3.7.94 \(\int \frac {x^{-1+3 n} (a+b x^n)^{3/2}}{\sqrt {c+d x^n}} \, dx\)

Optimal. Leaf size=291 \[ -\frac {(b c-a d) \left (3 a^2 d^2+10 a b c d+35 b^2 c^2\right ) \sqrt {a+b x^n} \sqrt {c+d x^n}}{64 b^2 d^4 n}+\frac {\left (3 a^2 d^2+10 a b c d+35 b^2 c^2\right ) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{96 b^2 d^3 n}+\frac {(b c-a d)^2 \left (3 a^2 d^2+10 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{64 b^{5/2} d^{9/2} n}-\frac {(3 a d+7 b c) \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{24 b^2 d^2 n}+\frac {x^n \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{4 b d n} \]

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Rubi [A]  time = 0.32, antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {446, 90, 80, 50, 63, 217, 206} \begin {gather*} \frac {\left (3 a^2 d^2+10 a b c d+35 b^2 c^2\right ) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{96 b^2 d^3 n}-\frac {(b c-a d) \left (3 a^2 d^2+10 a b c d+35 b^2 c^2\right ) \sqrt {a+b x^n} \sqrt {c+d x^n}}{64 b^2 d^4 n}+\frac {(b c-a d)^2 \left (3 a^2 d^2+10 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{64 b^{5/2} d^{9/2} n}-\frac {(3 a d+7 b c) \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{24 b^2 d^2 n}+\frac {x^n \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{4 b d n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(-1 + 3*n)*(a + b*x^n)^(3/2))/Sqrt[c + d*x^n],x]

[Out]

-((b*c - a*d)*(35*b^2*c^2 + 10*a*b*c*d + 3*a^2*d^2)*Sqrt[a + b*x^n]*Sqrt[c + d*x^n])/(64*b^2*d^4*n) + ((35*b^2
*c^2 + 10*a*b*c*d + 3*a^2*d^2)*(a + b*x^n)^(3/2)*Sqrt[c + d*x^n])/(96*b^2*d^3*n) - ((7*b*c + 3*a*d)*(a + b*x^n
)^(5/2)*Sqrt[c + d*x^n])/(24*b^2*d^2*n) + (x^n*(a + b*x^n)^(5/2)*Sqrt[c + d*x^n])/(4*b*d*n) + ((b*c - a*d)^2*(
35*b^2*c^2 + 10*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^n])/(Sqrt[b]*Sqrt[c + d*x^n])])/(64*b^(5/2)
*d^(9/2)*n)

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 80

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

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^{-1+3 n} \left (a+b x^n\right )^{3/2}}{\sqrt {c+d x^n}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2 (a+b x)^{3/2}}{\sqrt {c+d x}} \, dx,x,x^n\right )}{n}\\ &=\frac {x^n \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{4 b d n}+\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^{3/2} \left (-a c-\frac {1}{2} (7 b c+3 a d) x\right )}{\sqrt {c+d x}} \, dx,x,x^n\right )}{4 b d n}\\ &=-\frac {(7 b c+3 a d) \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{24 b^2 d^2 n}+\frac {x^n \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{4 b d n}+\frac {\left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx,x,x^n\right )}{48 b^2 d^2 n}\\ &=\frac {\left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{96 b^2 d^3 n}-\frac {(7 b c+3 a d) \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{24 b^2 d^2 n}+\frac {x^n \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{4 b d n}-\frac {\left ((b c-a d) \left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx,x,x^n\right )}{64 b^2 d^3 n}\\ &=-\frac {(b c-a d) \left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \sqrt {a+b x^n} \sqrt {c+d x^n}}{64 b^2 d^4 n}+\frac {\left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{96 b^2 d^3 n}-\frac {(7 b c+3 a d) \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{24 b^2 d^2 n}+\frac {x^n \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{4 b d n}+\frac {\left ((b c-a d)^2 \left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^n\right )}{128 b^2 d^4 n}\\ &=-\frac {(b c-a d) \left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \sqrt {a+b x^n} \sqrt {c+d x^n}}{64 b^2 d^4 n}+\frac {\left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{96 b^2 d^3 n}-\frac {(7 b c+3 a d) \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{24 b^2 d^2 n}+\frac {x^n \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{4 b d n}+\frac {\left ((b c-a d)^2 \left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x^n}\right )}{64 b^3 d^4 n}\\ &=-\frac {(b c-a d) \left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \sqrt {a+b x^n} \sqrt {c+d x^n}}{64 b^2 d^4 n}+\frac {\left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{96 b^2 d^3 n}-\frac {(7 b c+3 a d) \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{24 b^2 d^2 n}+\frac {x^n \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{4 b d n}+\frac {\left ((b c-a d)^2 \left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x^n}}{\sqrt {c+d x^n}}\right )}{64 b^3 d^4 n}\\ &=-\frac {(b c-a d) \left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \sqrt {a+b x^n} \sqrt {c+d x^n}}{64 b^2 d^4 n}+\frac {\left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{96 b^2 d^3 n}-\frac {(7 b c+3 a d) \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{24 b^2 d^2 n}+\frac {x^n \left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{4 b d n}+\frac {(b c-a d)^2 \left (35 b^2 c^2+10 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{64 b^{5/2} d^{9/2} n}\\ \end {align*}

