3.1629 \(\int (a+b x)^{3/2} \sqrt [4]{c+d x} \, dx\)

Optimal. Leaf size=185 \[ \frac {16 (b c-a d)^{13/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{77 b^{5/4} d^3 \sqrt {a+b x}}-\frac {8 \sqrt {a+b x} \sqrt [4]{c+d x} (b c-a d)^2}{77 b d^2}+\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x} (b c-a d)}{77 b d}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b} \]

[Out]

4/77*(-a*d+b*c)*(b*x+a)^(3/2)*(d*x+c)^(1/4)/b/d+4/11*(b*x+a)^(5/2)*(d*x+c)^(1/4)/b-8/77*(-a*d+b*c)^2*(d*x+c)^(
1/4)*(b*x+a)^(1/2)/b/d^2+16/77*(-a*d+b*c)^(13/4)*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^(5/4)/d^3/(b*x+a)^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {50, 63, 224, 221} \[ \frac {16 (b c-a d)^{13/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{77 b^{5/4} d^3 \sqrt {a+b x}}-\frac {8 \sqrt {a+b x} \sqrt [4]{c+d x} (b c-a d)^2}{77 b d^2}+\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x} (b c-a d)}{77 b d}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(b*c - a*d)^2*Sqrt[a + b*x]*(c + d*x)^(1/4))/(77*b*d^2) + (4*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(1/4))/
(77*b*d) + (4*(a + b*x)^(5/2)*(c + d*x)^(1/4))/(11*b) + (16*(b*c - a*d)^(13/4)*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])/(77*b^(5/4)*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 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]

Rubi steps

\begin {align*} \int (a+b x)^{3/2} \sqrt [4]{c+d x} \, dx &=\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}+\frac {(b c-a d) \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}} \, dx}{11 b}\\ &=\frac {4 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b d}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}-\frac {\left (6 (b c-a d)^2\right ) \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}} \, dx}{77 b d}\\ &=-\frac {8 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{77 b d^2}+\frac {4 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b d}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}+\frac {\left (4 (b c-a d)^3\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx}{77 b d^2}\\ &=-\frac {8 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{77 b d^2}+\frac {4 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b d}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}+\frac {\left (16 (b c-a d)^3\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 )}{77 b d^3}\\ &=-\frac {8 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{77 b d^2}+\frac {4 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b d}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}+\frac {\left (16 (b c-a d)^3 \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 )}{77 b d^3 \sqrt {a+b x}}\\ &=-\frac {8 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{77 b d^2}+\frac {4 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b d}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}+\frac {16 (b c-a d)^{13/4} \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 )}{77 b^{5/4} d^3 \sqrt {a+b x}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 73, normalized size = 0.39 \[ \frac {2 (a+b x)^{5/2} \sqrt [4]{c+d x} \, _2F_1\left (-\frac {1}{4},\frac {5}{2};\frac {7}{2};\frac {d (a+b x)}{a d-b c}\right )}{5 b \sqrt [4]{\frac {b (c+d x)}{b c-a d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)^(5/2)*(c + d*x)^(1/4)*Hypergeometric2F1[-1/4, 5/2, 7/2, (d*(a + b*x))/(-(b*c) + a*d)])/(5*b*((b*(
c + d*x))/(b*c - a*d))^(1/4))

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {1}{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {1}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {1}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {1}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{1/4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b x\right )^{\frac {3}{2}} \sqrt [4]{c + d x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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