3.16.4 \(\int \frac {(a+b x)^{5/4}}{(c+d x)^{5/4}} \, dx\)

Optimal. Leaf size=152 \[ -\frac {5 \sqrt [4]{b} (b c-a d) \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 d^{9/4}}-\frac {5 \sqrt [4]{b} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 d^{9/4}}+\frac {5 b \sqrt [4]{a+b x} (c+d x)^{3/4}}{d^2}-\frac {4 (a+b x)^{5/4}}{d \sqrt [4]{c+d x}} \]

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Rubi [A]  time = 0.09, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {47, 50, 63, 240, 212, 208, 205} \begin {gather*} \frac {5 b \sqrt [4]{a+b x} (c+d x)^{3/4}}{d^2}-\frac {5 \sqrt [4]{b} (b c-a d) \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 d^{9/4}}-\frac {5 \sqrt [4]{b} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 d^{9/4}}-\frac {4 (a+b x)^{5/4}}{d \sqrt [4]{c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 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 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]

Rubi steps

\begin {align*} \int \frac {(a+b x)^{5/4}}{(c+d x)^{5/4}} \, dx &=-\frac {4 (a+b x)^{5/4}}{d \sqrt [4]{c+d x}}+\frac {(5 b) \int \frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}} \, dx}{d}\\ &=-\frac {4 (a+b x)^{5/4}}{d \sqrt [4]{c+d x}}+\frac {5 b \sqrt [4]{a+b x} (c+d x)^{3/4}}{d^2}-\frac {(5 b (b c-a d)) \int \frac {1}{(a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx}{4 d^2}\\ &=-\frac {4 (a+b x)^{5/4}}{d \sqrt [4]{c+d x}}+\frac {5 b \sqrt [4]{a+b x} (c+d x)^{3/4}}{d^2}-\frac {(5 (b c-a d)) \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 )}{d^2}\\ &=-\frac {4 (a+b x)^{5/4}}{d \sqrt [4]{c+d x}}+\frac {5 b \sqrt [4]{a+b x} (c+d x)^{3/4}}{d^2}-\frac {(5 (b c-a d)) \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 )}{d^2}\\ &=-\frac {4 (a+b x)^{5/4}}{d \sqrt [4]{c+d x}}+\frac {5 b \sqrt [4]{a+b x} (c+d x)^{3/4}}{d^2}-\frac {\left (5 \sqrt {b} (b c-a d)\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 )}{2 d^2}-\frac {\left (5 \sqrt {b} (b c-a d)\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 )}{2 d^2}\\ &=-\frac {4 (a+b x)^{5/4}}{d \sqrt [4]{c+d x}}+\frac {5 b \sqrt [4]{a+b x} (c+d x)^{3/4}}{d^2}-\frac {5 \sqrt [4]{b} (b c-a d) \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 d^{9/4}}-\frac {5 \sqrt [4]{b} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{2 d^{9/4}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 73, normalized size = 0.48 \begin {gather*} \frac {4 (a+b x)^{9/4} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4} \, _2F_1\left (\frac {5}{4},\frac {9}{4};\frac {13}{4};\frac {d (a+b x)}{a d-b c}\right )}{9 b (c+d x)^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a + b*x)^(9/4)*((b*(c + d*x))/(b*c - a*d))^(5/4)*Hypergeometric2F1[5/4, 9/4, 13/4, (d*(a + b*x))/(-(b*c) +
a*d)])/(9*b*(c + d*x)^(5/4))