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Mathematica [A]  time = 1.01, size = 241, normalized size = 0.83 \begin {gather*} \frac {3 (b c-a d)^{5/2} \left (3 a^2 d^2+10 a b c d+35 b^2 c^2\right ) \sqrt {\frac {b \left (c+d x^n\right )}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b c-a d}}\right )-b \sqrt {d} \sqrt {a+b x^n} \left (c+d x^n\right ) \left (9 a^3 d^3+3 a^2 b d^2 \left (5 c-2 d x^n\right )-a b^2 d \left (145 c^2-92 c d x^n+72 d^2 x^{2 n}\right )+b^3 \left (105 c^3-70 c^2 d x^n+56 c d^2 x^{2 n}-48 d^3 x^{3 n}\right )\right )}{192 b^3 d^{9/2} n \sqrt {c+d x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(-1 + 3*n)*(a + b*x^n)^(3/2))/Sqrt[c + d*x^n],x]

[Out]

(-(b*Sqrt[d]*Sqrt[a + b*x^n]*(c + d*x^n)*(9*a^3*d^3 + 3*a^2*b*d^2*(5*c - 2*d*x^n) - a*b^2*d*(145*c^2 - 92*c*d*
x^n + 72*d^2*x^(2*n)) + b^3*(105*c^3 - 70*c^2*d*x^n + 56*c*d^2*x^(2*n) - 48*d^3*x^(3*n)))) + 3*(b*c - a*d)^(5/
2)*(35*b^2*c^2 + 10*a*b*c*d + 3*a^2*d^2)*Sqrt[(b*(c + d*x^n))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^n])/S
qrt[b*c - a*d]])/(192*b^3*d^(9/2)*n*Sqrt[c + d*x^n])

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IntegrateAlgebraic [F]  time = 0.42, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{-1+3 n} \left (a+b x^n\right )^{3/2}}{\sqrt {c+d x^n}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^(-1 + 3*n)*(a + b*x^n)^(3/2))/Sqrt[c + d*x^n],x]

[Out]

Defer[IntegrateAlgebraic][(x^(-1 + 3*n)*(a + b*x^n)^(3/2))/Sqrt[c + d*x^n], x]

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fricas [A]  time = 0.54, size = 607, normalized size = 2.09 \begin {gather*} \left [\frac {3 \, {\left (35 \, b^{4} c^{4} - 60 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} + 4 \, a^{3} b c d^{3} + 3 \, a^{4} d^{4}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2 \, n} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, \sqrt {b d} b d x^{n} + {\left (b c + a d\right )} \sqrt {b d}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{n}\right ) + 4 \, {\left (48 \, b^{4} d^{4} x^{3 \, n} - 105 \, b^{4} c^{3} d + 145 \, a b^{3} c^{2} d^{2} - 15 \, a^{2} b^{2} c d^{3} - 9 \, a^{3} b d^{4} - 8 \, {\left (7 \, b^{4} c d^{3} - 9 \, a b^{3} d^{4}\right )} x^{2 \, n} + 2 \, {\left (35 \, b^{4} c^{2} d^{2} - 46 \, a b^{3} c d^{3} + 3 \, a^{2} b^{2} d^{4}\right )} x^{n}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c}}{768 \, b^{3} d^{5} n}, -\frac {3 \, {\left (35 \, b^{4} c^{4} - 60 \, a b^{3} c^{3} d + 18 \, a^{2} b^{2} c^{2} d^{2} + 4 \, a^{3} b c d^{3} + 3 \, a^{4} d^{4}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, \sqrt {-b d} b d x^{n} + {\left (b c + a d\right )} \sqrt {-b d}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c}}{2 \, {\left (b^{2} d^{2} x^{2 \, n} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{n}\right )}}\right ) - 2 \, {\left (48 \, b^{4} d^{4} x^{3 \, n} - 105 \, b^{4} c^{3} d + 145 \, a b^{3} c^{2} d^{2} - 15 \, a^{2} b^{2} c d^{3} - 9 \, a^{3} b d^{4} - 8 \, {\left (7 \, b^{4} c d^{3} - 9 \, a b^{3} d^{4}\right )} x^{2 \, n} + 2 \, {\left (35 \, b^{4} c^{2} d^{2} - 46 \, a b^{3} c d^{3} + 3 \, a^{2} b^{2} d^{4}\right )} x^{n}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c}}{384 \, b^{3} d^{5} n}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)*(a+b*x^n)^(3/2)/(c+d*x^n)^(1/2),x, algorithm="fricas")

