3.17.78 \(\int (a+b x)^{3/4} (c+d x)^{5/4} \, dx\) [1678]
Optimal. Leaf size=205 \[ \frac {5 (b c-a d)^2 (a+b x)^{3/4} \sqrt [4]{c+d x}}{96 b^2 d}+\frac {5 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}{24 b^2}+\frac {(a+b x)^{7/4} (c+d x)^{5/4}}{3 b}+\frac {5 (b c-a d)^3 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{9/4} d^{7/4}}-\frac {5 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{9/4} d^{7/4}} \]
[Out]
5/96*(-a*d+b*c)^2*(b*x+a)^(3/4)*(d*x+c)^(1/4)/b^2/d+5/24*(-a*d+b*c)*(b*x+a)^(7/4)*(d*x+c)^(1/4)/b^2+1/3*(b*x+a
)^(7/4)*(d*x+c)^(5/4)/b+5/64*(-a*d+b*c)^3*arctan(d^(1/4)*(b*x+a)^(1/4)/b^(1/4)/(d*x+c)^(1/4))/b^(9/4)/d^(7/4)-
5/64*(-a*d+b*c)^3*arctanh(d^(1/4)*(b*x+a)^(1/4)/b^(1/4)/(d*x+c)^(1/4))/b^(9/4)/d^(7/4)
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Rubi [A]
time = 0.10, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {52, 65, 338,
304, 211, 214} \begin {gather*} \frac {5 (b c-a d)^3 \text {ArcTan}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{9/4} d^{7/4}}-\frac {5 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{9/4} d^{7/4}}+\frac {5 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)^2}{96 b^2 d}+\frac {5 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)}{24 b^2}+\frac {(a+b x)^{7/4} (c+d x)^{5/4}}{3 b} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + b*x)^(3/4)*(c + d*x)^(5/4),x]
[Out]
(5*(b*c - a*d)^2*(a + b*x)^(3/4)*(c + d*x)^(1/4))/(96*b^2*d) + (5*(b*c - a*d)*(a + b*x)^(7/4)*(c + d*x)^(1/4))
/(24*b^2) + ((a + b*x)^(7/4)*(c + d*x)^(5/4))/(3*b) + (5*(b*c - a*d)^3*ArcTan[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/
4)*(c + d*x)^(1/4))])/(64*b^(9/4)*d^(7/4)) - (5*(b*c - a*d)^3*ArcTanh[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c +
d*x)^(1/4))])/(64*b^(9/4)*d^(7/4))
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 211
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rule 214
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]
Rule 304
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !
GtQ[a/b, 0]
Rule 338
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]
Rubi steps
\begin {align*} \int (a+b x)^{3/4} (c+d x)^{5/4} \, dx &=\frac {(a+b x)^{7/4} (c+d x)^{5/4}}{3 b}+\frac {(5 (b c-a d)) \int (a+b x)^{3/4} \sqrt [4]{c+d x} \, dx}{12 b}\\ &=\frac {5 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}{24 b^2}+\frac {(a+b x)^{7/4} (c+d x)^{5/4}}{3 b}+\frac {\left (5 (b c-a d)^2\right ) \int \frac {(a+b x)^{3/4}}{(c+d x)^{3/4}} \, dx}{96 b^2}\\ &=\frac {5 (b c-a d)^2 (a+b x)^{3/4} \sqrt [4]{c+d x}}{96 b^2 d}+\frac {5 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}{24 b^2}+\frac {(a+b x)^{7/4} (c+d x)^{5/4}}{3 b}-\frac {\left (5 (b c-a d)^3\right ) \int \frac {1}{\sqrt [4]{a+b x} (c+d x)^{3/4}} \, dx}{128 b^2 d}\\ &=\frac {5 (b c-a d)^2 (a+b x)^{3/4} \sqrt [4]{c+d x}}{96 b^2 d}+\frac {5 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}{24 