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

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

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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, 240, 212, 208, 205} \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^{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}}-\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} \end {gather*}

Antiderivative was successfully verified.

[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 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}}{\sqrt [4]{c+d x}} \, dx &=\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 ) \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 )}{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 ) \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 )}{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 ) \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 \sqrt {b} d^2}+\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 \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*}

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Mathematica [C]  time = 0.03, size = 73, normalized size = 0.44 \begin {gather*} \frac {4 (a+b x)^{9/4} \sqrt [4]{\frac {b (c+d x)}{b c-a d}} \, _2F_1\left (\frac {1}{4},\frac {9}{4};\frac {13}{4};\frac {d (a+b x)}{a d-b c}\right )}{9 b \sqrt [4]{c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a + b*x)^(9/4)*((b*(c + d*x))/(b*c - a*d))^(1/4)*Hypergeometric2F1[1/4, 9/4, 13/4, (d*(a + b*x))/(-(b*c) +
a*d)])/(9*b*(c + d*x)^(1/4))

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IntegrateAlgebraic [A]  time = 0.40, size = 189, normalized size = 1.13 \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^{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}}+\frac {(a d-b c)^2 \left (\frac {9 d (a+b x)^{5/4}}{(c+d x)^{5/4}}-\frac {5 b \sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{8 d^2 \left (\frac {d (a+b x)}{c+d x}-b\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.15, size = 1468, normalized size = 8.79

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(20*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)*arctan(-((b^4*c^2*d^7 - 2*a*b^3
*c*d^8 + a^2*b^2*d^9)*(b*x + a)^(1/4)*(d*x + c)^(3/4)*((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^3*d^9
))^(3/4) - (b^2*d^8*x + b^2*c*d^7)*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^2*d^5*x + b^2*c*d^4)*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^3*d^9)))/(d*x + c))*((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))^(3/4))/(b^8*c^9 - 8*a*b^7*c^8
*d + 28*a^2*b^6*c^7*d^2 - 56*a^3*b^5*c^6*d^3 + 70*a^4*b^4*c^5*d^4 - 56*a^5*b^3*c^4*d^5 + 28*a^6*b^2*c^3*d^6 -
8*a^7*b*c^2*d^7 + a^8*c*d^8 + (b^8*c^8*d - 8*a*b^7*c^7*d^2 + 28*a^2*b^6*c^6*d^3 - 56*a^3*b^5*c^5*d^4 + 70*a^4*
b^4*c^4*d^5 - 56*a^5*b^3*c^3*d^6 + 28*a^6*b^2*c^2*d^7 - 8*a^7*b*c*d^8 + a^8*d^9)*x)) - 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*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)) - 4*(4*b*d*x - 5*b*c + 9*a*d)*(b*x + a)^(1/4)*(d*x + c)^(3/4))/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 {1}{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)^(1/4),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b x +a \right )^{\frac {5}{4}}}{\left (d x +c \right )^{\frac {1}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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 {1}{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)^(1/4),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(5/4)/(d*x + c)^(1/4), 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 )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a + b*x)^(5/4)/(c + d*x)^(1/4), 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}}}{\sqrt [4]{c + d x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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