3.1693 \(\int \frac{(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx\)
Optimal. Leaf size=167 \[ \frac{5 (b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{3/4} d^{9/4}}+\frac{5 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{3/4} d^{9/4}}-\frac{5 \sqrt [4]{a+b x} (c+d x)^{3/4} (b c-a d)}{8 d^2}+\frac{(a+b x)^{5/4} (c+d x)^{3/4}}{2 d} \]
[Out]
(-5*(b*c - a*d)*(a + b*x)^(1/4)*(c + d*x)^(3/4))/(8*d^2) + ((a + b*x)^(5/4)*(c + d*x)^(3/4))/(2*d) + (5*(b*c -
a*d)^2*ArcTan[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(16*b^(3/4)*d^(9/4)) + (5*(b*c - a*d)^2*A
rcTanh[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(16*b^(3/4)*d^(9/4))
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Rubi [A] time = 0.102244, antiderivative size = 167, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used =
{50, 63, 240, 212, 208, 205} \[ \frac{5 (b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{3/4} d^{9/4}}+\frac{5 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{3/4} d^{9/4}}-\frac{5 \sqrt [4]{a+b x} (c+d x)^{3/4} (b c-a d)}{8 d^2}+\frac{(a+b x)^{5/4} (c+d x)^{3/4}}{2 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*x)^(5/4)/(c + d*x)^(1/4),x]
[Out]
(-5*(b*c - a*d)*(a + b*x)^(1/4)*(c + d*x)^(3/4))/(8*d^2) + ((a + b*x)^(5/4)*(c + d*x)^(3/4))/(2*d) + (5*(b*c -
a*d)^2*ArcTan[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(16*b^(3/4)*d^(9/4)) + (5*(b*c - a*d)^2*A
rcTanh[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(16*b^(3/4)*d^(9/4))
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 240
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 1/n]
Rule 212
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx &=\frac{(a+b x)^{5/4} (c+d x)^{3/4}}{2 d}-\frac{(5 (b c-a d)) \int \frac{\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx}{8 d}\\ &=-\frac{5 (b c-a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 d^2}+\frac{(a+b x)^{5/4} (c+d x)^{3/4}}{2 d}+\frac{\left (5 (b c-a d)^2\right ) \int \frac{1}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx}{32 d^2}\\ &=-\frac{5 (b c-a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 d^2}+\frac{(a+b x)^{5/4} (c+d x)^{3/4}}{2 d}+\frac{\left (5 (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{c-\frac{a d}{b}+\frac{d x^4}{b}}} \, dx,x,\sqrt [4]{a+b x}\right )}{8 b d^2}\\ &=-\frac{5 (b c-a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 d^2}+\frac{(a+b x)^{5/4} (c+d x)^{3/4}}{2 d}+\frac{\left (5 (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^4}{b}} \, dx,x,\frac{\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{8 b d^2}\\ &=-\frac{5 (b c-a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 d^2}+\frac{(a+b x)^{5/4} (c+d x)^{3/4}}{2 d}+\frac{\left (5 (b c-a d)^2\right ) \operatorname{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 )}{16 \sqrt{b} d^2}+\frac{\left (5 (b c-a d)^2\right ) \operatorname{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 )}{16 \sqrt{b} d^2}\\ &=-\frac{5 (b c-a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 d^2}+\frac{(a+b x)^{5/4} (c+d x)^{3/4}}{2 d}+\frac{5 (b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{3/4} d^{9/4}}+\frac{5 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{3/4} d^{9/4}}\\ \end{align*}
Mathematica [C] time = 0.0288506, size = 73, normalized size = 0.44 \[ \frac{4 (a+b x)^{9/4} \sqrt [4]{\frac{b (c+d x)}{b c-a d}} \, _2F_1\left (\frac{1}{4},\frac{9}{4};\frac{13}{4};\frac{d (a+b x)}{a d-b c}\right )}{9 b \sqrt [4]{c+d x}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*x)^(5/4)/(c + d*x)^(1/4),x]
[Out]
(4*(a + b*x)^(9/4)*((b*(c + d*x))/(b*c - a*d))^(1/4)*Hypergeometric2F1[1/4, 9/4, 13/4, (d*(a + b*x))/(-(b*c) +
a*d)])/(9*b*(c + d*x)^(1/4))
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Maple [F] time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{{\frac{5}{4}}}{\frac{1}{\sqrt [4]{dx+c}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*x+a)^(5/4)/(d*x+c)^(1/4),x)
[Out]
int((b*x+a)^(5/4)/(d*x+c)^(1/4),x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{5}{4}}}{{\left (d x + c\right )}^{\frac{1}{4}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(5/4)/(d*x+c)^(1/4),x, algorithm="maxima")
[Out]
integrate((b*x + a)^(5/4)/(d*x + c)^(1/4), x)
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Fricas [B] time = 3.19403, size = 3090, normalized size = 18.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(5/4)/(d*x+c)^(1/4),x, algorithm="fricas")
[Out]
-1/32*(20*d^2*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^
5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b^3*d^9))^(1/4)*arctan(-((b^4*c^2*d^7 - 2*a*b^3
*c*d^8 + a^2*b^2*d^9)*(b*x + a)^(1/4)*(d*x + c)^(3/4)*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*
b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b^3*d^9
))^(3/4) - (b^2*d^8*x + b^2*c*d^7)*sqrt(((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^
4)*sqrt(b*x + a)*sqrt(d*x + c) + (b^2*d^5*x + b^2*c*d^4)*sqrt((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 -
56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(
b^3*d^9)))/(d*x + c))*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4
- 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b^3*d^9))^(3/4))/(b^8*c^9 - 8*a*b^7*c^8
*d + 28*a^2*b^6*c^7*d^2 - 56*a^3*b^5*c^6*d^3 + 70*a^4*b^4*c^5*d^4 - 56*a^5*b^3*c^4*d^5 + 28*a^6*b^2*c^3*d^6 -
8*a^7*b*c^2*d^7 + a^8*c*d^8 + (b^8*c^8*d - 8*a*b^7*c^7*d^2 + 28*a^2*b^6*c^6*d^3 - 56*a^3*b^5*c^5*d^4 + 70*a^4*
b^4*c^4*d^5 - 56*a^5*b^3*c^3*d^6 + 28*a^6*b^2*c^2*d^7 - 8*a^7*b*c*d^8 + a^8*d^9)*x)) - 5*d^2*((b^8*c^8 - 8*a*b
^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*
d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b^3*d^9))^(1/4)*log(5*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(b*x + a)^(1/4)*(d*x +
c)^(3/4) + (b*d^3*x + b*c*d^2)*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^
4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b^3*d^9))^(1/4))/(d*x + c)) +
5*d^2*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^
3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b^3*d^9))^(1/4)*log(5*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*
(b*x + a)^(1/4)*(d*x + c)^(3/4) - (b*d^3*x + b*c*d^2)*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*
b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b^3*d^9
))^(1/4))/(d*x + c)) - 4*(4*b*d*x - 5*b*c + 9*a*d)*(b*x + a)^(1/4)*(d*x + c)^(3/4))/d^2
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{\frac{5}{4}}}{\sqrt [4]{c + d x}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)**(5/4)/(d*x+c)**(1/4),x)
[Out]
Integral((a + b*x)**(5/4)/(c + d*x)**(1/4), x)
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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(5/4)/(d*x+c)^(1/4),x, algorithm="giac")
[Out]
Timed out