b^2}+\frac {(a+b x)^{7/4} (c+d x)^{5/4}}{3 b}-\frac {\left (5 (b c-a d)^3\right ) \text {Subst}\left (\int \frac {x^2}{\left (c-\frac {a d}{b}+\frac {d x^4}{b}\right )^{3/4}} \, dx,x,\sqrt [4]{a+b x}\right )}{32 b^3 d}\\ &=\frac {5 (b c-a d)^2 (a+b x)^{3/4} \sqrt [4]{c+d x}}{96 b^2 d}+\frac {5 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}{24 b^2}+\frac {(a+b x)^{7/4} (c+d x)^{5/4}}{3 b}-\frac {\left (5 (b c-a d)^3\right ) \text {Subst}\left (\int \frac {x^2}{1-\frac {d x^4}{b}} \, dx,x,\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{32 b^3 d}\\ &=\frac {5 (b c-a d)^2 (a+b x)^{3/4} \sqrt [4]{c+d x}}{96 b^2 d}+\frac {5 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}{24 b^2}+\frac {(a+b x)^{7/4} (c+d x)^{5/4}}{3 b}-\frac {\left (5 (b c-a d)^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b}-\sqrt {d} x^2} \, dx,x,\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{64 b^2 d^{3/2}}+\frac {\left (5 (b c-a d)^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b}+\sqrt {d} x^2} \, dx,x,\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{64 b^2 d^{3/2}}\\ &=\frac {5 (b c-a d)^2 (a+b x)^{3/4} \sqrt [4]{c+d x}}{96 b^2 d}+\frac {5 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}{24 b^2}+\frac {(a+b x)^{7/4} (c+d x)^{5/4}}{3 b}+\frac {5 (b c-a d)^3 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{9/4} d^{7/4}}-\frac {5 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{64 b^{9/4} d^{7/4}}\\ \end {align*}
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Mathematica [A]
time = 0.55, size = 174, normalized size = 0.85 \begin {gather*} \frac {(b c-a d)^3 \left (\frac {2 \sqrt [4]{b} d^{3/4} (a+b x)^{3/4} \sqrt [4]{c+d x} \left (-15 a^2 d^2+6 a b d (7 c+2 d x)+b^2 \left (5 c^2+52 c d x+32 d^2 x^2\right )\right )}{(b c-a d)^3}-15 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )-15 \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )\right )}{192 b^{9/4} d^{7/4}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(a + b*x)^(3/4)*(c + d*x)^(5/4),x]
[Out]
((b*c - a*d)^3*((2*b^(1/4)*d^(3/4)*(a + b*x)^(3/4)*(c + d*x)^(1/4)*(-15*a^2*d^2 + 6*a*b*d*(7*c + 2*d*x) + b^2*
(5*c^2 + 52*c*d*x + 32*d^2*x^2)))/(b*c - a*d)^3 - 15*ArcTan[(b^(1/4)*(c + d*x)^(1/4))/(d^(1/4)*(a + b*x)^(1/4)
)] - 15*ArcTanh[(b^(1/4)*(c + d*x)^(1/4))/(d^(1/4)*(a + b*x)^(1/4))]))/(192*b^(9/4)*d^(7/4))
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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (b x +a \right )^{\frac {3}{4}} \left (d x +c \right )^{\frac {5}{4}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*x+a)^(3/4)*(d*x+c)^(5/4),x)
[Out]
int((b*x+a)^(3/4)*(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)^(3/4)*(d*x+c)^(5/4),x, algorithm="maxima")
[Out]
integrate((b*x + a)^(3/4)*(d*x + c)^(5/4), x)
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 2151 vs.
\(2 (159) = 318\).
