Integrand size = 32, antiderivative size = 225 \[ \int \frac {x^4}{\sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{5/2}} \, dx=\frac {c (c+d x)^{3/2}}{12 b d^5 \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {c^2}{8 b^2 d^5 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}+\frac {23 c}{32 b^2 d^5 \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}-\frac {61 \sqrt {c+d x}}{64 b^2 d^5 \sqrt {b c^2-b d^2 x^2}}-\frac {67 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c} \sqrt {c+d x}}{\sqrt {b c^2-b d^2 x^2}}\right )}{64 \sqrt {2} b^{5/2} \sqrt {c} d^5} \] Output:
1/12*c*(d*x+c)^(3/2)/b/d^5/(-b*d^2*x^2+b*c^2)^(3/2)-1/8*c^2/b^2/d^5/(d*x+c )^(3/2)/(-b*d^2*x^2+b*c^2)^(1/2)+23/32*c/b^2/d^5/(d*x+c)^(1/2)/(-b*d^2*x^2 +b*c^2)^(1/2)-61/64*(d*x+c)^(1/2)/b^2/d^5/(-b*d^2*x^2+b*c^2)^(1/2)-67/128* arctanh(2^(1/2)*b^(1/2)*c^(1/2)*(d*x+c)^(1/2)/(-b*d^2*x^2+b*c^2)^(1/2))*2^ (1/2)/b^(5/2)/c^(1/2)/d^5
Time = 0.95 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.70 \[ \int \frac {x^4}{\sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \sqrt {c} \left (53 c^3+127 c^2 d x-61 c d^2 x^2-183 d^3 x^3\right )+201 \sqrt {2} \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{384 b^2 \sqrt {c} d^5 (c-d x) (c+d x)^{3/2} \sqrt {b \left (c^2-d^2 x^2\right )}} \] Input:
Integrate[x^4/(Sqrt[c + d*x]*(b*c^2 - b*d^2*x^2)^(5/2)),x]
Output:
-1/384*(2*Sqrt[c]*(53*c^3 + 127*c^2*d*x - 61*c*d^2*x^2 - 183*d^3*x^3) + 20 1*Sqrt[2]*Sqrt[c + d*x]*(c^2 - d^2*x^2)^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqr t[c + d*x])/Sqrt[c^2 - d^2*x^2]])/(b^2*Sqrt[c]*d^5*(c - d*x)*(c + d*x)^(3/ 2)*Sqrt[b*(c^2 - d^2*x^2)])
Time = 1.46 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.72, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {581, 27, 2170, 27, 2170, 27, 671, 467, 470, 467, 471, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^4}{\sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 581 |
\(\displaystyle \frac {2 (c+d x)^{5/2}}{b d^5 \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {2 \int \frac {5 c^4+16 d x c^3+18 d^2 x^2 c^2+8 d^3 x^3 c}{2 \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{5/2}}dx}{d^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 (c+d x)^{5/2}}{b d^5 \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\int \frac {5 c^4+16 d x c^3+18 d^2 x^2 c^2+8 d^3 x^3 c}{\sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{5/2}}dx}{d^4}\) |
\(\Big \downarrow \) 2170 |
\(\displaystyle \frac {2 (c+d x)^{5/2}}{b d^5 \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {2 \int -\frac {3 \left (6 b c^2 x^2 d^7+8 b c^3 x d^6+3 b c^4 d^5\right )}{2 \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{5/2}}dx}{3 b d^5}+\frac {16 c (c+d x)^{3/2}}{3 b d \left (b c^2-b d^2 x^2\right )^{3/2}}}{d^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 (c+d x)^{5/2}}{b d^5 \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {16 c (c+d x)^{3/2}}{3 b d \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\int \frac {6 b c^2 x^2 d^7+8 b c^3 x d^6+3 b c^4 d^5}{\sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{5/2}}dx}{b d^5}}{d^4}\) |
