Integrand size = 21, antiderivative size = 216 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{(d+e x)^5} \, dx=\frac {3 b^3 \sqrt {b x+c x^2}}{64 d^2 (c d-b e)^2 (d+e x)}+\frac {b^2 \left (b x+c x^2\right )^{3/2}}{32 d (c d-b e)^2 x (d+e x)^2}+\frac {\left (b x+c x^2\right )^{5/2}}{4 (c d-b e) x (d+e x)^4}-\frac {b \left (b x+c x^2\right )^{5/2}}{8 (c d-b e)^2 x^2 (d+e x)^3}+\frac {3 b^4 \text {arctanh}\left (\frac {\sqrt {c d-b e} x}{\sqrt {d} \sqrt {b x+c x^2}}\right )}{64 d^{5/2} (c d-b e)^{5/2}} \] Output:
3/64*b^3*(c*x^2+b*x)^(1/2)/d^2/(-b*e+c*d)^2/(e*x+d)+1/32*b^2*(c*x^2+b*x)^( 3/2)/d/(-b*e+c*d)^2/x/(e*x+d)^2+1/4*(c*x^2+b*x)^(5/2)/(-b*e+c*d)/x/(e*x+d) ^4-1/8*b*(c*x^2+b*x)^(5/2)/(-b*e+c*d)^2/x^2/(e*x+d)^3+3/64*b^4*arctanh((-b *e+c*d)^(1/2)*x/d^(1/2)/(c*x^2+b*x)^(1/2))/d^(5/2)/(-b*e+c*d)^(5/2)
Time = 10.33 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.96 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{(d+e x)^5} \, dx=\frac {(x (b+c x))^{3/2} \left (-\frac {2 (b+c x)}{(d+e x)^4}+\frac {b (b+c x)}{(c d-b e) x (d+e x)^3}+\frac {b^2}{4 d (-c d+b e) x (d+e x)^2}+\frac {\frac {3 b^3}{d x (b+c x) (d+e x)}+\frac {3 b^4 \text {arctanh}\left (\frac {\sqrt {c d-b e} \sqrt {x}}{\sqrt {d} \sqrt {b+c x}}\right )}{d^{3/2} \sqrt {c d-b e} x^{3/2} (b+c x)^{3/2}}}{-8 c d^2+8 b d e}\right )}{8 (-c d+b e)} \] Input:
Integrate[(b*x + c*x^2)^(3/2)/(d + e*x)^5,x]
Output:
((x*(b + c*x))^(3/2)*((-2*(b + c*x))/(d + e*x)^4 + (b*(b + c*x))/((c*d - b *e)*x*(d + e*x)^3) + b^2/(4*d*(-(c*d) + b*e)*x*(d + e*x)^2) + ((3*b^3)/(d* x*(b + c*x)*(d + e*x)) + (3*b^4*ArcTanh[(Sqrt[c*d - b*e]*Sqrt[x])/(Sqrt[d] *Sqrt[b + c*x])])/(d^(3/2)*Sqrt[c*d - b*e]*x^(3/2)*(b + c*x)^(3/2)))/(-8*c *d^2 + 8*b*d*e)))/(8*(-(c*d) + b*e))
Time = 0.57 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1152, 1152, 1154, 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.
