Integrand size = 22, antiderivative size = 227 \[ \int \frac {x^5}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx=-\frac {a^2 \left (b c^2-a d^2-2 b c d x\right )}{3 b^2 \left (b c^2+a d^2\right )^2 \left (a+b x^2\right )^{3/2}}+\frac {a \left (3 \left (2 b^2 c^4-3 a b c^2 d^2-a^2 d^4\right )-2 b c d \left (7 b c^2+a d^2\right ) x\right )}{3 b^2 \left (b c^2+a d^2\right )^3 \sqrt {a+b x^2}}+\frac {c^5 \sqrt {a+b x^2}}{\left (b c^2+a d^2\right )^3 (c+d x)}-\frac {5 a c^4 d \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{\left (b c^2+a d^2\right )^{7/2}} \] Output:
-1/3*a^2*(-2*b*c*d*x-a*d^2+b*c^2)/b^2/(a*d^2+b*c^2)^2/(b*x^2+a)^(3/2)+1/3* a*(-3*a^2*d^4-9*a*b*c^2*d^2+6*b^2*c^4-2*b*c*d*(a*d^2+7*b*c^2)*x)/b^2/(a*d^ 2+b*c^2)^3/(b*x^2+a)^(1/2)+c^5*(b*x^2+a)^(1/2)/(a*d^2+b*c^2)^3/(d*x+c)-5*a *c^4*d*arctanh((-b*c*x+a*d)/(a*d^2+b*c^2)^(1/2)/(b*x^2+a)^(1/2))/(a*d^2+b* c^2)^(7/2)
Time = 1.44 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.11 \[ \int \frac {x^5}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx=\frac {3 b^4 c^5 x^4-2 a^4 d^4 (c+d x)+2 a b^3 c^3 x^2 \left (6 c^2-4 c d x-7 d^2 x^2\right )-3 a^3 b d^2 \left (3 c^3+3 c^2 d x+c d^2 x^2+d^3 x^3\right )+a^2 b^2 c \left (8 c^4-7 c^3 d x-21 c^2 d^2 x^2-11 c d^3 x^3-2 d^4 x^4\right )}{3 b^2 \left (b c^2+a d^2\right )^3 (c+d x) \left (a+b x^2\right )^{3/2}}+\frac {10 a c^4 d \arctan \left (\frac {\sqrt {-b c^2-a d^2} x}{\sqrt {a} (c+d x)-c \sqrt {a+b x^2}}\right )}{\left (-b c^2-a d^2\right )^{7/2}} \] Input:
Integrate[x^5/((c + d*x)^2*(a + b*x^2)^(5/2)),x]
Output:
(3*b^4*c^5*x^4 - 2*a^4*d^4*(c + d*x) + 2*a*b^3*c^3*x^2*(6*c^2 - 4*c*d*x - 7*d^2*x^2) - 3*a^3*b*d^2*(3*c^3 + 3*c^2*d*x + c*d^2*x^2 + d^3*x^3) + a^2*b ^2*c*(8*c^4 - 7*c^3*d*x - 21*c^2*d^2*x^2 - 11*c*d^3*x^3 - 2*d^4*x^4))/(3*b ^2*(b*c^2 + a*d^2)^3*(c + d*x)*(a + b*x^2)^(3/2)) + (10*a*c^4*d*ArcTan[(Sq rt[-(b*c^2) - a*d^2]*x)/(Sqrt[a]*(c + d*x) - c*Sqrt[a + b*x^2])])/(-(b*c^2 ) - a*d^2)^(7/2)
Time = 1.49 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {601, 2178, 27, 679, 488, 219}
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.
