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

Optimal. Leaf size=199 \[ -\frac {(b c-a d)^2 (a d+5 b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{8 b^{3/2} d^{7/2} n}+\frac {(b c-a d) (a d+5 b c) \sqrt {a+b x^n} \sqrt {c+d x^n}}{8 b d^3 n}-\frac {(a d+5 b c) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{12 b d^2 n}+\frac {\left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{3 b d n} \]

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Rubi [A]  time = 0.16, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {446, 80, 50, 63, 217, 206} \begin {gather*} -\frac {(b c-a d)^2 (a d+5 b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{8 b^{3/2} d^{7/2} n}-\frac {(a d+5 b c) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{12 b d^2 n}+\frac {(b c-a d) (a d+5 b c) \sqrt {a+b x^n} \sqrt {c+d x^n}}{8 b d^3 n}+\frac {\left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{3 b d n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)*(5*b*c + a*d)*Sqrt[a + b*x^n]*Sqrt[c + d*x^n])/(8*b*d^3*n) - ((5*b*c + a*d)*(a + b*x^n)^(3/2)*Sqr
t[c + d*x^n])/(12*b*d^2*n) + ((a + b*x^n)^(5/2)*Sqrt[c + d*x^n])/(3*b*d*n) - ((b*c - a*d)^2*(5*b*c + a*d)*ArcT
anh[(Sqrt[d]*Sqrt[a + b*x^n])/(Sqrt[b]*Sqrt[c + d*x^n])])/(8*b^(3/2)*d^(7/2)*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 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+2 n} \left (a+b x^n\right )^{3/2}}{\sqrt {c+d x^n}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x (a+b x)^{3/2}}{\sqrt {c+d x}} \, dx,x,x^n\right )}{n}\\ &=\frac {\left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{3 b d n}-\frac {(5 b c+a d) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx,x,x^n\right )}{6 b d n}\\ &=-\frac {(5 b c+a d) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{12 b d^2 n}+\frac {\left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{3 b d n}+\frac {((b c-a d) (5 b c+a d)) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx,x,x^n\right )}{8 b d^2 n}\\ &=\frac {(b c-a d) (5 b c+a d) \sqrt {a+b x^n} \sqrt {c+d x^n}}{8 b d^3 n}-\frac {(5 b c+a d) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{12 b d^2 n}+\frac {\left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{3 b d n}-\frac {\left ((b c-a d)^2 (5 b c+a d)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^n\right )}{16 b d^3 n}\\ &=\frac {(b c-a d) (5 b c+a d) \sqrt {a+b x^n} \sqrt {c+d x^n}}{8 b d^3 n}-\frac {(5 b c+a d) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{12 b d^2 n}+\frac {\left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{3 b d n}-\frac {\left ((b c-a d)^2 (5 b c+a d)\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 )}{8 b^2 d^3 n}\\ &=\frac {(b c-a d) (5 b c+a d) \sqrt {a+b x^n} \sqrt {c+d x^n}}{8 b d^3 n}-\frac {(5 b c+a d) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{12 b d^2 n}+\frac {\left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{3 b d n}-\frac {\left ((b c-a d)^2 (5 b c+a d)\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 )}{8 b^2 d^3 n}\\ &=\frac {(b c-a d) (5 b c+a d) \sqrt {a+b x^n} \sqrt {c+d x^n}}{8 b d^3 n}-\frac {(5 b c+a d) \left (a+b x^n\right )^{3/2} \sqrt {c+d x^n}}{12 b d^2 n}+\frac {\left (a+b x^n\right )^{5/2} \sqrt {c+d x^n}}{3 b d n}-\frac {(b c-a d)^2 (5 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^n}}{\sqrt {b} \sqrt {c+d x^n}}\right )}{8 b^{3/2} d^{7/2} n}\\ \end {align*}

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Mathematica [A]  time = 0.67, size = 178, normalized size = 0.89 \begin {gather*} \frac {b \sqrt {d} \sqrt {a+b x^n} \left (c+d x^n\right ) \left (3 a^2 d^2+2 a b d \left (7 d x^n-11 c\right )+b^2 \left (15 c^2-10 c d x^n+8 d^2 x^{2 n}\right )\right )-3 (b c-a d)^{5/2} (a d+5 b c) \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 )}{24 b^2 d^{7/2} n \sqrt {c+d x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(-1 + 2*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)*(3*a^2*d^2 + 2*a*b*d*(-11*c + 7*d*x^n) + b^2*(15*c^2 - 10*c*d*x^n + 8*d
^2*x^(2*n))) - 3*(b*c - a*d)^(5/2)*(5*b*c + a*d)*Sqrt[(b*(c + d*x^n))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b
*x^n])/Sqrt[b*c - a*d]])/(24*b^2*d^(7/2)*n*Sqrt[c + d*x^n])

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IntegrateAlgebraic [F]  time = 0.43, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{-1+2 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 + 2*n)*(a + b*x^n)^(3/2))/Sqrt[c + d*x^n],x]

[Out]

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

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fricas [A]  time = 0.50, size = 469, normalized size = 2.36 \begin {gather*} \left [\frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\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 (8 \, b^{3} d^{3} x^{2 \, n} + 15 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{n}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c}}{96 \, b^{2} d^{4} n}, \frac {3 \, {\left (5 \, b^{3} c^{3} - 9 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3}\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 (8 \, b^{3} d^{3} x^{2 \, n} + 15 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{n}\right )} \sqrt {b x^{n} + a} \sqrt {d x^{n} + c}}{48 \, b^{2} d^{4} n}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*(5*b^3*c^3 - 9*a*b^2*c^2*d + 3*a^2*b*c*d^2 + a^3*d^3)*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)*sqrt(d*x^n + c) + 8*(b^2*c*
d + a*b*d^2)*x^n) + 4*(8*b^3*d^3*x^(2*n) + 15*b^3*c^2*d - 22*a*b^2*c*d^2 + 3*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 7*a*
b^2*d^3)*x^n)*sqrt(b*x^n + a)*sqrt(d*x^n + c))/(b^2*d^4*n), 1/48*(3*(5*b^3*c^3 - 9*a*b^2*c^2*d + 3*a^2*b*c*d^2
 + a^3*d^3)*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*(8*b^3*d^3*x^(2*n) + 15*b^3*c^2*d - 22*a*b^2*c*
d^2 + 3*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 7*a*b^2*d^3)*x^n)*sqrt(b*x^n + a)*sqrt(d*x^n + c))/(b^2*d^4*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^{2 \, 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+2*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^(2*n - 1)/sqrt(d*x^n + c), x)

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maple [F]  time = 0.96, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{n}+a \right )^{\frac {3}{2}} x^{2 n -1}}{\sqrt {d \,x^{n}+c}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^(2*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^{2 \, 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+2*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^(2*n - 1)/sqrt(d*x^n + c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{2\,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^(2*n - 1)*(a + b*x^n)^(3/2))/(c + d*x^n)^(1/2),x)

[Out]

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

[Out]

Timed out

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