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IntegrateAlgebraic [A]  time = 14.16, size = 200, normalized size = 1.32 \begin {gather*} \frac {d^{5/4} (a+b x)^{5/4} \left (\frac {5 \left (b^{5/4} c-a \sqrt [4]{b} d\right ) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{a d+b (c+d x)-b c}}\right )}{2 d^{9/4}}-\frac {5 \left (b^{5/4} c-a \sqrt [4]{b} d\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{a d+b (c+d x)-b c}}\right )}{2 d^{9/4}}+\frac {\sqrt [4]{a d+b (c+d x)-b c} (-4 a d+b (c+d x)+4 b c)}{d^{9/4} \sqrt [4]{c+d x}}\right )}{(a d+b d x)^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.22, size = 857, normalized size = 5.64 \begin {gather*} -\frac {20 \, {\left (d^{3} x + c d^{2}\right )} \left (\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}\right )^{\frac {1}{4}} \arctan \left (\frac {{\left (b c d^{7} - a d^{8}\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}} \left (\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}\right )^{\frac {3}{4}} + {\left (d^{8} x + c d^{7}\right )} \sqrt {\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c} + {\left (d^{5} x + c d^{4}\right )} \sqrt {\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}}}{d x + c}} \left (\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}\right )^{\frac {3}{4}}}{b^{5} c^{5} - 4 \, a b^{4} c^{4} d + 6 \, a^{2} b^{3} c^{3} d^{2} - 4 \, a^{3} b^{2} c^{2} d^{3} + a^{4} b c d^{4} + {\left (b^{5} c^{4} d - 4 \, a b^{4} c^{3} d^{2} + 6 \, a^{2} b^{3} c^{2} d^{3} - 4 \, a^{3} b^{2} c d^{4} + a^{4} b d^{5}\right )} x}\right ) + 5 \, {\left (d^{3} x + c d^{2}\right )} \left (\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {5 \, {\left ({\left (b c - a d\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}} + {\left (d^{3} x + c d^{2}\right )} \left (\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}\right )^{\frac {1}{4}}\right )}}{d x + c}\right ) - 5 \, {\left (d^{3} x + c d^{2}\right )} \left (\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}\right )^{\frac {1}{4}} \log \left (-\frac {5 \, {\left ({\left (b c - a d\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}} - {\left (d^{3} x + c d^{2}\right )} \left (\frac {b^{5} c^{4} - 4 \, a b^{4} c^{3} d + 6 \, a^{2} b^{3} c^{2} d^{2} - 4 \, a^{3} b^{2} c d^{3} + a^{4} b d^{4}}{d^{9}}\right )^{\frac {1}{4}}\right )}}{d x + c}\right ) - 4 \, {\left (b d x + 5 \, b c - 4 \, a d\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{4 \, {\left (d^{3} x + c d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(20*(d^3*x + c*d^2)*((b^5*c^4 - 4*a*b^4*c^3*d + 6*a^2*b^3*c^2*d^2 - 4*a^3*b^2*c*d^3 + a^4*b*d^4)/d^9)^(1/
4)*arctan(((b*c*d^7 - a*d^8)*(b*x + a)^(1/4)*(d*x + c)^(3/4)*((b^5*c^4 - 4*a*b^4*c^3*d + 6*a^2*b^3*c^2*d^2 - 4
*a^3*b^2*c*d^3 + a^4*b*d^4)/d^9)^(3/4) + (d^8*x + c*d^7)*sqrt(((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(b*x + a)*s
qrt(d*x + c) + (d^5*x + c*d^4)*sqrt((b^5*c^4 - 4*a*b^4*c^3*d + 6*a^2*b^3*c^2*d^2 - 4*a^3*b^2*c*d^3 + a^4*b*d^4
)/d^9))/(d*x + c))*((b^5*c^4 - 4*a*b^4*c^3*d + 6*a^2*b^3*c^2*d^2 - 4*a^3*b^2*c*d^3 + a^4*b*d^4)/d^9)^(3/4))/(b
^5*c^5 - 4*a*b^4*c^4*d + 6*a^2*b^3*c^3*d^2 - 4*a^3*b^2*c^2*d^3 + a^4*b*c*d^4 + (b^5*c^4*d - 4*a*b^4*c^3*d^2 +
6*a^2*b^3*c^2*d^3 - 4*a^3*b^2*c*d^4 + a^4*b*d^5)*x)) + 5*(d^3*x + c*d^2)*((b^5*c^4 - 4*a*b^4*c^3*d + 6*a^2*b^3
*c^2*d^2 - 4*a^3*b^2*c*d^3 + a^4*b*d^4)/d^9)^(1/4)*log(-5*((b*c - a*d)*(b*x + a)^(1/4)*(d*x + c)^(3/4) + (d^3*
x + c*d^2)*((b^5*c^4 - 4*a*b^4*c^3*d + 6*a^2*b^3*c^2*d^2 - 4*a^3*b^2*c*d^3 + a^4*b*d^4)/d^9)^(1/4))/(d*x + c))
- 5*(d^3*x + c*d^2)*((b^5*c^4 - 4*a*b^4*c^3*d + 6*a^2*b^3*c^2*d^2 - 4*a^3*b^2*c*d^3 + a^4*b*d^4)/d^9)^(1/4)*l
og(-5*((b*c - a*d)*(b*x + a)^(1/4)*(d*x + c)^(3/4) - (d^3*x + c*d^2)*((b^5*c^4 - 4*a*b^4*c^3*d + 6*a^2*b^3*c^2
*d^2 - 4*a^3*b^2*c*d^3 + a^4*b*d^4)/d^9)^(1/4))/(d*x + c)) - 4*(b*d*x + 5*b*c - 4*a*d)*(b*x + a)^(1/4)*(d*x +
c)^(3/4))/(d^3*x + c*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b x +a \right )^{\frac {5}{4}}}{\left (d x +c \right )^{\frac {5}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x)**(5/4)/(c + d*x)**(5/4), x)

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