\(\displaystyle \int \frac {x^5}{\left (a+b x^2\right )^{5/2} (c+d x)^2} \, dx\) |
\(\Big \downarrow \) 601 |
\(\displaystyle -\frac {\int \frac {\frac {2 a^3 d c^3}{b \left (b c^2+a d^2\right )^2}+\frac {a^2 \left (3 b c^2+a d^2\right ) x c^2}{b \left (b c^2+a d^2\right )^2}-\frac {4 a^3 d^3 x^2 c}{b \left (b c^2+a d^2\right )^2}-\frac {3 a x^3}{b}}{(c+d x)^2 \left (b x^2+a\right )^{3/2}}dx}{3 a}-\frac {a^2 \left (-a d^2+b c^2-2 b c d x\right )}{3 b^2 \left (a+b x^2\right )^{3/2} \left (a d^2+b c^2\right )^2}\) |
\(\Big \downarrow \) 2178 |
\(\displaystyle -\frac {-\frac {\int \frac {3 a^2 b c^4 \left (4 a c d+\left (b c^2+5 a d^2\right ) x\right )}{\left (b c^2+a d^2\right )^3 (c+d x)^2 \sqrt {b x^2+a}}dx}{a b}-\frac {a^2 \left (3 \left (-a^2 d^4-3 a b c^2 d^2+2 b^2 c^4\right )-2 b c d x \left (a d^2+7 b c^2\right )\right )}{b^2 \sqrt {a+b x^2} \left (a d^2+b c^2\right )^3}}{3 a}-\frac {a^2 \left (-a d^2+b c^2-2 b c d x\right )}{3 b^2 \left (a+b x^2\right )^{3/2} \left (a d^2+b c^2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {-\frac {3 a c^4 \int \frac {4 a c d+\left (b c^2+5 a d^2\right ) x}{(c+d x)^2 \sqrt {b x^2+a}}dx}{\left (a d^2+b c^2\right )^3}-\frac {a^2 \left (3 \left (-a^2 d^4-3 a b c^2 d^2+2 b^2 c^4\right )-2 b c d x \left (a d^2+7 b c^2\right )\right )}{b^2 \sqrt {a+b x^2} \left (a d^2+b c^2\right )^3}}{3 a}-\frac {a^2 \left (-a d^2+b c^2-2 b c d x\right )}{3 b^2 \left (a+b x^2\right )^{3/2} \left (a d^2+b c^2\right )^2}\) |
\(\Big \downarrow \) 679 |
\(\displaystyle -\frac {-\frac {3 a c^4 \left (5 a d \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx+\frac {c \sqrt {a+b x^2}}{c+d x}\right )}{\left (a d^2+b c^2\right )^3}-\frac {a^2 \left (3 \left (-a^2 d^4-3 a b c^2 d^2+2 b^2 c^4\right )-2 b c d x \left (a d^2+7 b c^2\right )\right )}{b^2 \sqrt {a+b x^2} \left (a d^2+b c^2\right )^3}}{3 a}-\frac {a^2 \left (-a d^2+b c^2-2 b c d x\right )}{3 b^2 \left (a+b x^2\right )^{3/2} \left (a d^2+b c^2\right )^2}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle -\frac {-\frac {3 a c^4 \left (\frac {c \sqrt {a+b x^2}}{c+d x}-5 a d \int \frac {1}{b c^2+a d^2-\frac {(a d-b c x)^2}{b x^2+a}}d\frac {a d-b c x}{\sqrt {b x^2+a}}\right )}{\left (a d^2+b c^2\right )^3}-\frac {a^2 \left (3 \left (-a^2 d^4-3 a b c^2 d^2+2 b^2 c^4\right )-2 b c d x \left (a d^2+7 b c^2\right )\right )}{b^2 \sqrt {a+b x^2} \left (a d^2+b c^2\right )^3}}{3 a}-\frac {a^2 \left (-a d^2+b c^2-2 b c d x\right )}{3 b^2 \left (a+b x^2\right )^{3/2} \left (a d^2+b c^2\right )^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {-\frac {a^2 \left (3 \left (-a^2 d^4-3 a b c^2 d^2+2 b^2 c^4\right )-2 b c d x \left (a d^2+7 b c^2\right )\right )}{b^2 \sqrt {a+b x^2} \left (a d^2+b c^2\right )^3}-\frac {3 a c^4 \left (\frac {c \sqrt {a+b x^2}}{c+d x}-\frac {5 a d \text {arctanh}\left (\frac {a d-b c x}{\sqrt {a+b x^2} \sqrt {a d^2+b c^2}}\right )}{\sqrt {a d^2+b c^2}}\right )}{\left (a d^2+b c^2\right )^3}}{3 a}-\frac {a^2 \left (-a d^2+b c^2-2 b c d x\right )}{3 b^2 \left (a+b x^2\right )^{3/2} \left (a d^2+b c^2\right )^2}\) |
Input:
Int[x^5/((c + d*x)^2*(a + b*x^2)^(5/2)),x]
Output:
-1/3*(a^2*(b*c^2 - a*d^2 - 2*b*c*d*x))/(b^2*(b*c^2 + a*d^2)^2*(a + b*x^2)^ (3/2)) - (-((a^2*(3*(2*b^2*c^4 - 3*a*b*c^2*d^2 - a^2*d^4) - 2*b*c*d*(7*b*c ^2 + a*d^2)*x))/(b^2*(b*c^2 + a*d^2)^3*Sqrt[a + b*x^2])) - (3*a*c^4*((c*Sq rt[a + b*x^2])/(c + d*x) - (5*a*d*ArcTanh[(a*d - b*c*x)/(Sqrt[b*c^2 + a*d^ 2]*Sqrt[a + b*x^2])])/Sqrt[b*c^2 + a*d^2]))/(b*c^2 + a*d^2)^3)/(3*a)
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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) *((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* (2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1 )/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[(c*d*f + a*e*g)/(c*d^2 + a*e^2) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[Simplify[m + 2*p + 3], 0]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x^2, x], R = Coeff[Po lynomialRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 0], S = Coeff[Polynomia lRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*S - b*R*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(d + e*x )^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*b*(p + 1)*Qx)/(d + e*x)^m + (b*R*( 2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x ] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1662\) vs. \(2(208)=416\).
Time = 0.42 (sec) , antiderivative size = 1663, normalized size of antiderivative = 7.33
Input:
int(x^5/(d*x+c)^2/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/d^2*(-x^2/b/(b*x^2+a)^(3/2)-2/3*a/b^2/(b*x^2+a)^(3/2))-4*c^3/d^5*(1/3*x/ a/(b*x^2+a)^(3/2)+2/3/a^2/(b*x^2+a)^(1/2)*x)-2*c/d^3*(-1/2*x/b/(b*x^2+a)^( 3/2)+1/2*a/b*(1/3*x/a/(b*x^2+a)^(3/2)+2/3/a^2/(b*x^2+a)^(1/2)*x))-c^2/d^4/ b/(b*x^2+a)^(3/2)+5/d^6*c^4*(1/3/(a*d^2+b*c^2)*d^2/(b*(x+c/d)^2-2*b*c/d*(x +c/d)+(a*d^2+b*c^2)/d^2)^(3/2)+b*c*d/(a*d^2+b*c^2)*(2/3*(2*b*(x+c/d)-2*b*c /d)/(4*b*(a*d^2+b*c^2)/d^2-4*b^2*c^2/d^2)/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a* d^2+b*c^2)/d^2)^(3/2)+16/3*b/(4*b*(a*d^2+b*c^2)/d^2-4*b^2*c^2/d^2)^2*(2*b* (x+c/d)-2*b*c/d)/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2))+1/ (a*d^2+b*c^2)*d^2*(1/(a*d^2+b*c^2)*d^2/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2 +b*c^2)/d^2)^(1/2)+2*b*c*d/(a*d^2+b*c^2)*(2*b*(x+c/d)-2*b*c/d)/(4*b*(a*d^2 +b*c^2)/d^2-4*b^2*c^2/d^2)/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2) ^(1/2)-1/(a*d^2+b*c^2)*d^2/((a*d^2+b*c^2)/d^2)^(1/2)*ln((2*(a*d^2+b*c^2)/d ^2-2*b*c/d*(x+c/d)+2*((a*d^2+b*c^2)/d^2)^(1/2)*(b*(x+c/d)^2-2*b*c/d*(x+c/d )+(a*d^2+b*c^2)/d^2)^(1/2))/(x+c/d))))-c^5/d^7*(-1/(a*d^2+b*c^2)*d^2/(x+c/ d)/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(3/2)+5*b*c*d/(a*d^2+b* c^2)*(1/3/(a*d^2+b*c^2)*d^2/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2 )^(3/2)+b*c*d/(a*d^2+b*c^2)*(2/3*(2*b*(x+c/d)-2*b*c/d)/(4*b*(a*d^2+b*c^2)/ d^2-4*b^2*c^2/d^2)/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(3/2)+1 6/3*b/(4*b*(a*d^2+b*c^2)/d^2-4*b^2*c^2/d^2)^2*(2*b*(x+c/d)-2*b*c/d)/(b*(x+ c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2))+1/(a*d^2+b*c^2)*d^2*(1...