time = 0.87, size = 2151, normalized size = 10.49 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(3/4)*(d*x+c)^(5/4),x, algorithm="fricas")
[Out]
-1/384*(60*b^2*d*((b^12*c^12 - 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8
*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6*b^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3
*d^9 + 66*a^10*b^2*c^2*d^10 - 12*a^11*b*c*d^11 + a^12*d^12)/(b^9*d^7))^(1/4)*arctan(((b^10*c^3*d^5 - 3*a*b^9*c
^2*d^6 + 3*a^2*b^8*c*d^7 - a^3*b^7*d^8)*(b*x + a)^(3/4)*(d*x + c)^(1/4)*((b^12*c^12 - 12*a*b^11*c^11*d + 66*a^
2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6*b^6*c^6*d^6 - 792*
a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 - 12*a^11*b*c*d^11 + a^12*d
^12)/(b^9*d^7))^(3/4) + (b^8*d^5*x + a*b^7*d^5)*sqrt(((b^6*c^6 - 6*a*b^5*c^5*d + 15*a^2*b^4*c^4*d^2 - 20*a^3*b
^3*c^3*d^3 + 15*a^4*b^2*c^2*d^4 - 6*a^5*b*c*d^5 + a^6*d^6)*sqrt(b*x + a)*sqrt(d*x + c) + (b^5*d^4*x + a*b^4*d^
4)*sqrt((b^12*c^12 - 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792
*a^5*b^7*c^7*d^5 + 924*a^6*b^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*
a^10*b^2*c^2*d^10 - 12*a^11*b*c*d^11 + a^12*d^12)/(b^9*d^7)))/(b*x + a))*((b^12*c^12 - 12*a*b^11*c^11*d + 66*a
^2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6*b^6*c^6*d^6 - 792
*a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 - 12*a^11*b*c*d^11 + a^12*
d^12)/(b^9*d^7))^(3/4))/(a*b^12*c^12 - 12*a^2*b^11*c^11*d + 66*a^3*b^10*c^10*d^2 - 220*a^4*b^9*c^9*d^3 + 495*a
^5*b^8*c^8*d^4 - 792*a^6*b^7*c^7*d^5 + 924*a^7*b^6*c^6*d^6 - 792*a^8*b^5*c^5*d^7 + 495*a^9*b^4*c^4*d^8 - 220*a
^10*b^3*c^3*d^9 + 66*a^11*b^2*c^2*d^10 - 12*a^12*b*c*d^11 + a^13*d^12 + (b^13*c^12 - 12*a*b^12*c^11*d + 66*a^2
*b^11*c^10*d^2 - 220*a^3*b^10*c^9*d^3 + 495*a^4*b^9*c^8*d^4 - 792*a^5*b^8*c^7*d^5 + 924*a^6*b^7*c^6*d^6 - 792*
a^7*b^6*c^5*d^7 + 495*a^8*b^5*c^4*d^8 - 220*a^9*b^4*c^3*d^9 + 66*a^10*b^3*c^2*d^10 - 12*a^11*b^2*c*d^11 + a^12
*b*d^12)*x)) + 15*b^2*d*((b^12*c^12 - 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*
b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6*b^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*
b^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 - 12*a^11*b*c*d^11 + a^12*d^12)/(b^9*d^7))^(1/4)*log(-5*((b^3*c^3 - 3*a*b^2
*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*(b*x + a)^(3/4)*(d*x + c)^(1/4) + (b^3*d^2*x + a*b^2*d^2)*((b^12*c^12 - 12*a
*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^
6*b^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 - 12*a^
11*b*c*d^11 + a^12*d^12)/(b^9*d^7))^(1/4))/(b*x + a)) - 15*b^2*d*((b^12*c^12 - 12*a*b^11*c^11*d + 66*a^2*b^10*
c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6*b^6*c^6*d^6 - 792*a^7*b^5
*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 - 12*a^11*b*c*d^11 + a^12*d^12)/(b
^9*d^7))^(1/4)*log(-5*((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*(b*x + a)^(3/4)*(d*x + c)^(1/4) - (
b^3*d^2*x + a*b^2*d^2)*((b^12*c^12 - 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b
^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6*b^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b
^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 - 12*a^11*b*c*d^11 + a^12*d^12)/(b^9*d^7))^(1/4))/(b*x + a)) - 4*(32*b^2*d^2
*x^2 + 5*b^2*c^2 + 42*a*b*c*d - 15*a^2*d^2 + 4*(13*b^2*c*d + 3*a*b*d^2)*x)*(b*x + a)^(3/4)*(d*x + c)^(1/4))/(b
^2*d)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b x\right )^{\frac {3}{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)**(3/4)*(d*x+c)**(5/4),x)
[Out]
Integral((a + b*x)**(3/4)*(c + d*x)**(5/4), x)
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(3/4)*(d*x+c)^(5/4),x, algorithm="giac")
[Out]
Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, need to
choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was done assuming
[sageVARa,sag
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+b\,x\right )}^{3/4}\,{\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)^(3/4)*(c + d*x)^(5/4),x)
[Out]
int((a + b*x)^(3/4)*(c + d*x)^(5/4), x)
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