Integrand size = 17, antiderivative size = 472 \[ \int \frac {(d+e x)^m}{\left (a+c x^2\right )^3} \, dx=\frac {(a e+c d x) (d+e x)^{1+m}}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}+\frac {(d+e x)^{1+m} \left (a e \left (a e^2 (3-m)+c d^2 (1+m)\right )+c d \left (3 c d^2+a e^2 (5-2 m)\right ) x\right )}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac {\left (a \sqrt {c} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m-\sqrt {-a} \left (3 c^2 d^4+a c d^2 e^2 \left (6-2 m-m^2\right )+a^2 e^4 \left (3-4 m+m^2\right )\right )\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{16 a^3 \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right )^2 (1+m)}+\frac {\left (a \sqrt {c} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m+\sqrt {-a} \left (3 c^2 d^4+a c d^2 e^2 \left (6-2 m-m^2\right )+a^2 e^4 \left (3-4 m+m^2\right )\right )\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{16 a^3 \left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right )^2 (1+m)} \]
1/4*(c*d*x+a*e)*(e*x+d)^(1+m)/a/(a*e^2+c*d^2)/(c*x^2+a)^2+1/8*(e*x+d)^(1+m )*(a*e*(a*e^2*(3-m)+c*d^2*(1+m))+c*d*(3*c*d^2+a*e^2*(5-2*m))*x)/a^2/(a*e^2 +c*d^2)^2/(c*x^2+a)+1/16*(e*x+d)^(1+m)*hypergeom([1, 1+m],[2+m],(e*x+d)*c^ (1/2)/(-e*(-a)^(1/2)+d*c^(1/2)))*(-(3*c^2*d^4+a*c*d^2*e^2*(-m^2-2*m+6)+a^2 *e^4*(m^2-4*m+3))*(-a)^(1/2)+a*d*e*(3*c*d^2+a*e^2*(5-2*m))*m*c^(1/2))/a^3/ (a*e^2+c*d^2)^2/(1+m)/(-e*(-a)^(1/2)+d*c^(1/2))+1/16*(e*x+d)^(1+m)*hyperge om([1, 1+m],[2+m],(e*x+d)*c^(1/2)/(e*(-a)^(1/2)+d*c^(1/2)))*((3*c^2*d^4+a* c*d^2*e^2*(-m^2-2*m+6)+a^2*e^4*(m^2-4*m+3))*(-a)^(1/2)+a*d*e*(3*c*d^2+a*e^ 2*(5-2*m))*m*c^(1/2))/a^3/(a*e^2+c*d^2)^2/(1+m)/(e*(-a)^(1/2)+d*c^(1/2))
Time = 1.03 (sec) , antiderivative size = 396, normalized size of antiderivative = 0.84 \[ \int \frac {(d+e x)^m}{\left (a+c x^2\right )^3} \, dx=\frac {(d+e x)^{1+m} \left (\frac {4 a \left (c d^2+a e^2\right ) (a e+c d x)}{\left (a+c x^2\right )^2}+\frac {2 \left (-a^2 e^3 (-3+m)+3 c^2 d^3 x+a c d e (d (1+m)+e (5-2 m) x)\right )}{a+c x^2}+\frac {\frac {\left (a \sqrt {c} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m+\sqrt {-a} \left (-3 c^2 d^4-a^2 e^4 \left (3-4 m+m^2\right )+a c d^2 e^2 \left (-6+2 m+m^2\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\sqrt {c} d-\sqrt {-a} e}+\frac {\left (a \sqrt {c} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m+\sqrt {-a} \left (3 c^2 d^4+a^2 e^4 \left (3-4 m+m^2\right )-a c d^2 e^2 \left (-6+2 m+m^2\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\sqrt {c} d+\sqrt {-a} e}}{a (1+m)}\right )}{16 a^2 \left (c d^2+a e^2\right )^2} \]
((d + e*x)^(1 + m)*((4*a*(c*d^2 + a*e^2)*(a*e + c*d*x))/(a + c*x^2)^2 + (2 *(-(a^2*e^3*(-3 + m)) + 3*c^2*d^3*x + a*c*d*e*(d*(1 + m) + e*(5 - 2*m)*x)) )/(a + c*x^2) + (((a*Sqrt[c]*d*e*(3*c*d^2 + a*e^2*(5 - 2*m))*m + Sqrt[-a]* (-3*c^2*d^4 - a^2*e^4*(3 - 4*m + m^2) + a*c*d^2*e^2*(-6 + 2*m + m^2)))*Hyp ergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]* e)])/(Sqrt[c]*d - Sqrt[-a]*e) + ((a*Sqrt[c]*d*e*(3*c*d^2 + a*e^2*(5 - 2*m) )*m + Sqrt[-a]*(3*c^2*d^4 + a^2*e^4*(3 - 4*m + m^2) - a*c*d^2*e^2*(-6 + 2* m + m^2)))*Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c] *d + Sqrt[-a]*e)])/(Sqrt[c]*d + Sqrt[-a]*e))/(a*(1 + m))))/(16*a^2*(c*d^2 + a*e^2)^2)
Time = 0.84 (sec) , antiderivative size = 488, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {496, 25, 686, 25, 27, 657, 2009}
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.