Leaf count of result is larger than twice the leaf count of optimal. 830 vs. \(2 (209) = 418\).
Time = 0.34 (sec) , antiderivative size = 1686, normalized size of antiderivative = 7.43 \[ \int \frac {x^5}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate(x^5/(d*x+c)^2/(b*x^2+a)^(5/2),x, algorithm="fricas")
Output:
[1/6*(15*(a*b^4*c^4*d^2*x^5 + a*b^4*c^5*d*x^4 + 2*a^2*b^3*c^4*d^2*x^3 + 2* a^2*b^3*c^5*d*x^2 + a^3*b^2*c^4*d^2*x + a^3*b^2*c^5*d)*sqrt(b*c^2 + a*d^2) *log((2*a*b*c*d*x - a*b*c^2 - 2*a^2*d^2 - (2*b^2*c^2 + a*b*d^2)*x^2 - 2*sq rt(b*c^2 + a*d^2)*(b*c*x - a*d)*sqrt(b*x^2 + a))/(d^2*x^2 + 2*c*d*x + c^2) ) + 2*(8*a^2*b^3*c^7 - a^3*b^2*c^5*d^2 - 11*a^4*b*c^3*d^4 - 2*a^5*c*d^6 + (3*b^5*c^7 - 11*a*b^4*c^5*d^2 - 16*a^2*b^3*c^3*d^4 - 2*a^3*b^2*c*d^6)*x^4 - (8*a*b^4*c^6*d + 19*a^2*b^3*c^4*d^3 + 14*a^3*b^2*c^2*d^5 + 3*a^4*b*d^7)* x^3 + 3*(4*a*b^4*c^7 - 3*a^2*b^3*c^5*d^2 - 8*a^3*b^2*c^3*d^4 - a^4*b*c*d^6 )*x^2 - (7*a^2*b^3*c^6*d + 16*a^3*b^2*c^4*d^3 + 11*a^4*b*c^2*d^5 + 2*a^5*d ^7)*x)*sqrt(b*x^2 + a))/(a^2*b^6*c^9 + 4*a^3*b^5*c^7*d^2 + 6*a^4*b^4*c^5*d ^4 + 4*a^5*b^3*c^3*d^6 + a^6*b^2*c*d^8 + (b^8*c^8*d + 4*a*b^7*c^6*d^3 + 6* a^2*b^6*c^4*d^5 + 4*a^3*b^5*c^2*d^7 + a^4*b^4*d^9)*x^5 + (b^8*c^9 + 4*a*b^ 7*c^7*d^2 + 6*a^2*b^6*c^5*d^4 + 4*a^3*b^5*c^3*d^6 + a^4*b^4*c*d^8)*x^4 + 2 *(a*b^7*c^8*d + 4*a^2*b^6*c^6*d^3 + 6*a^3*b^5*c^4*d^5 + 4*a^4*b^4*c^2*d^7 + a^5*b^3*d^9)*x^3 + 2*(a*b^7*c^9 + 4*a^2*b^6*c^7*d^2 + 6*a^3*b^5*c^5*d^4 + 4*a^4*b^4*c^3*d^6 + a^5*b^3*c*d^8)*x^2 + (a^2*b^6*c^8*d + 4*a^3*b^5*c^6* d^3 + 6*a^4*b^4*c^4*d^5 + 4*a^5*b^3*c^2*d^7 + a^6*b^2*d^9)*x), -1/3*(15*(a *b^4*c^4*d^2*x^5 + a*b^4*c^5*d*x^4 + 2*a^2*b^3*c^4*d^2*x^3 + 2*a^2*b^3*c^5 *d*x^2 + a^3*b^2*c^4*d^2*x + a^3*b^2*c^5*d)*sqrt(-b*c^2 - a*d^2)*arctan(sq rt(-b*c^2 - a*d^2)*(b*c*x - a*d)*sqrt(b*x^2 + a)/(a*b*c^2 + a^2*d^2 + (...