[Out]

[1/768*(3*(35*b^4*c^4 - 60*a*b^3*c^3*d + 18*a^2*b^2*c^2*d^2 + 4*a^3*b*c*d^3 + 3*a^4*d^4)*sqrt(b*d)*log(8*b^2*d
^2*x^(2*n) + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*sqrt(b*d)*b*d*x^n + (b*c + a*d)*sqrt(b*d))*sqrt(b*x^n + a)*s
qrt(d*x^n + c) + 8*(b^2*c*d + a*b*d^2)*x^n) + 4*(48*b^4*d^4*x^(3*n) - 105*b^4*c^3*d + 145*a*b^3*c^2*d^2 - 15*a
^2*b^2*c*d^3 - 9*a^3*b*d^4 - 8*(7*b^4*c*d^3 - 9*a*b^3*d^4)*x^(2*n) + 2*(35*b^4*c^2*d^2 - 46*a*b^3*c*d^3 + 3*a^
2*b^2*d^4)*x^n)*sqrt(b*x^n + a)*sqrt(d*x^n + c))/(b^3*d^5*n), -1/384*(3*(35*b^4*c^4 - 60*a*b^3*c^3*d + 18*a^2*
b^2*c^2*d^2 + 4*a^3*b*c*d^3 + 3*a^4*d^4)*sqrt(-b*d)*arctan(1/2*(2*sqrt(-b*d)*b*d*x^n + (b*c + a*d)*sqrt(-b*d))
*sqrt(b*x^n + a)*sqrt(d*x^n + c)/(b^2*d^2*x^(2*n) + a*b*c*d + (b^2*c*d + a*b*d^2)*x^n)) - 2*(48*b^4*d^4*x^(3*n
) - 105*b^4*c^3*d + 145*a*b^3*c^2*d^2 - 15*a^2*b^2*c*d^3 - 9*a^3*b*d^4 - 8*(7*b^4*c*d^3 - 9*a*b^3*d^4)*x^(2*n)
+ 2*(35*b^4*c^2*d^2 - 46*a*b^3*c*d^3 + 3*a^2*b^2*d^4)*x^n)*sqrt(b*x^n + a)*sqrt(d*x^n + c))/(b^3*d^5*n)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x^{n} + a\right )}^{\frac {3}{2}} x^{3 \, n - 1}}{\sqrt {d x^{n} + c}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)*(a+b*x^n)^(3/2)/(c+d*x^n)^(1/2),x, algorithm="giac")

[Out]

integrate((b*x^n + a)^(3/2)*x^(3*n - 1)/sqrt(d*x^n + c), x)

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maple [F]  time = 0.93, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{n}+a \right )^{\frac {3}{2}} x^{3 n -1}}{\sqrt {d \,x^{n}+c}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3*n-1)*(b*x^n+a)^(3/2)/(d*x^n+c)^(1/2),x)

[Out]

int(x^(3*n-1)*(b*x^n+a)^(3/2)/(d*x^n+c)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x^{n} + a\right )}^{\frac {3}{2}} x^{3 \, n - 1}}{\sqrt {d x^{n} + c}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)*(a+b*x^n)^(3/2)/(c+d*x^n)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^n + a)^(3/2)*x^(3*n - 1)/sqrt(d*x^n + c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^{3\,n-1}\,{\left (a+b\,x^n\right )}^{3/2}}{\sqrt {c+d\,x^n}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(3*n - 1)*(a + b*x^n)^(3/2))/(c + d*x^n)^(1/2),x)

[Out]

int((x^(3*n - 1)*(a + b*x^n)^(3/2))/(c + d*x^n)^(1/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+3*n)*(a+b*x**n)**(3/2)/(c+d*x**n)**(1/2),x)

[Out]

Timed out

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