\(\displaystyle \int \frac {\left (b x+c x^2\right )^{3/2}}{(d+e x)^5} \, dx\) |
\(\Big \downarrow \) 1152 |
\(\displaystyle \frac {\left (b x+c x^2\right )^{3/2} (x (2 c d-b e)+b d)}{8 d (d+e x)^4 (c d-b e)}-\frac {3 b^2 \int \frac {\sqrt {c x^2+b x}}{(d+e x)^3}dx}{16 d (c d-b e)}\) |
\(\Big \downarrow \) 1152 |
\(\displaystyle \frac {\left (b x+c x^2\right )^{3/2} (x (2 c d-b e)+b d)}{8 d (d+e x)^4 (c d-b e)}-\frac {3 b^2 \left (\frac {\sqrt {b x+c x^2} (x (2 c d-b e)+b d)}{4 d (d+e x)^2 (c d-b e)}-\frac {b^2 \int \frac {1}{(d+e x) \sqrt {c x^2+b x}}dx}{8 d (c d-b e)}\right )}{16 d (c d-b e)}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {\left (b x+c x^2\right )^{3/2} (x (2 c d-b e)+b d)}{8 d (d+e x)^4 (c d-b e)}-\frac {3 b^2 \left (\frac {b^2 \int \frac {1}{4 d (c d-b e)-\frac {(b d+(2 c d-b e) x)^2}{c x^2+b x}}d\left (-\frac {b d+(2 c d-b e) x}{\sqrt {c x^2+b x}}\right )}{4 d (c d-b e)}+\frac {\sqrt {b x+c x^2} (x (2 c d-b e)+b d)}{4 d (d+e x)^2 (c d-b e)}\right )}{16 d (c d-b e)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\left (b x+c x^2\right )^{3/2} (x (2 c d-b e)+b d)}{8 d (d+e x)^4 (c d-b e)}-\frac {3 b^2 \left (\frac {\sqrt {b x+c x^2} (x (2 c d-b e)+b d)}{4 d (d+e x)^2 (c d-b e)}-\frac {b^2 \text {arctanh}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{8 d^{3/2} (c d-b e)^{3/2}}\right )}{16 d (c d-b e)}\) |
Input:
Int[(b*x + c*x^2)^(3/2)/(d + e*x)^5,x]
Output:
((b*d + (2*c*d - b*e)*x)*(b*x + c*x^2)^(3/2))/(8*d*(c*d - b*e)*(d + e*x)^4 ) - (3*b^2*(((b*d + (2*c*d - b*e)*x)*Sqrt[b*x + c*x^2])/(4*d*(c*d - b*e)*( d + e*x)^2) - (b^2*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b *e]*Sqrt[b*x + c*x^2])])/(8*d^(3/2)*(c*d - b*e)^(3/2))))/(16*d*(c*d - b*e) )
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[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-(d + e*x)^(m + 1))*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b *x + c*x^2)^p/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[p*((b^2 - 4*a *c)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[m + 2*p + 2, 0] && GtQ[p, 0]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Time = 0.86 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.72
method | result | size |
pseudoelliptic | \(-\frac {3 \left (2 c d x +b \left (-e x +d \right )\right ) \left (\left (e^{2} x^{2}+\frac {14}{3} d e x +d^{2}\right ) b^{2}-\frac {8 d c x \left (-e x +d \right ) b}{3}-\frac {8 d^{2} c^{2} x^{2}}{3}\right ) \sqrt {d \left (b e -c d \right )}\, \sqrt {x \left (c x +b \right )}+3 b^{4} \left (e x +d \right )^{4} \arctan \left (\frac {\sqrt {x \left (c x +b \right )}\, d}{x \sqrt {d \left (b e -c d \right )}}\right )}{64 \sqrt {d \left (b e -c d \right )}\, \left (e x +d \right )^{4} d^{2} \left (b e -c d \right )^{2}}\) | \(156\) |
default | \(\text {Expression too large to display}\) | \(4448\) |
Input:
int((c*x^2+b*x)^(3/2)/(e*x+d)^5,x,method=_RETURNVERBOSE)
Output:
-1/64/(d*(b*e-c*d))^(1/2)*(3*(2*c*d*x+b*(-e*x+d))*((e^2*x^2+14/3*d*e*x+d^2 )*b^2-8/3*d*c*x*(-e*x+d)*b-8/3*d^2*c^2*x^2)*(d*(b*e-c*d))^(1/2)*(x*(c*x+b) )^(1/2)+3*b^4*(e*x+d)^4*arctan((x*(c*x+b))^(1/2)/x*d/(d*(b*e-c*d))^(1/2))) /(e*x+d)^4/d^2/(b*e-c*d)^2
Leaf count of result is larger than twice the leaf count of optimal. 525 vs. \(2 (188) = 376\).
