Optimal. Leaf size=242 \[ \frac {\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^q \left (a^2 d^2 \left (q^2+3 q+2\right )+2 a b c d (p+1) (q+1)+b^2 c^2 \left (p^2+3 p+2\right )\right ) \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \, _2F_1\left (p+1,-q;p+2;-\frac {d \left (b x^2+a\right )}{b c-a d}\right )}{2 b^3 d^2 (p+1) (p+q+2) (p+q+3)}-\frac {\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^{q+1} (a d (q+2)+b c (p+2))}{2 b^2 d^2 (p+q+2) (p+q+3)}+\frac {x^2 \left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^{q+1}}{2 b d (p+q+3)} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.31, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {446, 90, 80, 70, 69} \[ \frac {\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^q \left (a^2 d^2 \left (q^2+3 q+2\right )+2 a b c d (p+1) (q+1)+b^2 c^2 \left (p^2+3 p+2\right )\right ) \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \, _2F_1\left (p+1,-q;p+2;-\frac {d \left (b x^2+a\right )}{b c-a d}\right )}{2 b^3 d^2 (p+1) (p+q+2) (p+q+3)}-\frac {\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^{q+1} (a d (q+2)+b c (p+2))}{2 b^2 d^2 (p+q+2) (p+q+3)}+\frac {x^2 \left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^{q+1}}{2 b d (p+q+3)} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 69
Rule 70
Rule 80
Rule 90
Rule 446
Rubi steps
\begin {align*} \int x^5 \left (a+b x^2\right )^p \left (c+d x^2\right )^q \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x^2 (a+b x)^p (c+d x)^q \, dx,x,x^2\right )\\ &=\frac {x^2 \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b d (3+p+q)}+\frac {\operatorname {Subst}\left (\int (a+b x)^p (c+d x)^q (-a c-(b c (2+p)+a d (2+q)) x) \, dx,x,x^2\right )}{2 b d (3+p+q)}\\ &=-\frac {(b c (2+p)+a d (2+q)) \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b^2 d^2 (2+p+q) (3+p+q)}+\frac {x^2 \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b d (3+p+q)}+\frac {\left (b^2 c^2 \left (2+3 p+p^2\right )+2 a b c d (1+p) (1+q)+a^2 d^2 \left (2+3 q+q^2\right )\right ) \operatorname {Subst}\left (\int (a+b x)^p (c+d x)^q \, dx,x,x^2\right )}{2 b^2 d^2 (2+p+q) (3+p+q)}\\ &=-\frac {(b c (2+p)+a d (2+q)) \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b^2 d^2 (2+p+q) (3+p+q)}+\frac {x^2 \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b d (3+p+q)}+\frac {\left (\left (b^2 c^2 \left (2+3 p+p^2\right )+2 a b c d (1+p) (1+q)+a^2 d^2 \left (2+3 q+q^2\right )\right ) \left (c+d x^2\right )^q \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q}\right ) \operatorname {Subst}\left (\int (a+b x)^p \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^q \, dx,x,x^2\right )}{2 b^2 d^2 (2+p+q) (3+p+q)}\\ &=-\frac {(b c (2+p)+a d (2+q)) \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b^2 d^2 (2+p+q) (3+p+q)}+\frac {x^2 \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b d (3+p+q)}+\frac {\left (b^2 c^2 \left (2+3 p+p^2\right )+2 a b c d (1+p) (1+q)+a^2 d^2 \left (2+3 q+q^2\right )\right ) \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^q \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \, _2F_1\left (1+p,-q;2+p;-\frac {d \left (a+b x^2\right )}{b c-a d}\right )}{2 b^3 d^2 (1+p) (2+p+q) (3+p+q)}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.30, size = 195, normalized size = 0.81 \[ \frac {\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^q \left (\frac {\left (a^2 d^2 \left (q^2+3 q+2\right )+2 a b c d (p+1) (q+1)+b^2 c^2 \left (p^2+3 p+2\right )\right ) \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \, _2F_1\left (p+1,-q;p+2;\frac {d \left (b x^2+a\right )}{a d-b c}\right )}{b^2 d (p+1) (p+q+2)}-\frac {\left (c+d x^2\right ) (a d (q+2)+b c (p+2))}{b d (p+q+2)}+x^2 \left (c+d x^2\right )\right )}{2 b d (p+q+3)} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [F] time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q} x^{5}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q} x^{5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [F] time = 0.06, size = 0, normalized size = 0.00 \[ \int x^{5} \left (b \,x^{2}+a \right )^{p} \left (d \,x^{2}+c \right )^{q}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q} x^{5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int x^5\,{\left (b\,x^2+a\right )}^p\,{\left (d\,x^2+c\right )}^q \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________