\[ \int \frac {x^5}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx=\int \frac {x^{5}}{\left (a + b x^{2}\right )^{\frac {5}{2}} \left (c + d x\right )^{2}}\, dx \] Input:
integrate(x**5/(d*x+c)**2/(b*x**2+a)**(5/2),x)
Output:
Integral(x**5/((a + b*x**2)**(5/2)*(c + d*x)**2), x)
Leaf count of result is larger than twice the leaf count of optimal. 966 vs. \(2 (209) = 418\).
Time = 0.15 (sec) , antiderivative size = 966, normalized size of antiderivative = 4.26 \[ \int \frac {x^5}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate(x^5/(d*x+c)^2/(b*x^2+a)^(5/2),x, algorithm="maxima")
Output:
-5*b^2*c^7*x/(sqrt(b*x^2 + a)*a*b^3*c^6*d^3 + 3*sqrt(b*x^2 + a)*a^2*b^2*c^ 4*d^5 + 3*sqrt(b*x^2 + a)*a^3*b*c^2*d^7 + sqrt(b*x^2 + a)*a^4*d^9) - 5/3*b ^2*c^7*x/((b*x^2 + a)^(3/2)*a*b^2*c^4*d^5 + 2*(b*x^2 + a)^(3/2)*a^2*b*c^2* d^7 + (b*x^2 + a)^(3/2)*a^3*d^9) - 10/3*b^2*c^7*x/(sqrt(b*x^2 + a)*a^2*b^2 *c^4*d^5 + 2*sqrt(b*x^2 + a)*a^3*b*c^2*d^7 + sqrt(b*x^2 + a)*a^4*d^9) - 5* b*c^6/(sqrt(b*x^2 + a)*b^3*c^6*d^2 + 3*sqrt(b*x^2 + a)*a*b^2*c^4*d^4 + 3*s qrt(b*x^2 + a)*a^2*b*c^2*d^6 + sqrt(b*x^2 + a)*a^3*d^8) - 5/3*b*c^6/((b*x^ 2 + a)^(3/2)*b^2*c^4*d^4 + 2*(b*x^2 + a)^(3/2)*a*b*c^2*d^6 + (b*x^2 + a)^( 3/2)*a^2*d^8) + 5*b*c^5*x/(sqrt(b*x^2 + a)*a*b^2*c^4*d^3 + 2*sqrt(b*x^2 + a)*a^2*b*c^2*d^5 + sqrt(b*x^2 + a)*a^3*d^7) + 3*b*c^5*x/((b*x^2 + a)^(3/2) *a*b*c^2*d^5 + (b*x^2 + a)^(3/2)*a^2*d^7) + 6*b*c^5*x/(sqrt(b*x^2 + a)*a^2 *b*c^2*d^5 + sqrt(b*x^2 + a)*a^3*d^7) + c^5/((b*x^2 + a)^(3/2)*b*c^2*d^5*x + (b*x^2 + a)^(3/2)*a*d^7*x + (b*x^2 + a)^(3/2)*b*c^3*d^4 + (b*x^2 + a)^( 3/2)*a*c*d^6) + 5*c^4/(sqrt(b*x^2 + a)*b^2*c^4*d^2 + 2*sqrt(b*x^2 + a)*a*b *c^2*d^4 + sqrt(b*x^2 + a)*a^2*d^6) + 5/3*c^4/((b*x^2 + a)^(3/2)*b*c^2*d^4 + (b*x^2 + a)^(3/2)*a*d^6) - x^2/((b*x^2 + a)^(3/2)*b*d^2) - 8/3*c^3*x/(s qrt(b*x^2 + a)*a^2*d^5) - 4/3*c^3*x/((b*x^2 + a)^(3/2)*a*d^5) + 2/3*c*x/(( b*x^2 + a)^(3/2)*b*d^3) - 2/3*c*x/(sqrt(b*x^2 + a)*a*b*d^3) - 5*b*c^6*arcs inh(b*c*x/(sqrt(a*b)*abs(d*x + c)) - a*d/(sqrt(a*b)*abs(d*x + c)))/((a + b *c^2/d^2)^(7/2)*d^8) + 5*c^4*arcsinh(b*c*x/(sqrt(a*b)*abs(d*x + c)) - a...