Time = 0.11 (sec) , antiderivative size = 1066, normalized size of antiderivative = 4.94 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{(d+e x)^5} \, dx =\text {Too large to display} \] Input:
integrate((c*x^2+b*x)^(3/2)/(e*x+d)^5,x, algorithm="fricas")
Output:
[1/128*(3*(b^4*e^4*x^4 + 4*b^4*d*e^3*x^3 + 6*b^4*d^2*e^2*x^2 + 4*b^4*d^3*e *x + b^4*d^4)*sqrt(c*d^2 - b*d*e)*log((b*d + (2*c*d - b*e)*x + 2*sqrt(c*d^ 2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d)) - 2*(3*b^3*c*d^5 - 3*b^4*d^4*e - (16*c^4*d^5 - 40*b*c^3*d^4*e + 26*b^2*c^2*d^3*e^2 + b^3*c*d^2*e^3 - 3*b^4* d*e^4)*x^3 - (24*b*c^3*d^5 - 68*b^2*c^2*d^4*e + 55*b^3*c*d^3*e^2 - 11*b^4* d^2*e^3)*x^2 - (2*b^2*c^2*d^5 - 13*b^3*c*d^4*e + 11*b^4*d^3*e^2)*x)*sqrt(c *x^2 + b*x))/(c^3*d^10 - 3*b*c^2*d^9*e + 3*b^2*c*d^8*e^2 - b^3*d^7*e^3 + ( c^3*d^6*e^4 - 3*b*c^2*d^5*e^5 + 3*b^2*c*d^4*e^6 - b^3*d^3*e^7)*x^4 + 4*(c^ 3*d^7*e^3 - 3*b*c^2*d^6*e^4 + 3*b^2*c*d^5*e^5 - b^3*d^4*e^6)*x^3 + 6*(c^3* d^8*e^2 - 3*b*c^2*d^7*e^3 + 3*b^2*c*d^6*e^4 - b^3*d^5*e^5)*x^2 + 4*(c^3*d^ 9*e - 3*b*c^2*d^8*e^2 + 3*b^2*c*d^7*e^3 - b^3*d^6*e^4)*x), -1/64*(3*(b^4*e ^4*x^4 + 4*b^4*d*e^3*x^3 + 6*b^4*d^2*e^2*x^2 + 4*b^4*d^3*e*x + b^4*d^4)*sq rt(-c*d^2 + b*d*e)*arctan(sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/(c*d*x + b*d)) + (3*b^3*c*d^5 - 3*b^4*d^4*e - (16*c^4*d^5 - 40*b*c^3*d^4*e + 26*b^2 *c^2*d^3*e^2 + b^3*c*d^2*e^3 - 3*b^4*d*e^4)*x^3 - (24*b*c^3*d^5 - 68*b^2*c ^2*d^4*e + 55*b^3*c*d^3*e^2 - 11*b^4*d^2*e^3)*x^2 - (2*b^2*c^2*d^5 - 13*b^ 3*c*d^4*e + 11*b^4*d^3*e^2)*x)*sqrt(c*x^2 + b*x))/(c^3*d^10 - 3*b*c^2*d^9* e + 3*b^2*c*d^8*e^2 - b^3*d^7*e^3 + (c^3*d^6*e^4 - 3*b*c^2*d^5*e^5 + 3*b^2 *c*d^4*e^6 - b^3*d^3*e^7)*x^4 + 4*(c^3*d^7*e^3 - 3*b*c^2*d^6*e^4 + 3*b^2*c *d^5*e^5 - b^3*d^4*e^6)*x^3 + 6*(c^3*d^8*e^2 - 3*b*c^2*d^7*e^3 + 3*b^2*...
\[ \int \frac {\left (b x+c x^2\right )^{3/2}}{(d+e x)^5} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{\left (d + e x\right )^{5}}\, dx \] Input:
integrate((c*x**2+b*x)**(3/2)/(e*x+d)**5,x)
Output:
Integral((x*(b + c*x))**(3/2)/(d + e*x)**5, x)
Exception generated. \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{(d+e x)^5} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((c*x^2+b*x)^(3/2)/(e*x+d)^5,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 1136 vs. \(2 (188) = 376\).
