\(\int \frac {(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx\) [1693]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.09 (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, 246, 218, 214, 211} \[ \int \frac {(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx=\frac {5 (b c-a d)^2 \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{3/4} d^{9/4}}+\frac {5 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{3/4} d^{9/4}}-\frac {5 \sqrt [4]{a+b x} (c+d x)^{3/4} (b c-a d)}{8 d^2}+\frac {(a+b x)^{5/4} (c+d x)^{3/4}}{2 d} \]

[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 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 218

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] &&  !Gt
Q[a/b, 0]

Rule 246

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]

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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 ) \text {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 ) \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 \sqrt {b} d^2}+\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 \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 [A] (verified)

Time = 0.33 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx=\frac {2 \sqrt [4]{d} \sqrt [4]{a+b x} (c+d x)^{3/4} (-5 b c+9 a d+4 b d x)-\frac {5 (b c-a d)^2 \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{b^{3/4}}+\frac {5 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{b^{3/4}}}{16 d^{9/4}} \]

[In]

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

[Out]

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

Maple [F]

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

[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)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 1219, normalized size of antiderivative = 7.30 \[ \int \frac {(a+b x)^{5/4}}{\sqrt [4]{c+d x}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/32*(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 - 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
^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)) - 5*I*d^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
^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) - (I*b*d^3*x + I*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*I*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) - (-I*b*d^3*x - I*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

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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