\(\Big \downarrow \) 2170 |
\(\displaystyle \frac {2 (c+d x)^{5/2}}{b d^5 \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {16 c (c+d x)^{3/2}}{3 b d \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {2 \int \frac {b^2 c^3 d^9 (9 c+4 d x)}{2 \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{5/2}}dx}{5 b d^4}+\frac {12 c^2 d^4 \sqrt {c+d x}}{5 \left (b c^2-b d^2 x^2\right )^{3/2}}}{b d^5}}{d^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 (c+d x)^{5/2}}{b d^5 \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {16 c (c+d x)^{3/2}}{3 b d \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {1}{5} b c^3 d^5 \int \frac {9 c+4 d x}{\sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{5/2}}dx+\frac {12 c^2 d^4 \sqrt {c+d x}}{5 \left (b c^2-b d^2 x^2\right )^{3/2}}}{b d^5}}{d^4}\) |
\(\Big \downarrow \) 671 |
\(\displaystyle \frac {2 (c+d x)^{5/2}}{b d^5 \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {16 c (c+d x)^{3/2}}{3 b d \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {1}{5} b c^3 d^5 \left (\frac {67}{8} \int \frac {\sqrt {c+d x}}{\left (b c^2-b d^2 x^2\right )^{5/2}}dx-\frac {5}{4 b d \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}\right )+\frac {12 c^2 d^4 \sqrt {c+d x}}{5 \left (b c^2-b d^2 x^2\right )^{3/2}}}{b d^5}}{d^4}\) |
\(\Big \downarrow \) 467 |
\(\displaystyle \frac {2 (c+d x)^{5/2}}{b d^5 \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {16 c (c+d x)^{3/2}}{3 b d \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {1}{5} b c^3 d^5 \left (\frac {67}{8} \left (\frac {5 \int \frac {1}{\sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}dx}{6 b c}+\frac {\sqrt {c+d x}}{3 b c d \left (b c^2-b d^2 x^2\right )^{3/2}}\right )-\frac {5}{4 b d \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}\right )+\frac {12 c^2 d^4 \sqrt {c+d x}}{5 \left (b c^2-b d^2 x^2\right )^{3/2}}}{b d^5}}{d^4}\) |
\(\Big \downarrow \) 470 |
\(\displaystyle \frac {2 (c+d x)^{5/2}}{b d^5 \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {16 c (c+d x)^{3/2}}{3 b d \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {1}{5} b c^3 d^5 \left (\frac {67}{8} \left (\frac {5 \left (\frac {3 \int \frac {\sqrt {c+d x}}{\left (b c^2-b d^2 x^2\right )^{3/2}}dx}{4 c}-\frac {1}{2 b c d \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}\right )}{6 b c}+\frac {\sqrt {c+d x}}{3 b c d \left (b c^2-b d^2 x^2\right )^{3/2}}\right )-\frac {5}{4 b d \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}\right )+\frac {12 c^2 d^4 \sqrt {c+d x}}{5 \left (b c^2-b d^2 x^2\right )^{3/2}}}{b d^5}}{d^4}\) |
\(\Big \downarrow \) 467 |
\(\displaystyle \frac {2 (c+d x)^{5/2}}{b d^5 \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {16 c (c+d x)^{3/2}}{3 b d \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {1}{5} b c^3 d^5 \left (\frac {67}{8} \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}dx}{2 b c}+\frac {\sqrt {c+d x}}{b c d \sqrt {b c^2-b d^2 x^2}}\right )}{4 c}-\frac {1}{2 b c d \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}\right )}{6 b c}+\frac {\sqrt {c+d x}}{3 b c d \left (b c^2-b d^2 x^2\right )^{3/2}}\right )-\frac {5}{4 b d \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}\right )+\frac {12 c^2 d^4 \sqrt {c+d x}}{5 \left (b c^2-b d^2 x^2\right )^{3/2}}}{b d^5}}{d^4}\) |