Time = 0.29 (sec) , antiderivative size = 1136, normalized size of antiderivative = 5.26 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{(d+e x)^5} \, dx=\text {Too large to display} \] Input:
integrate((c*x^2+b*x)^(3/2)/(e*x+d)^5,x, algorithm="giac")
Output:
-1/128*(3*b^4*log(abs(2*c*d*e - b*e^2 - 2*sqrt(c*d^2 - b*d*e)*(sqrt(c - 2* c*d/(e*x + d) + c*d^2/(e*x + d)^2 + b*e/(e*x + d) - b*d*e/(e*x + d)^2) + s qrt(c*d^2*e^2 - b*d*e^3)/((e*x + d)*e))*abs(e)))*sgn(1/(e*x + d))*sgn(e)/( (c^2*d^4*e - 2*b*c*d^3*e^2 + b^2*d^2*e^3)*sqrt(c*d^2 - b*d*e)*abs(e)) - (3 *b^4*e^5*log(abs(2*c*d*e - b*e^2 - 2*sqrt(c*d^2 - b*d*e)*sqrt(c)*abs(e))) - 32*sqrt(c*d^2 - b*d*e)*c^(7/2)*d^3*abs(e) + 48*sqrt(c*d^2 - b*d*e)*b*c^( 5/2)*d^2*e*abs(e) - 4*sqrt(c*d^2 - b*d*e)*b^2*c^(3/2)*d*e^2*abs(e) - 6*sqr t(c*d^2 - b*d*e)*b^3*sqrt(c)*e^3*abs(e))*sgn(1/(e*x + d))*sgn(e)/(sqrt(c*d ^2 - b*d*e)*c^2*d^4*e^6*abs(e) - 2*sqrt(c*d^2 - b*d*e)*b*c*d^3*e^7*abs(e) + sqrt(c*d^2 - b*d*e)*b^2*d^2*e^8*abs(e)) - 2*sqrt(c - 2*c*d/(e*x + d) + c *d^2/(e*x + d)^2 + b*e/(e*x + d) - b*d*e/(e*x + d)^2)*((16*c^4*d^5*e^17*sg n(1/(e*x + d))*sgn(e) - 40*b*c^3*d^4*e^18*sgn(1/(e*x + d))*sgn(e) + 26*b^2 *c^2*d^3*e^19*sgn(1/(e*x + d))*sgn(e) + b^3*c*d^2*e^20*sgn(1/(e*x + d))*sg n(e) - 3*b^4*d*e^21*sgn(1/(e*x + d))*sgn(e))/(c^3*d^6*e^23 - 3*b*c^2*d^5*e ^24 + 3*b^2*c*d^4*e^25 - b^3*d^3*e^26) - 2*((24*c^4*d^6*e^18*sgn(1/(e*x + d))*sgn(e) - 72*b*c^3*d^5*e^19*sgn(1/(e*x + d))*sgn(e) + 73*b^2*c^2*d^4*e^ 20*sgn(1/(e*x + d))*sgn(e) - 26*b^3*c*d^3*e^21*sgn(1/(e*x + d))*sgn(e) + b ^4*d^2*e^22*sgn(1/(e*x + d))*sgn(e))/(c^3*d^6*e^23 - 3*b*c^2*d^5*e^24 + 3* b^2*c*d^4*e^25 - b^3*d^3*e^26) - 4*(3*(2*c^4*d^7*e^19*sgn(1/(e*x + d))*sgn (e) - 7*b*c^3*d^6*e^20*sgn(1/(e*x + d))*sgn(e) + 9*b^2*c^2*d^5*e^21*sgn...
Timed out. \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{(d+e x)^5} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{{\left (d+e\,x\right )}^5} \,d x \] Input:
int((b*x + c*x^2)^(3/2)/(d + e*x)^5,x)
Output:
int((b*x + c*x^2)^(3/2)/(d + e*x)^5, x)
\[ \int \frac {\left (b x+c x^2\right )^{3/2}}{(d+e x)^5} \, dx=\int \frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{\left (e x +d \right )^{5}}d x \] Input:
int((c*x^2+b*x)^(3/2)/(e*x+d)^5,x)
Output:
int((c*x^2+b*x)^(3/2)/(e*x+d)^5,x)