Timed out. \[ \int \frac {x^5}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(x^5/(d*x+c)^2/(b*x^2+a)^(5/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {x^5}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx=\int \frac {x^5}{{\left (b\,x^2+a\right )}^{5/2}\,{\left (c+d\,x\right )}^2} \,d x \] Input:
int(x^5/((a + b*x^2)^(5/2)*(c + d*x)^2),x)
Output:
int(x^5/((a + b*x^2)^(5/2)*(c + d*x)^2), x)
Time = 0.28 (sec) , antiderivative size = 1400, normalized size of antiderivative = 6.17 \[ \int \frac {x^5}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:
int(x^5/(d*x+c)^2/(b*x^2+a)^(5/2),x)
Output:
(15*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**3*b**2*c**5*d + 15*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2) *sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**3*b**2*c**4*d**2*x + 30*sqrt(a*d* *2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a** 2*b**3*c**5*d*x**2 + 30*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a* d**2 + b*c**2) - a*d + b*c*x)*a**2*b**3*c**4*d**2*x**3 + 15*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a*b**4*c **5*d*x**4 + 15*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b *c**2) - a*d + b*c*x)*a*b**4*c**4*d**2*x**5 - 15*sqrt(a*d**2 + b*c**2)*log (c + d*x)*a**3*b**2*c**5*d - 15*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a**3*b* *2*c**4*d**2*x - 30*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a**2*b**3*c**5*d*x* *2 - 30*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a**2*b**3*c**4*d**2*x**3 - 15*s qrt(a*d**2 + b*c**2)*log(c + d*x)*a*b**4*c**5*d*x**4 - 15*sqrt(a*d**2 + b* c**2)*log(c + d*x)*a*b**4*c**4*d**2*x**5 - 2*sqrt(a + b*x**2)*a**5*c*d**6 - 2*sqrt(a + b*x**2)*a**5*d**7*x - 11*sqrt(a + b*x**2)*a**4*b*c**3*d**4 - 11*sqrt(a + b*x**2)*a**4*b*c**2*d**5*x - 3*sqrt(a + b*x**2)*a**4*b*c*d**6* x**2 - 3*sqrt(a + b*x**2)*a**4*b*d**7*x**3 - sqrt(a + b*x**2)*a**3*b**2*c* *5*d**2 - 16*sqrt(a + b*x**2)*a**3*b**2*c**4*d**3*x - 24*sqrt(a + b*x**2)* a**3*b**2*c**3*d**4*x**2 - 14*sqrt(a + b*x**2)*a**3*b**2*c**2*d**5*x**3 - 2*sqrt(a + b*x**2)*a**3*b**2*c*d**6*x**4 + 8*sqrt(a + b*x**2)*a**2*b**3...