Optimal. Leaf size=91 \[ \frac {a \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac {b \cos \left (c+d x^2\right )}{2 d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^2\right )\right )} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.10, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3379, 2664, 12, 2660, 618, 204} \[ \frac {a \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac {b \cos \left (c+d x^2\right )}{2 d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^2\right )\right )} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 12
Rule 204
Rule 618
Rule 2660
Rule 2664
Rule 3379
Rubi steps
\begin {align*} \int \frac {x}{\left (a+b \sin \left (c+d x^2\right )\right )^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(a+b \sin (c+d x))^2} \, dx,x,x^2\right )\\ &=\frac {b \cos \left (c+d x^2\right )}{2 \left (a^2-b^2\right ) d \left (a+b \sin \left (c+d x^2\right )\right )}+\frac {\operatorname {Subst}\left (\int \frac {a}{a+b \sin (c+d x)} \, dx,x,x^2\right )}{2 \left (a^2-b^2\right )}\\ &=\frac {b \cos \left (c+d x^2\right )}{2 \left (a^2-b^2\right ) d \left (a+b \sin \left (c+d x^2\right )\right )}+\frac {a \operatorname {Subst}\left (\int \frac {1}{a+b \sin (c+d x)} \, dx,x,x^2\right )}{2 \left (a^2-b^2\right )}\\ &=\frac {b \cos \left (c+d x^2\right )}{2 \left (a^2-b^2\right ) d \left (a+b \sin \left (c+d x^2\right )\right )}+\frac {a \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} \left (c+d x^2\right )\right )\right )}{\left (a^2-b^2\right ) d}\\ &=\frac {b \cos \left (c+d x^2\right )}{2 \left (a^2-b^2\right ) d \left (a+b \sin \left (c+d x^2\right )\right )}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )\right )}{\left (a^2-b^2\right ) d}\\ &=\frac {a \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {b \cos \left (c+d x^2\right )}{2 \left (a^2-b^2\right ) d \left (a+b \sin \left (c+d x^2\right )\right )}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.20, size = 91, normalized size = 1.00 \[ \frac {\frac {2 a \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {b \cos \left (c+d x^2\right )}{a+b \sin \left (c+d x^2\right )}}{2 d (a-b) (a+b)} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.67, size = 366, normalized size = 4.02 \[ \left [\frac {{\left (a b \sin \left (d x^{2} + c\right ) + a^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x^{2} + c\right )^{2} - 2 \, a b \sin \left (d x^{2} + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x^{2} + c\right ) \sin \left (d x^{2} + c\right ) + b \cos \left (d x^{2} + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x^{2} + c\right )^{2} - 2 \, a b \sin \left (d x^{2} + c\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} \cos \left (d x^{2} + c\right )}{4 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x^{2} + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}}, -\frac {{\left (a b \sin \left (d x^{2} + c\right ) + a^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x^{2} + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x^{2} + c\right )}\right ) - {\left (a^{2} b - b^{3}\right )} \cos \left (d x^{2} + c\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x^{2} + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.43, size = 144, normalized size = 1.58 \[ \frac {{\left (\pi \left \lfloor \frac {d x^{2} + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a}{{\left (a^{2} d - b^{2} d\right )} \sqrt {a^{2} - b^{2}}} + \frac {b^{2} \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right ) + a b}{{\left (a^{3} d - a b^{2} d\right )} {\left (a \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right ) + a\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.10, size = 164, normalized size = 1.80 \[ \frac {b^{2} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{d \left (\left (\tan ^{2}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right ) b +a \right ) a \left (a^{2}-b^{2}\right )}+\frac {b}{d \left (\left (\tan ^{2}\left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right ) b +a \right ) \left (a^{2}-b^{2}\right )}+\frac {a \arctan \left (\frac {2 a \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a^{2}-b^{2}\right )^{\frac {3}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [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]
________________________________________________________________________________________
mupad [B] time = 5.10, size = 178, normalized size = 1.96 \[ \frac {\frac {b}{a^2-b^2}+\frac {b^2\,\mathrm {tan}\left (\frac {d\,x^2}{2}+\frac {c}{2}\right )}{a\,\left (a^2-b^2\right )}}{d\,\left (a\,{\mathrm {tan}\left (\frac {d\,x^2}{2}+\frac {c}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {d\,x^2}{2}+\frac {c}{2}\right )+a\right )}+\frac {a\,\mathrm {atan}\left (\frac {\left (a^2-b^2\right )\,\left (\frac {a^2\,\mathrm {tan}\left (\frac {d\,x^2}{2}+\frac {c}{2}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}+\frac {a\,\left (2\,a^2\,b-2\,b^3\right )}{2\,{\left (a+b\right )}^{3/2}\,\left (a^2-b^2\right )\,{\left (a-b\right )}^{3/2}}\right )}{a}\right )}{d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (a + b \sin {\left (c + d x^{2} \right )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________