3.906 \(\int \frac {1}{x^3 (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx\)

Optimal. Leaf size=206 \[ \frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (5 a d+7 b c)}{8 a^2 c^2 x}-\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{16 a^{11/4} c^{9/4}}-\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{16 a^{11/4} c^{9/4}}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{2 a c x^2} \]

[Out]

-1/2*(b*x+a)^(1/4)*(d*x+c)^(3/4)/a/c/x^2+1/8*(5*a*d+7*b*c)*(b*x+a)^(1/4)*(d*x+c)^(3/4)/a^2/c^2/x-1/16*(5*a^2*d
^2+6*a*b*c*d+21*b^2*c^2)*arctan(c^(1/4)*(b*x+a)^(1/4)/a^(1/4)/(d*x+c)^(1/4))/a^(11/4)/c^(9/4)-1/16*(5*a^2*d^2+
6*a*b*c*d+21*b^2*c^2)*arctanh(c^(1/4)*(b*x+a)^(1/4)/a^(1/4)/(d*x+c)^(1/4))/a^(11/4)/c^(9/4)

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Rubi [A]  time = 0.11, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {129, 151, 12, 93, 212, 208, 205} \[ -\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{16 a^{11/4} c^{9/4}}-\frac {\left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{16 a^{11/4} c^{9/4}}+\frac {\sqrt [4]{a+b x} (c+d x)^{3/4} (5 a d+7 b c)}{8 a^2 c^2 x}-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{2 a c x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x)^(3/4)*(c + d*x)^(1/4)),x]

[Out]

-((a + b*x)^(1/4)*(c + d*x)^(3/4))/(2*a*c*x^2) + ((7*b*c + 5*a*d)*(a + b*x)^(1/4)*(c + d*x)^(3/4))/(8*a^2*c^2*
x) - ((21*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)*ArcTan[(c^(1/4)*(a + b*x)^(1/4))/(a^(1/4)*(c + d*x)^(1/4))])/(16*a^
(11/4)*c^(9/4)) - ((21*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)*ArcTanh[(c^(1/4)*(a + b*x)^(1/4))/(a^(1/4)*(c + d*x)^(
1/4))])/(16*a^(11/4)*c^(9/4))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 129

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && ILtQ[m + n
+ p + 2, 0] && NeQ[m, -1] && (SumSimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && S
umSimplerQ[p, 1])))

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

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]

Rubi steps

\begin {align*} \int \frac {1}{x^3 (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx &=-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{2 a c x^2}-\frac {\int \frac {\frac {1}{4} (7 b c+5 a d)+b d x}{x^2 (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx}{2 a c}\\ &=-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{2 a c x^2}+\frac {(7 b c+5 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 a^2 c^2 x}+\frac {\int \frac {21 b^2 c^2+6 a b c d+5 a^2 d^2}{16 x (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx}{2 a^2 c^2}\\ &=-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{2 a c x^2}+\frac {(7 b c+5 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 a^2 c^2 x}+\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \int \frac {1}{x (a+b x)^{3/4} \sqrt [4]{c+d x}} \, dx}{32 a^2 c^2}\\ &=-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{2 a c x^2}+\frac {(7 b c+5 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 a^2 c^2 x}+\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^4} \, dx,x,\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{8 a^2 c^2}\\ &=-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{2 a c x^2}+\frac {(7 b c+5 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 a^2 c^2 x}-\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}-\sqrt {c} x^2} \, dx,x,\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{16 a^{5/2} c^2}-\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a}+\sqrt {c} x^2} \, dx,x,\frac {\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{16 a^{5/2} c^2}\\ &=-\frac {\sqrt [4]{a+b x} (c+d x)^{3/4}}{2 a c x^2}+\frac {(7 b c+5 a d) \sqrt [4]{a+b x} (c+d x)^{3/4}}{8 a^2 c^2 x}-\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{16 a^{11/4} c^{9/4}}-\frac {\left (21 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt [4]{a+b x}}{\sqrt [4]{a} \sqrt [4]{c+d x}}\right )}{16 a^{11/4} c^{9/4}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 106, normalized size = 0.51 \[ -\frac {\sqrt [4]{a+b x} \left (x^2 \left (5 a^2 d^2+6 a b c d+21 b^2 c^2\right ) \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};\frac {c (a+b x)}{a (c+d x)}\right )+a (c+d x) (4 a c-5 a d x-7 b c x)\right )}{8 a^3 c^2 x^2 \sqrt [4]{c+d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x)^(3/4)*(c + d*x)^(1/4)),x]

[Out]

-1/8*((a + b*x)^(1/4)*(a*(c + d*x)*(4*a*c - 7*b*c*x - 5*a*d*x) + (21*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)*x^2*Hype
rgeometric2F1[1/4, 1, 5/4, (c*(a + b*x))/(a*(c + d*x))]))/(a^3*c^2*x^2*(c + d*x)^(1/4))

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fricas [B]  time = 1.21, size = 1528, normalized size = 7.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x+a)^(3/4)/(d*x+c)^(1/4),x, algorithm="fricas")

[Out]

