*d + 2*b*d*x)^2]*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))*Sqrt[(a*d + b*(c + 2*d*x))^2/
((b*c - a*d)^2*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))^2)]*EllipticE[2*ArcTan[(Sqrt[2]
*b^(1/4)*d^(1/4)*((a + b*x)*(c + d*x))^(1/4))/Sqrt[b*c - a*d]], 1/2])/(Sqrt[2]*b^(3/4)*d^(7/4)*(a + b*x)^(1/4)
*(c + d*x)^(1/4)*(b*c + a*d + 2*b*d*x)*Sqrt[(a*d + b*(c + 2*d*x))^2]) - ((b*c - a*d)^(5/2)*((a + b*x)*(c + d*x
))^(1/4)*Sqrt[(b*c + a*d + 2*b*d*x)^2]*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))*Sqrt[(a
*d + b*(c + 2*d*x))^2/((b*c - a*d)^2*(1 + (2*Sqrt[b]*Sqrt[d]*Sqrt[(a + b*x)*(c + d*x)])/(b*c - a*d))^2)]*Ellip
ticF[2*ArcTan[(Sqrt[2]*b^(1/4)*d^(1/4)*((a + b*x)*(c + d*x))^(1/4))/Sqrt[b*c - a*d]], 1/2])/(2*Sqrt[2]*b^(3/4)
*d^(7/4)*(a + b*x)^(1/4)*(c + d*x)^(1/4)*(b*c + a*d + 2*b*d*x)*Sqrt[(a*d + b*(c + 2*d*x))^2])

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 62

Int[((a_.) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Dist[((a + b*x)^m*(c + d*x)^m)/((a + b*x)
*(c + d*x))^m, Int[(a*c + (b*c + a*d)*x + b*d*x^2)^m, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] &&
LtQ[-1, m, 0] && LeQ[3, Denominator[m], 4]

Rule 623

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{d = Denominator[p]}, Dist[(d*Sqrt[(b + 2*c*x)
^2])/(b + 2*c*x), Subst[Int[x^(d*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4*c*x^d], x], x, (a + b*x + c*x^2)^(1/d)], x]
/; 3 <= d <= 4] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && RationalQ[p]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, 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] && PosQ[b/a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{(a+b x)^{3/4}}{\sqrt [4]{c+d x}} \, dx &=\frac{2 (a+b x)^{3/4} (c+d x)^{3/4}}{3 d}-\frac{(b c-a d) \int \frac{1}{\sqrt [4]{a+b x} \sqrt [4]{c+d x}} \, dx}{2 d}\\ &=\frac{2 (a+b x)^{3/4} (c+d x)^{3/4}}{3 d}-\frac{\left ((b c-a d) \sqrt [4]{(a+b x) (c+d x)}\right ) \int \frac{1}{\sqrt [4]{a c+(b c+a d) x+b d x^2}} \, dx}{2 d \sqrt [4]{a+b x} \sqrt [4]{c+d x}}\\ &=\frac{2 (a+b x)^{3/4} (c+d x)^{3/4}}{3 d}-\frac{\left (2 (b c-a d) \sqrt [4]{(a+b x) (c+d x)} \sqrt{(b c+a d+2 b d x)^2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{-4 a b c d+(b c+a d)^2+4 b d x^4}} \, dx,x,\sqrt [4]{(a+b x) (c+d x)}\right )}{d \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}\\ &=\frac{2 (a+b x)^{3/4} (c+d x)^{3/4}}{3 d}-\frac{\left ((b c-a d)^2 \sqrt [4]{(a+b x) (c+d x)} \sqrt{(b c+a d+2 b d x)^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-4 a b c d+(b c+a d)^2+4 b d x^4}} \, dx,x,\sqrt [4]{(a+b x) (c+d x)}\right )}{\sqrt{b} d^{3/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}+\frac{\left ((b c-a d)^2 \sqrt [4]{(a+b x) (c+d x)} \sqrt{(b c+a d+2 b d x)^2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{2 \sqrt{b} \sqrt{d} x^2}{b c-a d}}{\sqrt{-4 a b c d+(b c+a d)^2+4 b d x^4}} \, dx,x,\sqrt [4]{(a+b x) (c+d x)}\right )}{\sqrt{b} d^{3/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}\\ &=\frac{2 (a+b x)^{3/4} (c+d x)^{3/4}}{3 d}-\frac{\sqrt{(a+b x) (c+d x)} \sqrt{(b c+a d+2 b d x)^2} \sqrt{(a d+b (c+2 d x))^2}}{\sqrt{b} d^{3/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \left (1+\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}\right )}+\frac{(b c-a d)^{5/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(b c+a d+2 b d x)^2} \left (1+\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{\sqrt{2} b^{3/4} d^{7/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}-\frac{(b c-a d)^{5/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(b c+a d+2 b d x)^2} \left (1+\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{2 \sqrt{2} b^{3/4} d^{7/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}\\ \end{align*}

Mathematica [C]  time = 0.0255792, size = 73, normalized size = 0.1 \[ \frac{4 (a+b x)^{7/4} \sqrt [4]{\frac{b (c+d x)}{b c-a d}} \, _2F_1\left (\frac{1}{4},\frac{7}{4};\frac{11}{4};\frac{d (a+b x)}{a d-b c}\right )}{7 b \sqrt [4]{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a + b*x)^(7/4)*((b*(c + d*x))/(b*c - a*d))^(1/4)*Hypergeometric2F1[1/4, 7/4, 11/4, (d*(a + b*x))/(-(b*c) +
a*d)])/(7*b*(c + d*x)^(1/4))

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Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{{\frac{3}{4}}}{\frac{1}{\sqrt [4]{dx+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{\frac{3}{4}}}{{\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)^(3/4)/(d*x+c)^(1/4),x, algorithm="fricas")

[Out]

integral((b*x + a)^(3/4)/(d*x + c)^(1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x)**(3/4)/(c + d*x)**(1/4), x)

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Giac [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)^(3/4)/(d*x+c)^(1/4),x, algorithm="giac")

[Out]

Timed out