\(\Big \downarrow \) 471 |
\(\displaystyle \frac {2 (c+d x)^{5/2}}{b d^5 \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {16 c (c+d x)^{3/2}}{3 b d \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {1}{5} b c^3 d^5 \left (\frac {67}{8} \left (\frac {5 \left (\frac {3 \left (\frac {d \int \frac {1}{\frac {d^2 \left (b c^2-b d^2 x^2\right )}{c+d x}-2 b c d^2}d\frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}}}{b c}+\frac {\sqrt {c+d x}}{b c d \sqrt {b c^2-b d^2 x^2}}\right )}{4 c}-\frac {1}{2 b c d \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}\right )}{6 b c}+\frac {\sqrt {c+d x}}{3 b c d \left (b c^2-b d^2 x^2\right )^{3/2}}\right )-\frac {5}{4 b d \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}\right )+\frac {12 c^2 d^4 \sqrt {c+d x}}{5 \left (b c^2-b d^2 x^2\right )^{3/2}}}{b d^5}}{d^4}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 (c+d x)^{5/2}}{b d^5 \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {16 c (c+d x)^{3/2}}{3 b d \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {\frac {1}{5} b c^3 d^5 \left (\frac {67}{8} \left (\frac {5 \left (\frac {3 \left (\frac {\sqrt {c+d x}}{b c d \sqrt {b c^2-b d^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {2} \sqrt {b} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {2} b^{3/2} c^{3/2} d}\right )}{4 c}-\frac {1}{2 b c d \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}\right )}{6 b c}+\frac {\sqrt {c+d x}}{3 b c d \left (b c^2-b d^2 x^2\right )^{3/2}}\right )-\frac {5}{4 b d \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}\right )+\frac {12 c^2 d^4 \sqrt {c+d x}}{5 \left (b c^2-b d^2 x^2\right )^{3/2}}}{b d^5}}{d^4}\) |
Input:
Int[x^4/(Sqrt[c + d*x]*(b*c^2 - b*d^2*x^2)^(5/2)),x]
Output:
(2*(c + d*x)^(5/2))/(b*d^5*(b*c^2 - b*d^2*x^2)^(3/2)) - ((16*c*(c + d*x)^( 3/2))/(3*b*d*(b*c^2 - b*d^2*x^2)^(3/2)) - ((12*c^2*d^4*Sqrt[c + d*x])/(5*( b*c^2 - b*d^2*x^2)^(3/2)) + (b*c^3*d^5*(-5/(4*b*d*Sqrt[c + d*x]*(b*c^2 - b *d^2*x^2)^(3/2)) + (67*(Sqrt[c + d*x]/(3*b*c*d*(b*c^2 - b*d^2*x^2)^(3/2)) + (5*(-1/2*1/(b*c*d*Sqrt[c + d*x]*Sqrt[b*c^2 - b*d^2*x^2]) + (3*(Sqrt[c + d*x]/(b*c*d*Sqrt[b*c^2 - b*d^2*x^2]) - ArcTanh[Sqrt[b*c^2 - b*d^2*x^2]/(Sq rt[2]*Sqrt[b]*Sqrt[c]*Sqrt[c + d*x])]/(Sqrt[2]*b^(3/2)*c^(3/2)*d)))/(4*c)) )/(6*b*c)))/8))/5)/(b*d^5))/d^4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-c)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*a*d*(p + 1))), x] + Simp[c*((n + 2 *p + 2)/(2*a*(p + 1))) Int[(c + d*x)^(n - 1)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[p, -1] && LtQ[0, n, 1] && IntegerQ[2*p]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[(n + 2*p + 2)/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + p + 1, 0] && IntegerQ[2*p]
Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim p[2*d Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] ], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[(c + d*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*d^(m - 1)*(m + n + 2*p + 1))), x] + Simp[1/(d^m*(m + n + 2*p + 1)) Int[(c + d*x)^n*(a + b*x^ 2)^p*ExpandToSum[d^m*(m + n + 2*p + 1)*x^m - (m + n + 2*p + 1)*(c + d*x)^m + c*(c + d*x)^(m - 2)*(c*(m + n - 1) + c*(m + n + 2*p + 1) + 2*d*(m + n + p )*x), x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] & & IGtQ[m, 1] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[2*p] || ILtQ[m + n, 0] )
Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ ), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m + p + 1)) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) ^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si mp[1/(b*e^q*(m + q + 2*p + 1)) Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - b*d*x), x], x], x] /; NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && !IGtQ[m, 0]
Time = 0.28 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.16
method | result | size |
default | \(-\frac {\sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, \left (-201 \sqrt {\left (-d x +c \right ) b}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}\, \sqrt {2}}{2 \sqrt {b c}}\right ) d^{3} x^{3}-201 \sqrt {\left (-d x +c \right ) b}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}\, \sqrt {2}}{2 \sqrt {b c}}\right ) c \,d^{2} x^{2}+201 \sqrt {\left (-d x +c \right ) b}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}\, \sqrt {2}}{2 \sqrt {b c}}\right ) c^{2} d x -366 \sqrt {b c}\, d^{3} x^{3}+201 \sqrt {\left (-d x +c \right ) b}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}\, \sqrt {2}}{2 \sqrt {b c}}\right ) c^{3}-122 \sqrt {b c}\, c \,d^{2} x^{2}+254 \sqrt {b c}\, c^{2} d x +106 \sqrt {b c}\, c^{3}\right )}{384 b^{3} \left (d x +c \right )^{\frac {5}{2}} \left (-d x +c \right )^{2} d^{5} \sqrt {b c}}\) | \(260\) |
Input:
int(x^4/(d*x+c)^(1/2)/(-b*d^2*x^2+b*c^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/384*(b*(-d^2*x^2+c^2))^(1/2)/b^3*(-201*((-d*x+c)*b)^(1/2)*2^(1/2)*arcta nh(1/2*((-d*x+c)*b)^(1/2)*2^(1/2)/(b*c)^(1/2))*d^3*x^3-201*((-d*x+c)*b)^(1 /2)*2^(1/2)*arctanh(1/2*((-d*x+c)*b)^(1/2)*2^(1/2)/(b*c)^(1/2))*c*d^2*x^2+ 201*((-d*x+c)*b)^(1/2)*2^(1/2)*arctanh(1/2*((-d*x+c)*b)^(1/2)*2^(1/2)/(b*c )^(1/2))*c^2*d*x-366*(b*c)^(1/2)*d^3*x^3+201*((-d*x+c)*b)^(1/2)*2^(1/2)*ar ctanh(1/2*((-d*x+c)*b)^(1/2)*2^(1/2)/(b*c)^(1/2))*c^3-122*(b*c)^(1/2)*c*d^ 2*x^2+254*(b*c)^(1/2)*c^2*d*x+106*(b*c)^(1/2)*c^3)/(d*x+c)^(5/2)/(-d*x+c)^ 2/d^5/(b*c)^(1/2)
