3.17.86 \(\int (a+b x)^{5/4} (c+d x)^{5/4} \, dx\) [1686]

Optimal. Leaf size=408 \[ -\frac {5 (b c-a d)^3 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{84 b^2 d^2}+\frac {(b c-a d)^2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{42 b^2 d}+\frac {(b c-a d) (a+b x)^{9/4} \sqrt [4]{c+d x}}{7 b^2}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}+\frac {5 (b c-a d)^{9/2} ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2} \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt {\frac {(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac {2 \sqrt {b} \sqrt {d} \sqrt {(a+b x) (c+d x)}}{b c-a d}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt {b c-a d}}\right )|\frac {1}{2}\right )}{168 \sqrt {2} b^{9/4} d^{9/4} (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}} \]

[Out]

-5/84*(-a*d+b*c)^3*(b*x+a)^(1/4)*(d*x+c)^(1/4)/b^2/d^2+1/42*(-a*d+b*c)^2*(b*x+a)^(5/4)*(d*x+c)^(1/4)/b^2/d+1/7
*(-a*d+b*c)*(b*x+a)^(9/4)*(d*x+c)^(1/4)/b^2+2/7*(b*x+a)^(9/4)*(d*x+c)^(5/4)/b+5/336*(-a*d+b*c)^(9/2)*((b*x+a)*
(d*x+c))^(3/4)*(cos(2*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2)))^2)^(1/2)/cos(2
*arctan(b^(1/4)*d^(1/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2)))*EllipticF(sin(2*arctan(b^(1/4)*d^(1
/4)*((b*x+a)*(d*x+c))^(1/4)*2^(1/2)/(-a*d+b*c)^(1/2))),1/2*2^(1/2))*(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*x+c))^(1/
2)/(-a*d+b*c))*((2*b*d*x+a*d+b*c)^2)^(1/2)*((a*d+b*(2*d*x+c))^2/(-a*d+b*c)^2/(1+2*b^(1/2)*d^(1/2)*((b*x+a)*(d*
x+c))^(1/2)/(-a*d+b*c))^2)^(1/2)/b^(9/4)/d^(9/4)/(b*x+a)^(3/4)/(d*x+c)^(3/4)/(2*b*d*x+a*d+b*c)*2^(1/2)/((a*d+b
*(2*d*x+c))^2)^(1/2)

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Rubi [A]
time = 0.41, antiderivative size = 408, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {52, 64, 637, 226} \begin {gather*} \frac {5 (b c-a d)^{9/2} ((a+b x) (c+d x))^{3/4} \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 \text {ArcTan}\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 )}{168 \sqrt {2} b^{9/4} d^{9/4} (a+b x)^{3/4} (c+d x)^{3/4} (a d+b c+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}-\frac {5 \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c-a d)^3}{84 b^2 d^2}+\frac {(a+b x)^{5/4} \sqrt [4]{c+d x} (b c-a d)^2}{42 b^2 d}+\frac {(a+b x)^{9/4} \sqrt [4]{c+d x} (b c-a d)}{7 b^2}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(b*c - a*d)^3*(a + b*x)^(1/4)*(c + d*x)^(1/4))/(84*b^2*d^2) + ((b*c - a*d)^2*(a + b*x)^(5/4)*(c + d*x)^(1/
4))/(42*b^2*d) + ((b*c - a*d)*(a + b*x)^(9/4)*(c + d*x)^(1/4))/(7*b^2) + (2*(a + b*x)^(9/4)*(c + d*x)^(5/4))/(
7*b) + (5*(b*c - a*d)^(9/2)*((a + b*x)*(c + d*x))^(3/4)*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]*Sq
rt[(a + b*x)*(c + d*x)])/(b*c - a*d))^2)]*EllipticF[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])/(168*Sqrt[2]*b^(9/4)*d^(9/4)*(a + b*x)^(3/4)*(c + d*x)^(3/4)*(b*c + a*d + 2*b*d*x
)*Sqrt[(a*d + b*(c + 2*d*x))^2])

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 64

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 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 637

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]

