3.1679 \(\int \frac{(c+d x)^{5/4}}{\sqrt [4]{a+b x}} \, dx\)
Optimal. Leaf size=167 \[ -\frac{5 (b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{9/4} d^{3/4}}+\frac{5 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{9/4} d^{3/4}}+\frac{5 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)}{8 b^2}+\frac{(a+b x)^{3/4} (c+d x)^{5/4}}{2 b} \]
[Out]
(5*(b*c - a*d)*(a + b*x)^(3/4)*(c + d*x)^(1/4))/(8*b^2) + ((a + b*x)^(3/4)*(c + d*x)^(5/4))/(2*b) - (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^(9/4)*d^(3/4)) + (5*(b*c - a*d)^2*Ar
cTanh[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(16*b^(9/4)*d^(3/4))
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Rubi [A] time = 0.102389, 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, 331, 298, 205, 208} \[ -\frac{5 (b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{9/4} d^{3/4}}+\frac{5 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{9/4} d^{3/4}}+\frac{5 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)}{8 b^2}+\frac{(a+b x)^{3/4} (c+d x)^{5/4}}{2 b} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x)^(5/4)/(a + b*x)^(1/4),x]
[Out]
(5*(b*c - a*d)*(a + b*x)^(3/4)*(c + d*x)^(1/4))/(8*b^2) + ((a + b*x)^(3/4)*(c + d*x)^(5/4))/(2*b) - (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^(9/4)*d^(3/4)) + (5*(b*c - a*d)^2*Ar
cTanh[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(16*b^(9/4)*d^(3/4))
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 331
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]
Rule 298
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 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]
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]
Rubi steps
\begin{align*} \int \frac{(c+d x)^{5/4}}{\sqrt [4]{a+b x}} \, dx &=\frac{(a+b x)^{3/4} (c+d x)^{5/4}}{2 b}+\frac{(5 (b c-a d)) \int \frac{\sqrt [4]{c+d x}}{\sqrt [4]{a+b x}} \, dx}{8 b}\\ &=\frac{5 (b c-a d) (a+b x)^{3/4} \sqrt [4]{c+d x}}{8 b^2}+\frac{(a+b x)^{3/4} (c+d x)^{5/4}}{2 b}+\frac{\left (5 (b c-a d)^2\right ) \int \frac{1}{\sqrt [4]{a+b x} (c+d x)^{3/4}} \, dx}{32 b^2}\\ &=\frac{5 (b c-a d) (a+b x)^{3/4} \sqrt [4]{c+d x}}{8 b^2}+\frac{(a+b x)^{3/4} (c+d x)^{5/4}}{2 b}+\frac{\left (5 (b c-a d)^2\right ) \operatorname{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 )}{8 b^3}\\ &=\frac{5 (b c-a d) (a+b x)^{3/4} \sqrt [4]{c+d x}}{8 b^2}+\frac{(a+b x)^{3/4} (c+d x)^{5/4}}{2 b}+\frac{\left (5 (b c-a d)^2\right ) \operatorname{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 )}{8 b^3}\\ &=\frac{5 (b c-a d) (a+b x)^{3/4} \sqrt [4]{c+d x}}{8 b^2}+\frac{(a+b x)^{3/4} (c+d x)^{5/4}}{2 b}+\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 b^2 \sqrt{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 b^2 \sqrt{d}}\\ &=\frac{5 (b c-a d) (a+b x)^{3/4} \sqrt [4]{c+d x}}{8 b^2}+\frac{(a+b x)^{3/4} (c+d x)^{5/4}}{2 b}-\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^{9/4} d^{3/4}}+\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^{9/4} d^{3/4}}\\ \end{align*}
Mathematica [C] time = 0.0438787, size = 73, normalized size = 0.44 \[ \frac{4 (a+b x)^{3/4} (c+d x)^{5/4} \, _2F_1\left (-\frac{5}{4},\frac{3}{4};\frac{7}{4};\frac{d (a+b x)}{a d-b c}\right )}{3 b \left (\frac{b (c+d x)}{b c-a d}\right )^{5/4}} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x)^(5/4)/(a + b*x)^(1/4),x]
[Out]
(4*(a + b*x)^(3/4)*(c + d*x)^(5/4)*Hypergeometric2F1[-5/4, 3/4, 7/4, (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 = 0.017, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dx+c \right ) ^{{\frac{5}{4}}}{\frac{1}{\sqrt [4]{bx+a}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*x+c)^(5/4)/(b*x+a)^(1/4),x)
[Out]
int((d*x+c)^(5/4)/(b*x+a)^(1/4),x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{\frac{5}{4}}}{{\left (b x + a\right )}^{\frac{1}{4}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^(5/4)/(b*x+a)^(1/4),x, algorithm="maxima")
[Out]
integrate((d*x + c)^(5/4)/(b*x + a)^(1/4), x)
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Fricas [B] time = 3.64042, 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((d*x+c)^(5/4)/(b*x+a)^(1/4),x, algorithm="fricas")
[Out]
-1/32*(20*b^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^9*d^3))^(1/4)*arctan(-((b^9*c^2*d^2 - 2*a*b^8
*c*d^3 + a^2*b^7*d^4)*(b*x + a)^(3/4)*(d*x + c)^(1/4)*((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^9*d^3
))^(3/4) - (b^8*d^2*x + a*b^7*d^2)*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^5*d^2*x + a*b^4*d^2)*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^9*d^3)))/(b*x + a))*((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^9*d^3))^(3/4))/(a*b^8*c^8 - 8*a^2*b^7
*c^7*d + 28*a^3*b^6*c^6*d^2 - 56*a^4*b^5*c^5*d^3 + 70*a^5*b^4*c^4*d^4 - 56*a^6*b^3*c^3*d^5 + 28*a^7*b^2*c^2*d^
6 - 8*a^8*b*c*d^7 + a^9*d^8 + (b^9*c^8 - 8*a*b^8*c^7*d + 28*a^2*b^7*c^6*d^2 - 56*a^3*b^6*c^5*d^3 + 70*a^4*b^5*
c^4*d^4 - 56*a^5*b^4*c^3*d^5 + 28*a^6*b^3*c^2*d^6 - 8*a^7*b^2*c*d^7 + a^8*b*d^8)*x)) - 5*b^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^9*d^3))^(1/4)*log(5*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(b*x + a)^(3/4)*(d*x +
c)^(1/4) + (b^3*d*x + a*b^2*d)*((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^9*d^3))^(1/4))/(b*x + a)) +
5*b^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^9*d^3))^(1/4)*log(5*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*
(b*x + a)^(3/4)*(d*x + c)^(1/4) - (b^3*d*x + a*b^2*d)*((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^9*d^3
))^(1/4))/(b*x + a)) - 4*(4*b*d*x + 9*b*c - 5*a*d)*(b*x + a)^(3/4)*(d*x + c)^(1/4))/b^2
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d x\right )^{\frac{5}{4}}}{\sqrt [4]{a + b x}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)**(5/4)/(b*x+a)**(1/4),x)
[Out]
Integral((c + d*x)**(5/4)/(a + b*x)**(1/4), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{\frac{5}{4}}}{{\left (b x + a\right )}^{\frac{1}{4}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^(5/4)/(b*x+a)^(1/4),x, algorithm="giac")
[Out]
integrate((d*x + c)^(5/4)/(b*x + a)^(1/4), x)