Time = 0.11 (sec) , antiderivative size = 517, normalized size of antiderivative = 2.30 \[ \int \frac {x^4}{\sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{5/2}} \, dx=\left [\frac {201 \, \sqrt {2} {\left (d^{5} x^{5} + c d^{4} x^{4} - 2 \, c^{2} d^{3} x^{3} - 2 \, c^{3} d^{2} x^{2} + c^{4} d x + c^{5}\right )} \sqrt {b c} \log \left (-\frac {b d^{2} x^{2} - 2 \, b c d x - 3 \, b c^{2} + 2 \, \sqrt {2} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {b c} \sqrt {d x + c}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 4 \, {\left (183 \, c d^{3} x^{3} + 61 \, c^{2} d^{2} x^{2} - 127 \, c^{3} d x - 53 \, c^{4}\right )} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c}}{768 \, {\left (b^{3} c d^{10} x^{5} + b^{3} c^{2} d^{9} x^{4} - 2 \, b^{3} c^{3} d^{8} x^{3} - 2 \, b^{3} c^{4} d^{7} x^{2} + b^{3} c^{5} d^{6} x + b^{3} c^{6} d^{5}\right )}}, \frac {201 \, \sqrt {2} {\left (d^{5} x^{5} + c d^{4} x^{4} - 2 \, c^{2} d^{3} x^{3} - 2 \, c^{3} d^{2} x^{2} + c^{4} d x + c^{5}\right )} \sqrt {-b c} \arctan \left (\frac {\sqrt {2} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {-b c} \sqrt {d x + c}}{2 \, {\left (b c d x + b c^{2}\right )}}\right ) + 2 \, {\left (183 \, c d^{3} x^{3} + 61 \, c^{2} d^{2} x^{2} - 127 \, c^{3} d x - 53 \, c^{4}\right )} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c}}{384 \, {\left (b^{3} c d^{10} x^{5} + b^{3} c^{2} d^{9} x^{4} - 2 \, b^{3} c^{3} d^{8} x^{3} - 2 \, b^{3} c^{4} d^{7} x^{2} + b^{3} c^{5} d^{6} x + b^{3} c^{6} d^{5}\right )}}\right ] \] Input:
integrate(x^4/(d*x+c)^(1/2)/(-b*d^2*x^2+b*c^2)^(5/2),x, algorithm="fricas" )
Output:
[1/768*(201*sqrt(2)*(d^5*x^5 + c*d^4*x^4 - 2*c^2*d^3*x^3 - 2*c^3*d^2*x^2 + c^4*d*x + c^5)*sqrt(b*c)*log(-(b*d^2*x^2 - 2*b*c*d*x - 3*b*c^2 + 2*sqrt(2 )*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(b*c)*sqrt(d*x + c))/(d^2*x^2 + 2*c*d*x + c ^2)) + 4*(183*c*d^3*x^3 + 61*c^2*d^2*x^2 - 127*c^3*d*x - 53*c^4)*sqrt(-b*d ^2*x^2 + b*c^2)*sqrt(d*x + c))/(b^3*c*d^10*x^5 + b^3*c^2*d^9*x^4 - 2*b^3*c ^3*d^8*x^3 - 2*b^3*c^4*d^7*x^2 + b^3*c^5*d^6*x + b^3*c^6*d^5), 1/384*(201* sqrt(2)*(d^5*x^5 + c*d^4*x^4 - 2*c^2*d^3*x^3 - 2*c^3*d^2*x^2 + c^4*d*x + c ^5)*sqrt(-b*c)*arctan(1/2*sqrt(2)*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(-b*c)*sqrt (d*x + c)/(b*c*d*x + b*c^2)) + 2*(183*c*d^3*x^3 + 61*c^2*d^2*x^2 - 127*c^3 *d*x - 53*c^4)*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c))/(b^3*c*d^10*x^5 + b ^3*c^2*d^9*x^4 - 2*b^3*c^3*d^8*x^3 - 2*b^3*c^4*d^7*x^2 + b^3*c^5*d^6*x + b ^3*c^6*d^5)]
\[ \int \frac {x^4}{\sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{5/2}} \, dx=\int \frac {x^{4}}{\left (- b \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}} \sqrt {c + d x}}\, dx \] Input:
integrate(x**4/(d*x+c)**(1/2)/(-b*d**2*x**2+b*c**2)**(5/2),x)
Output:
Integral(x**4/((-b*(-c + d*x)*(c + d*x))**(5/2)*sqrt(c + d*x)), x)
\[ \int \frac {x^4}{\sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{5/2}} \, dx=\int { \frac {x^{4}}{{\left (-b d^{2} x^{2} + b c^{2}\right )}^{\frac {5}{2}} \sqrt {d x + c}} \,d x } \] Input:
integrate(x^4/(d*x+c)^(1/2)/(-b*d^2*x^2+b*c^2)^(5/2),x, algorithm="maxima" )
Output:
integrate(x^4/((-b*d^2*x^2 + b*c^2)^(5/2)*sqrt(d*x + c)), x)
