\(\int \frac {(c+d x)^{5/4}}{\sqrt [4]{a+b x}} \, dx\) [1679]

Optimal result

Integrand size = 19, antiderivative size = 167 \[ \int \frac {(c+d x)^{5/4}}{\sqrt [4]{a+b x}} \, dx=\frac {5 (b c-a d) (a+b x)^{3/4} \sqrt [4]{c+d x}}{8 b^2}+\frac {(a+b x)^{3/4} (c+d x)^{5/4}}{2 b}-\frac {5 (b c-a d)^2 \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{9/4} d^{3/4}}+\frac {5 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{9/4} d^{3/4}} \]

[Out]

5/8*(-a*d+b*c)*(b*x+a)^(3/4)*(d*x+c)^(1/4)/b^2+1/2*(b*x+a)^(3/4)*(d*x+c)^(5/4)/b-5/16*(-a*d+b*c)^2*arctan(d^(1
/4)*(b*x+a)^(1/4)/b^(1/4)/(d*x+c)^(1/4))/b^(9/4)/d^(3/4)+5/16*(-a*d+b*c)^2*arctanh(d^(1/4)*(b*x+a)^(1/4)/b^(1/
4)/(d*x+c)^(1/4))/b^(9/4)/d^(3/4)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {52, 65, 338, 304, 211, 214} \[ \int \frac {(c+d x)^{5/4}}{\sqrt [4]{a+b x}} \, dx=-\frac {5 (b c-a d)^2 \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{9/4} d^{3/4}}+\frac {5 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{9/4} d^{3/4}}+\frac {5 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)}{8 b^2}+\frac {(a+b x)^{3/4} (c+d x)^{5/4}}{2 b} \]

[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 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*} \text {integral}& = \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 ) \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 )}{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 ) \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 )}{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 ) \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 )}{16 b^2 \sqrt {d}}-\frac {\left (5 (b c-a d)^2\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 )}{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 [A] (verified)

Time = 0.26 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.86 \[ \int \frac {(c+d x)^{5/4}}{\sqrt [4]{a+b x}} \, dx=\frac {2 \sqrt [4]{b} (a+b x)^{3/4} \sqrt [4]{c+d x} (9 b c-5 a d+4 b d x)+\frac {5 (b c-a d)^2 \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{d^{3/4}}+\frac {5 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{d^{3/4}}}{16 b^{9/4}} \]

[In]

Integrate[(c + d*x)^(5/4)/(a + b*x)^(1/4),x]

[Out]

(2*b^(1/4)*(a + b*x)^(3/4)*(c + d*x)^(1/4)*(9*b*c - 5*a*d + 4*b*d*x) + (5*(b*c - a*d)^2*ArcTan[(b^(1/4)*(c + d
*x)^(1/4))/(d^(1/4)*(a + b*x)^(1/4))])/d^(3/4) + (5*(b*c - a*d)^2*ArcTanh[(b^(1/4)*(c + d*x)^(1/4))/(d^(1/4)*(
a + b*x)^(1/4))])/d^(3/4))/(16*b^(9/4))

Maple [F]

\[\int \frac {\left (d x +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {1}{4}}}d x\]

[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)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 1219, normalized size of antiderivative = 7.30 \[ \int \frac {(c+d x)^{5/4}}{\sqrt [4]{a+b x}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/32*(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 - 5
6*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)) - 5*I*b^2*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 5
6*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) - (I*b^3*d*x + I*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*I*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) - (-I*b^3*d*x - I*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

Sympy [F]

\[ \int \frac {(c+d x)^{5/4}}{\sqrt [4]{a+b x}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{4}}}{\sqrt [4]{a + b x}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {(c+d x)^{5/4}}{\sqrt [4]{a+b x}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {1}{4}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {(c+d x)^{5/4}}{\sqrt [4]{a+b x}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {1}{4}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/4}}{\sqrt [4]{a+b x}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{1/4}} \,d x \]

[In]

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

[Out]

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