Rubi steps

\begin {align*} \int (a+b x)^{5/4} (c+d x)^{5/4} \, dx &=\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}+\frac {(5 (b c-a d)) \int (a+b x)^{5/4} \sqrt [4]{c+d x} \, dx}{14 b}\\ &=\frac {(b c-a d) (a+b x)^{9/4} \sqrt [4]{c+d x}}{7 b^2}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}+\frac {(b c-a d)^2 \int \frac {(a+b x)^{5/4}}{(c+d x)^{3/4}} \, dx}{28 b^2}\\ &=\frac {(b c-a d)^2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{42 b^2 d}+\frac {(b c-a d) (a+b x)^{9/4} \sqrt [4]{c+d x}}{7 b^2}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}-\frac {\left (5 (b c-a d)^3\right ) \int \frac {\sqrt [4]{a+b x}}{(c+d x)^{3/4}} \, dx}{168 b^2 d}\\ &=-\frac {5 (b c-a d)^3 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{84 b^2 d^2}+\frac {(b c-a d)^2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{42 b^2 d}+\frac {(b c-a d) (a+b x)^{9/4} \sqrt [4]{c+d x}}{7 b^2}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}+\frac {\left (5 (b c-a d)^4\right ) \int \frac {1}{(a+b x)^{3/4} (c+d x)^{3/4}} \, dx}{336 b^2 d^2}\\ &=-\frac {5 (b c-a d)^3 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{84 b^2 d^2}+\frac {(b c-a d)^2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{42 b^2 d}+\frac {(b c-a d) (a+b x)^{9/4} \sqrt [4]{c+d x}}{7 b^2}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}+\frac {\left (5 (b c-a d)^4 ((a+b x) (c+d x))^{3/4}\right ) \int \frac {1}{\left (a c+(b c+a d) x+b d x^2\right )^{3/4}} \, dx}{336 b^2 d^2 (a+b x)^{3/4} (c+d x)^{3/4}}\\ &=-\frac {5 (b c-a d)^3 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{84 b^2 d^2}+\frac {(b c-a d)^2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{42 b^2 d}+\frac {(b c-a d) (a+b x)^{9/4} \sqrt [4]{c+d x}}{7 b^2}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}+\frac {\left (5 (b c-a d)^4 ((a+b x) (c+d x))^{3/4} \sqrt {(b c+a d+2 b d x)^2}\right ) \text {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 )}{84 b^2 d^2 (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x)}\\ &=-\frac {5 (b c-a d)^3 \sqrt [4]{a+b x} \sqrt [4]{c+d x}}{84 b^2 d^2}+\frac {(b c-a d)^2 (a+b x)^{5/4} \sqrt [4]{c+d x}}{42 b^2 d}+\frac {(b c-a d) (a+b x)^{9/4} \sqrt [4]{c+d x}}{7 b^2}+\frac {2 (a+b x)^{9/4} (c+d x)^{5/4}}{7 b}+\frac {5 (b c-a d)^{9/2} ((a+b x) (c+d x))^{3/4} \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 )}{168 \sqrt {2} b^{9/4} d^{9/4} (a+b x)^{3/4} (c+d x)^{3/4} (b c+a d+2 b d x) \sqrt {(a d+b (c+2 d x))^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.07, size = 73, normalized size = 0.18 \begin {gather*} \frac {4 (a+b x)^{9/4} (c+d x)^{5/4} \, _2F_1\left (-\frac {5}{4},\frac {9}{4};\frac {13}{4};\frac {d (a+b x)}{-b c+a d}\right )}{9 b \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a + b*x)^(9/4)*(c + d*x)^(5/4)*Hypergeometric2F1[-5/4, 9/4, 13/4, (d*(a + b*x))/(-(b*c) + a*d)])/(9*b*((b*
(c + d*x))/(b*c - a*d))^(5/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (b x +a \right )^{\frac {5}{4}} \left (d x +c \right )^{\frac {5}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x)**(5/4)*(c + d*x)**(5/4), x)

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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)^(5/4)*(d*x+c)^(5/4),x, algorithm="giac")

[Out]

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

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

[Out]

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

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