\(\displaystyle \int \frac {(d+e x)^m}{\left (a+c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 496 |
\(\displaystyle \frac {(d+e x)^{m+1} (a e+c d x)}{4 a \left (a+c x^2\right )^2 \left (a e^2+c d^2\right )}-\frac {\int -\frac {(d+e x)^m \left (3 c d^2+c e (2-m) x d+a e^2 (3-m)\right )}{\left (c x^2+a\right )^2}dx}{4 a \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {(d+e x)^m \left (3 c d^2+c e (2-m) x d+a e^2 (3-m)\right )}{\left (c x^2+a\right )^2}dx}{4 a \left (a e^2+c d^2\right )}+\frac {(d+e x)^{m+1} (a e+c d x)}{4 a \left (a+c x^2\right )^2 \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 686 |
\(\displaystyle \frac {\frac {(d+e x)^{m+1} \left (c d x \left (a e^2 (5-2 m)+3 c d^2\right )+a e \left (a e^2 (3-m)+c d^2 (m+1)\right )\right )}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}-\frac {\int -\frac {c (d+e x)^m \left (3 c^2 d^4+a c e^2 \left (-m^2-2 m+6\right ) d^2-c e \left (3 c d^2+a e^2 (5-2 m)\right ) m x d+a^2 e^4 \left (m^2-4 m+3\right )\right )}{c x^2+a}dx}{2 a c \left (a e^2+c d^2\right )}}{4 a \left (a e^2+c d^2\right )}+\frac {(d+e x)^{m+1} (a e+c d x)}{4 a \left (a+c x^2\right )^2 \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {c (d+e x)^m \left (3 c^2 d^4+a c e^2 \left (-m^2-2 m+6\right ) d^2-c e \left (3 c d^2+a e^2 (5-2 m)\right ) m x d+a^2 e^4 \left (m^2-4 m+3\right )\right )}{c x^2+a}dx}{2 a c \left (a e^2+c d^2\right )}+\frac {(d+e x)^{m+1} \left (c d x \left (a e^2 (5-2 m)+3 c d^2\right )+a e \left (a e^2 (3-m)+c d^2 (m+1)\right )\right )}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}}{4 a \left (a e^2+c d^2\right )}+\frac {(d+e x)^{m+1} (a e+c d x)}{4 a \left (a+c x^2\right )^2 \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {(d+e x)^m \left (3 c^2 d^4+a c e^2 \left (-m^2-2 m+6\right ) d^2-c e \left (3 c d^2+a e^2 (5-2 m)\right ) m x d+a^2 e^4 \left (m^2-4 m+3\right )\right )}{c x^2+a}dx}{2 a \left (a e^2+c d^2\right )}+\frac {(d+e x)^{m+1} \left (c d x \left (a e^2 (5-2 m)+3 c d^2\right )+a e \left (a e^2 (3-m)+c d^2 (m+1)\right )\right )}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}}{4 a \left (a e^2+c d^2\right )}+\frac {(d+e x)^{m+1} (a e+c d x)}{4 a \left (a+c x^2\right )^2 \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 657 |
\(\displaystyle \frac {\frac {\int \left (\frac {\left (a \sqrt {c} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m+\sqrt {-a} \left (3 c^2 d^4+a c e^2 \left (-m^2-2 m+6\right ) d^2+a^2 e^4 \left (m^2-4 m+3\right )\right )\right ) (d+e x)^m}{2 a \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {\left (\sqrt {-a} \left (3 c^2 d^4+a c e^2 \left (-m^2-2 m+6\right ) d^2+a^2 e^4 \left (m^2-4 m+3\right )\right )-a \sqrt {c} d e \left (3 c d^2+a e^2 (5-2 m)\right ) m\right ) (d+e x)^m}{2 a \left (\sqrt {c} x+\sqrt {-a}\right )}\right )dx}{2 a \left (a e^2+c d^2\right )}+\frac {(d+e x)^{m+1} \left (c d x \left (a e^2 (5-2 m)+3 c d^2\right )+a e \left (a e^2 (3-m)+c d^2 (m+1)\right )\right )}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}}{4 a \left (a e^2+c d^2\right )}+\frac {(d+e x)^{m+1} (a e+c d x)}{4 a \left (a+c x^2\right )^2 \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\frac {(d+e x)^{m+1} \left (a \sqrt {c} d e m \left (a e^2 (5-2 m)+3 c d^2\right )-\sqrt {-a} \left (a^2 e^4 \left (m^2-4 m+3\right )+a c d^2 e^2 \left (-m^2-2 m+6\right )+3 c^2 d^4\right )\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 a (m+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}+\frac {(d+e x)^{m+1} \left (\sqrt {-a} \left (a^2 e^4 \left (m^2-4 m+3\right )+a c d^2 e^2 \left (-m^2-2 m+6\right )+3 c^2 d^4\right )+a \sqrt {c} d e m \left (a e^2 (5-2 m)+3 c d^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 a (m+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}}{2 a \left (a e^2+c d^2\right )}+\frac {(d+e x)^{m+1} \left (c d x \left (a e^2 (5-2 m)+3 c d^2\right )+a e \left (a e^2 (3-m)+c d^2 (m+1)\right )\right )}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )}}{4 a \left (a e^2+c d^2\right )}+\frac {(d+e x)^{m+1} (a e+c d x)}{4 a \left (a+c x^2\right )^2 \left (a e^2+c d^2\right )}\) |
((a*e + c*d*x)*(d + e*x)^(1 + m))/(4*a*(c*d^2 + a*e^2)*(a + c*x^2)^2) + (( (d + e*x)^(1 + m)*(a*e*(a*e^2*(3 - m) + c*d^2*(1 + m)) + c*d*(3*c*d^2 + a* e^2*(5 - 2*m))*x))/(2*a*(c*d^2 + a*e^2)*(a + c*x^2)) + (((a*Sqrt[c]*d*e*(3 *c*d^2 + a*e^2*(5 - 2*m))*m - Sqrt[-a]*(3*c^2*d^4 + a*c*d^2*e^2*(6 - 2*m - m^2) + a^2*e^4*(3 - 4*m + m^2)))*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(2*a*(Sqrt[c]* d - Sqrt[-a]*e)*(1 + m)) + ((a*Sqrt[c]*d*e*(3*c*d^2 + a*e^2*(5 - 2*m))*m + Sqrt[-a]*(3*c^2*d^4 + a*c*d^2*e^2*(6 - 2*m - m^2) + a^2*e^4*(3 - 4*m + m^ 2)))*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (Sqrt[c]*(d + e* x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(2*a*(Sqrt[c]*d + Sqrt[-a]*e)*(1 + m)))/(2* a*(c*d^2 + a*e^2)))/(4*a*(c*d^2 + a*e^2))
3.8.26.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 *p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad raticQ[a, 0, b, c, d, n, p, x]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[ 1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)) Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Sim p[f*(c^2*d^2*(2*p + 3) + a*c*e^2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ [p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
\[\int \frac {\left (e x +d \right )^{m}}{\left (c \,x^{2}+a \right )^{3}}d x\]
\[ \int \frac {(d+e x)^m}{\left (a+c x^2\right )^3} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^m}{\left (a+c x^2\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {(d+e x)^m}{\left (a+c x^2\right )^3} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + a\right )}^{3}} \,d x } \]
\[ \int \frac {(d+e x)^m}{\left (a+c x^2\right )^3} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^m}{\left (a+c x^2\right )^3} \, dx=\int \frac {{\left (d+e\,x\right )}^m}{{\left (c\,x^2+a\right )}^3} \,d x \]