Time = 0.14 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.68 \[ \int \frac {x^4}{\sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{5/2}} \, dx=\frac {\frac {201 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (d x + c\right )} b + 2 \, b c}}{2 \, \sqrt {-b c}}\right )}{\sqrt {-b c} b^{2} d^{4}} - \frac {16 \, {\left (15 \, {\left (d x + c\right )} b - 28 \, b c\right )}}{{\left ({\left (d x + c\right )} b - 2 \, b c\right )} \sqrt {-{\left (d x + c\right )} b + 2 \, b c} b^{2} d^{4}} + \frac {6 \, {\left (38 \, \sqrt {-{\left (d x + c\right )} b + 2 \, b c} b c - 21 \, {\left (-{\left (d x + c\right )} b + 2 \, b c\right )}^{\frac {3}{2}}\right )}}{{\left (d x + c\right )}^{2} b^{4} d^{4}}}{384 \, d} \] Input:
integrate(x^4/(d*x+c)^(1/2)/(-b*d^2*x^2+b*c^2)^(5/2),x, algorithm="giac")
Output:
1/384*(201*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-(d*x + c)*b + 2*b*c)/sqrt(-b*c ))/(sqrt(-b*c)*b^2*d^4) - 16*(15*(d*x + c)*b - 28*b*c)/(((d*x + c)*b - 2*b *c)*sqrt(-(d*x + c)*b + 2*b*c)*b^2*d^4) + 6*(38*sqrt(-(d*x + c)*b + 2*b*c) *b*c - 21*(-(d*x + c)*b + 2*b*c)^(3/2))/((d*x + c)^2*b^4*d^4))/d
Timed out. \[ \int \frac {x^4}{\sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{5/2}} \, dx=\int \frac {x^4}{{\left (b\,c^2-b\,d^2\,x^2\right )}^{5/2}\,\sqrt {c+d\,x}} \,d x \] Input:
int(x^4/((b*c^2 - b*d^2*x^2)^(5/2)*(c + d*x)^(1/2)),x)
Output:
int(x^4/((b*c^2 - b*d^2*x^2)^(5/2)*(c + d*x)^(1/2)), x)
Time = 0.23 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.54 \[ \int \frac {x^4}{\sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {b}\, \left (201 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) c^{3}+201 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) c^{2} d x -201 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) c \,d^{2} x^{2}-201 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) d^{3} x^{3}-201 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) c^{3}-201 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) c^{2} d x +201 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) c \,d^{2} x^{2}+201 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) d^{3} x^{3}-212 c^{4}-508 c^{3} d x +244 c^{2} d^{2} x^{2}+732 c \,d^{3} x^{3}\right )}{768 \sqrt {-d x +c}\, b^{3} c \,d^{5} \left (-d^{3} x^{3}-c \,d^{2} x^{2}+c^{2} d x +c^{3}\right )} \] Input:
int(x^4/(d*x+c)^(1/2)/(-b*d^2*x^2+b*c^2)^(5/2),x)
Output:
(sqrt(b)*(201*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sq rt(2))*c**3 + 201*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c )*sqrt(2))*c**2*d*x - 201*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*c*d**2*x**2 - 201*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqr t(c - d*x) - sqrt(c)*sqrt(2))*d**3*x**3 - 201*sqrt(c)*sqrt(c - d*x)*sqrt(2 )*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*c**3 - 201*sqrt(c)*sqrt(c - d*x)*sq rt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*c**2*d*x + 201*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*c*d**2*x**2 + 201*sqrt( c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*d**3*x**3 - 212*c**4 - 508*c**3*d*x + 244*c**2*d**2*x**2 + 732*c*d**3*x**3))/(768*sqrt (c - d*x)*b**3*c*d**5*(c**3 + c**2*d*x - c*d**2*x**2 - d**3*x**3))