3.1636 \(\int \sqrt {a+b x} (c+d x)^{3/4} \, dx\)

Optimal. Leaf size=232 \[ \frac {8 (b c-a d)^{11/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{15 b^{7/4} d^2 \sqrt {a+b x}}-\frac {8 (b c-a d)^{11/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{15 b^{7/4} d^2 \sqrt {a+b x}}+\frac {4 \sqrt {a+b x} (c+d x)^{3/4} (b c-a d)}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b} \]

[Out]

4/9*(b*x+a)^(3/2)*(d*x+c)^(3/4)/b+4/15*(-a*d+b*c)*(d*x+c)^(3/4)*(b*x+a)^(1/2)/b/d-8/15*(-a*d+b*c)^(11/4)*Ellip
ticE(b^(1/4)*(d*x+c)^(1/4)/(-a*d+b*c)^(1/4),I)*(-d*(b*x+a)/(-a*d+b*c))^(1/2)/b^(7/4)/d^2/(b*x+a)^(1/2)+8/15*(-
a*d+b*c)^(11/4)*EllipticF(b^(1/4)*(d*x+c)^(1/4)/(-a*d+b*c)^(1/4),I)*(-d*(b*x+a)/(-a*d+b*c))^(1/2)/b^(7/4)/d^2/
(b*x+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {50, 63, 307, 224, 221, 1200, 1199, 424} \[ \frac {8 (b c-a d)^{11/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{15 b^{7/4} d^2 \sqrt {a+b x}}-\frac {8 (b c-a d)^{11/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{15 b^{7/4} d^2 \sqrt {a+b x}}+\frac {4 \sqrt {a+b x} (c+d x)^{3/4} (b c-a d)}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x]*(c + d*x)^(3/4),x]

[Out]

(4*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/4))/(15*b*d) + (4*(a + b*x)^(3/2)*(c + d*x)^(3/4))/(9*b) - (8*(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*b^(7/4)*d^2*Sqrt[a + b*x]) + (8*(b*c - a*d)^(11/4)*Sqrt[-((d*(a + b*x))/(b*c - a*d))]*EllipticF[ArcS
in[(b^(1/4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)], -1])/(15*b^(7/4)*d^2*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 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 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 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 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]

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 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]

Rubi steps

\begin {align*} \int \sqrt {a+b x} (c+d x)^{3/4} \, dx &=\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b}+\frac {(b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt [4]{c+d x}} \, dx}{3 b}\\ &=\frac {4 (b c-a d) \sqrt {a+b x} (c+d x)^{3/4}}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b}-\frac {\left (2 (b c-a d)^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt [4]{c+d x}} \, dx}{15 b d}\\ &=\frac {4 (b c-a d) \sqrt {a+b x} (c+d x)^{3/4}}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b}-\frac {\left (8 (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 b d^2}\\ &=\frac {4 (b c-a d) \sqrt {a+b x} (c+d x)^{3/4}}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b}+\frac {\left (8 (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 b^{3/2} d^2}-\frac {\left (8 (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 b^{3/2} d^2}\\ &=\frac {4 (b c-a d) \sqrt {a+b x} (c+d x)^{3/4}}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b}+\frac {\left (8 (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 b^{3/2} d^2 \sqrt {a+b x}}-\frac {\left (8 (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 b^{3/2} d^2 \sqrt {a+b x}}\\ &=\frac {4 (b c-a d) \sqrt {a+b x} (c+d x)^{3/4}}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b}+\frac {8 (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 b^{7/4} d^2 \sqrt {a+b x}}-\frac {\left (8 (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 b^{3/2} d^2 \sqrt {a+b x}}\\ &=\frac {4 (b c-a d) \sqrt {a+b x} (c+d x)^{3/4}}{15 b d}+\frac {4 (a+b x)^{3/2} (c+d x)^{3/4}}{9 b}-\frac {8 (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 b^{7/4} d^2 \sqrt {a+b x}}+\frac {8 (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 b^{7/4} d^2 \sqrt {a+b x}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.03, size = 73, normalized size = 0.31 \[ \frac {2 (a+b x)^{3/2} (c+d x)^{3/4} \, _2F_1\left (-\frac {3}{4},\frac {3}{2};\frac {5}{2};\frac {d (a+b x)}{a d-b c}\right )}{3 b \left (\frac {b (c+d x)}{b c-a d}\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \sqrt {b x +a}\, \left (d x +c \right )^{\frac {3}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{3/4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b x} \left (c + d x\right )^{\frac {3}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________