1/32*(4*a^2*c^2*x^2*((194481*b^8*c^8 + 222264*a*b^7*c^7*d + 280476*a^2*b^6*c^6*d^2 + 176904*a^3*b^5*c^5*d^3 +
112806*a^4*b^4*c^4*d^4 + 42120*a^5*b^3*c^3*d^5 + 15900*a^6*b^2*c^2*d^6 + 3000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^11
*c^9))^(1/4)*arctan(-((21*a^8*b^2*c^9 + 6*a^9*b*c^8*d + 5*a^10*c^7*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4)*((1944
81*b^8*c^8 + 222264*a*b^7*c^7*d + 280476*a^2*b^6*c^6*d^2 + 176904*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 4
2120*a^5*b^3*c^3*d^5 + 15900*a^6*b^2*c^2*d^6 + 3000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^11*c^9))^(3/4) - (a^8*c^7*d*
x + a^8*c^8)*sqrt(((441*b^4*c^4 + 252*a*b^3*c^3*d + 246*a^2*b^2*c^2*d^2 + 60*a^3*b*c*d^3 + 25*a^4*d^4)*sqrt(b*
x + a)*sqrt(d*x + c) + (a^6*c^4*d*x + a^6*c^5)*sqrt((194481*b^8*c^8 + 222264*a*b^7*c^7*d + 280476*a^2*b^6*c^6*
d^2 + 176904*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 42120*a^5*b^3*c^3*d^5 + 15900*a^6*b^2*c^2*d^6 + 3000*a
^7*b*c*d^7 + 625*a^8*d^8)/(a^11*c^9)))/(d*x + c))*((194481*b^8*c^8 + 222264*a*b^7*c^7*d + 280476*a^2*b^6*c^6*d
^2 + 176904*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4*d^4 + 42120*a^5*b^3*c^3*d^5 + 15900*a^6*b^2*c^2*d^6 + 3000*a^
7*b*c*d^7 + 625*a^8*d^8)/(a^11*c^9))^(3/4))/(194481*b^8*c^9 + 222264*a*b^7*c^8*d + 280476*a^2*b^6*c^7*d^2 + 17
6904*a^3*b^5*c^6*d^3 + 112806*a^4*b^4*c^5*d^4 + 42120*a^5*b^3*c^4*d^5 + 15900*a^6*b^2*c^3*d^6 + 3000*a^7*b*c^2
*d^7 + 625*a^8*c*d^8 + (194481*b^8*c^8*d + 222264*a*b^7*c^7*d^2 + 280476*a^2*b^6*c^6*d^3 + 176904*a^3*b^5*c^5*
d^4 + 112806*a^4*b^4*c^4*d^5 + 42120*a^5*b^3*c^3*d^6 + 15900*a^6*b^2*c^2*d^7 + 3000*a^7*b*c*d^8 + 625*a^8*d^9)
*x)) - a^2*c^2*x^2*((194481*b^8*c^8 + 222264*a*b^7*c^7*d + 280476*a^2*b^6*c^6*d^2 + 176904*a^3*b^5*c^5*d^3 + 1
12806*a^4*b^4*c^4*d^4 + 42120*a^5*b^3*c^3*d^5 + 15900*a^6*b^2*c^2*d^6 + 3000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^11*
c^9))^(1/4)*log(((21*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4) + (a^3*c^2*d*x + a^3*c^3
)*((194481*b^8*c^8 + 222264*a*b^7*c^7*d + 280476*a^2*b^6*c^6*d^2 + 176904*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c^4
*d^4 + 42120*a^5*b^3*c^3*d^5 + 15900*a^6*b^2*c^2*d^6 + 3000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^11*c^9))^(1/4))/(d*x
+ c)) + a^2*c^2*x^2*((194481*b^8*c^8 + 222264*a*b^7*c^7*d + 280476*a^2*b^6*c^6*d^2 + 176904*a^3*b^5*c^5*d^3 +
112806*a^4*b^4*c^4*d^4 + 42120*a^5*b^3*c^3*d^5 + 15900*a^6*b^2*c^2*d^6 + 3000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^1
1*c^9))^(1/4)*log(((21*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)*(b*x + a)^(1/4)*(d*x + c)^(3/4) - (a^3*c^2*d*x + a^3*c
^3)*((194481*b^8*c^8 + 222264*a*b^7*c^7*d + 280476*a^2*b^6*c^6*d^2 + 176904*a^3*b^5*c^5*d^3 + 112806*a^4*b^4*c
^4*d^4 + 42120*a^5*b^3*c^3*d^5 + 15900*a^6*b^2*c^2*d^6 + 3000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^11*c^9))^(1/4))/(d
*x + c)) - 4*(4*a*c - (7*b*c + 5*a*d)*x)*(b*x + a)^(1/4)*(d*x + c)^(3/4))/(a^2*c^2*x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x+a)^(3/4)/(d*x+c)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(3/4)*(d*x + c)^(1/4)*x^3), x)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b x +a \right )^{\frac {3}{4}} \left (d x +c \right )^{\frac {1}{4}} x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x+a)^(3/4)/(d*x+c)^(1/4),x)

[Out]

int(1/x^3/(b*x+a)^(3/4)/(d*x+c)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {3}{4}} {\left (d x + c\right )}^{\frac {1}{4}} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x+a)^(3/4)/(d*x+c)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(3/4)*(d*x + c)^(1/4)*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{x^3\,{\left (a+b\,x\right )}^{3/4}\,{\left (c+d\,x\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x)^(3/4)*(c + d*x)^(1/4)),x)

[Out]

int(1/(x^3*(a + b*x)^(3/4)*(c + d*x)^(1/4)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \left (a + b x\right )^{\frac {3}{4}} \sqrt [4]{c + d x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x+a)**(3/4)/(d*x+c)**(1/4),x)

[Out]

Integral(1/(x**3*(a + b*x)**(3/4)*(c + d*x)**(1/4)), x)

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