3.17.42 \(\int \sqrt {a+b x} (c+d x)^{5/4} \, dx\) [1642]

Optimal. Leaf size=182 \[ \frac {20 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{231 b^2 d}+\frac {20 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b^2}+\frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b}-\frac {40 (b c-a d)^{13/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right )\right |-1\right )}{231 b^{9/4} d^2 \sqrt {a+b x}} \]

[Out]

20/77*(-a*d+b*c)*(b*x+a)^(3/2)*(d*x+c)^(1/4)/b^2+4/11*(b*x+a)^(3/2)*(d*x+c)^(5/4)/b+20/231*(-a*d+b*c)^2*(d*x+c
)^(1/4)*(b*x+a)^(1/2)/b^2/d-40/231*(-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^(9/4)/d^2/(b*x+a)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 182, 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 = {52, 65, 230, 227} \begin {gather*} -\frac {40 (b c-a d)^{13/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} F\left (\left .\text {ArcSin}\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^2 \sqrt {a+b x}}+\frac {20 \sqrt {a+b x} \sqrt [4]{c+d x} (b c-a d)^2}{231 b^2 d}+\frac {20 (a+b x)^{3/2} \sqrt [4]{c+d x} (b c-a d)}{77 b^2}+\frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x]*(c + d*x)^(5/4),x]

[Out]

(20*(b*c - a*d)^2*Sqrt[a + b*x]*(c + d*x)^(1/4))/(231*b^2*d) + (20*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(1/4)
)/(77*b^2) + (4*(a + b*x)^(3/2)*(c + d*x)^(5/4))/(11*b) - (40*(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])/(231*b^(9/4)*d^2*Sqrt[a + b*x])

Rule 52

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 65

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 227

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 230

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 \sqrt {a+b x} (c+d x)^{5/4} \, dx &=\frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b}+\frac {(5 (b c-a d)) \int \sqrt {a+b x} \sqrt [4]{c+d x} \, dx}{11 b}\\ &=\frac {20 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b^2}+\frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b}+\frac {\left (5 (b c-a d)^2\right ) \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}} \, dx}{77 b^2}\\ &=\frac {20 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{231 b^2 d}+\frac {20 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b^2}+\frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b}-\frac {\left (10 (b c-a d)^3\right ) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}} \, dx}{231 b^2 d}\\ &=\frac {20 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{231 b^2 d}+\frac {20 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b^2}+\frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b}-\frac {\left (40 (b c-a d)^3\right ) \text {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^2}\\ &=\frac {20 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{231 b^2 d}+\frac {20 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b^2}+\frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b}-\frac {\left (40 (b c-a d)^3 \sqrt {\frac {d (a+b x)}{-b c+a d}}\right ) \text {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^2 \sqrt {a+b x}}\\ &=\frac {20 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{231 b^2 d}+\frac {20 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b^2}+\frac {4 (a+b x)^{3/2} (c+d x)^{5/4}}{11 b}-\frac {40 (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 )}{231 b^{9/4} d^2 \sqrt {a+b x}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.04, size = 73, normalized size = 0.40 \begin {gather*} \frac {2 (a+b x)^{3/2} (c+d x)^{5/4} \, _2F_1\left (-\frac {5}{4},\frac {3}{2};\frac {5}{2};\frac {d (a+b x)}{-b c+a d}\right )}{3 b \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \sqrt {b x +a}\, \left (d x +c \right )^{\frac {5}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(1/2)*(d*x+c)^(5/4),x)

[Out]

int((b*x+a)^(1/2)*(d*x+c)^(5/4),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)*(d*x+c)^(5/4),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)*(d*x+c)^(5/4),x, algorithm="fricas")

[Out]

integral(sqrt(b*x + a)*(d*x + c)^(5/4), x)

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Sympy [A]
time = 8.33, size = 218, normalized size = 1.20 \begin {gather*} - \frac {2 a d \left (a + b x\right )^{\frac {3}{2}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {a d e^{i \pi }}{b \operatorname {polar\_lift}{\left (- \frac {a d}{b} + c \right )}} + \frac {d x e^{i \pi }}{\operatorname {polar\_lift}{\left (- \frac {a d}{b} + c \right )}}} \right )} \sqrt [4]{\operatorname {polar\_lift}{\left (- \frac {a d}{b} + c \right )}}}{3 b^{2}} + \frac {2 c \left (a + b x\right )^{\frac {3}{2}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {a d e^{i \pi }}{b \operatorname {polar\_lift}{\left (- \frac {a d}{b} + c \right )}} + \frac {d x e^{i \pi }}{\operatorname {polar\_lift}{\left (- \frac {a d}{b} + c \right )}}} \right )} \sqrt [4]{\operatorname {polar\_lift}{\left (- \frac {a d}{b} + c \right )}}}{3 b} + \frac {2 d \left (a + b x\right )^{\frac {5}{2}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {a d e^{i \pi }}{b \operatorname {polar\_lift}{\left (- \frac {a d}{b} + c \right )}} + \frac {d x e^{i \pi }}{\operatorname {polar\_lift}{\left (- \frac {a d}{b} + c \right )}}} \right )} \sqrt [4]{\operatorname {polar\_lift}{\left (- \frac {a d}{b} + c \right )}}}{5 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(1/2)*(d*x+c)**(5/4),x)

[Out]

-2*a*d*(a + b*x)**(3/2)*hyper((-1/4, 3/2), (5/2,), a*d*exp_polar(I*pi)/(b*polar_lift(-a*d/b + c)) + d*x*exp_po
lar(I*pi)/polar_lift(-a*d/b + c))*polar_lift(-a*d/b + c)**(1/4)/(3*b**2) + 2*c*(a + b*x)**(3/2)*hyper((-1/4, 3
/2), (5/2,), a*d*exp_polar(I*pi)/(b*polar_lift(-a*d/b + c)) + d*x*exp_polar(I*pi)/polar_lift(-a*d/b + c))*pola
r_lift(-a*d/b + c)**(1/4)/(3*b) + 2*d*(a + b*x)**(5/2)*hyper((-1/4, 5/2), (7/2,), a*d*exp_polar(I*pi)/(b*polar
_lift(-a*d/b + c)) + d*x*exp_polar(I*pi)/polar_lift(-a*d/b + c))*polar_lift(-a*d/b + c)**(1/4)/(5*b**2)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)*(d*x+c)^(5/4),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{5/4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(1/2)*(c + d*x)^(5/4),x)

[Out]

int((a + b*x)^(1/2)*(c + d*x)^(5/4), x)

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