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

Optimal. Leaf size=167 \[ -\frac {5 (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{9/4} d^{3/4}}+\frac {5 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 b^{9/4} d^{3/4}}+\frac {5 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)}{8 b^2}+\frac {(a+b x)^{3/4} (c+d x)^{5/4}}{2 b} \]

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 167, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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]

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 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]

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 = 0.04, size = 73, normalized size = 0.44 \begin {gather*} \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}} \end {gather*}

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))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.38, size = 189, normalized size = 1.13 \begin {gather*} \frac {5 (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{16 b^{9/4} d^{3/4}}+\frac {5 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{d} \sqrt [4]{a+b x}}\right )}{16 b^{9/4} d^{3/4}}+\frac {(b c-a d)^2 \left (\frac {9 b (c+d x)^{5/4}}{(a+b x)^{5/4}}-\frac {5 d \sqrt [4]{c+d x}}{\sqrt [4]{a+b x}}\right )}{8 b^2 \left (\frac {b (c+d x)}{a+b x}-d\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*((-5*d*(c + d*x)^(1/4))/(a + b*x)^(1/4) + (9*b*(c + d*x)^(5/4))/(a + b*x)^(5/4)))/(8*b^2*(-d +
(b*(c + d*x))/(a + b*x))^2) + (5*(b*c - a*d)^2*ArcTan[(b^(1/4)*(c + d*x)^(1/4))/(d^(1/4)*(a + b*x)^(1/4))])/(1
6*b^(9/4)*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))])/(16*b^(9/4
)*d^(3/4))

________________________________________________________________________________________

fricas [B]  time = 1.34, size = 1468, normalized size = 8.79

result too large to display

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

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)

________________________________________________________________________________________

maple [F]  time = 0.04, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d x +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {1}{4}}}\, dx \end {gather*}

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)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c + d x\right )^{\frac {5}{4}}}{\sqrt [4]{a + b x}}\, dx \end {gather*}

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)

________________________________________________________________________________________