Optimal. Leaf size=94 \[ -\frac {\cos ^2(c+d x) (a \tan (c+d x)+b)}{2 d \left (a^2+b^2\right )}+\frac {a^2 b \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {a x \left (a^2-b^2\right )}{2 \left (a^2+b^2\right )^2} \]
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Rubi [A] time = 0.16, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3516, 1647, 801, 635, 203, 260} \[ -\frac {\cos ^2(c+d x) (a \tan (c+d x)+b)}{2 d \left (a^2+b^2\right )}+\frac {a^2 b \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {a x \left (a^2-b^2\right )}{2 \left (a^2+b^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 203
Rule 260
Rule 635
Rule 801
Rule 1647
Rule 3516
Rubi steps
\begin {align*} \int \frac {\sin ^2(c+d x)}{a+b \tan (c+d x)} \, dx &=\frac {b \operatorname {Subst}\left (\int \frac {x^2}{(a+x) \left (b^2+x^2\right )^2} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d}-\frac {\operatorname {Subst}\left (\int \frac {-\frac {a^2 b^2}{a^2+b^2}+\frac {a b^2 x}{a^2+b^2}}{(a+x) \left (b^2+x^2\right )} \, dx,x,b \tan (c+d x)\right )}{2 b d}\\ &=-\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d}-\frac {\operatorname {Subst}\left (\int \left (-\frac {2 a^2 b^2}{\left (a^2+b^2\right )^2 (a+x)}-\frac {a b^2 \left (a^2-b^2-2 a x\right )}{\left (a^2+b^2\right )^2 \left (b^2+x^2\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{2 b d}\\ &=\frac {a^2 b \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d}+\frac {(a b) \operatorname {Subst}\left (\int \frac {a^2-b^2-2 a x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d}\\ &=\frac {a^2 b \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d}-\frac {\left (a^2 b\right ) \operatorname {Subst}\left (\int \frac {x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{\left (a^2+b^2\right )^2 d}+\frac {\left (a b \left (a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d}\\ &=\frac {a \left (a^2-b^2\right ) x}{2 \left (a^2+b^2\right )^2}+\frac {a^2 b \log (\cos (c+d x))}{\left (a^2+b^2\right )^2 d}+\frac {a^2 b \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d}\\ \end {align*}
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Mathematica [A] time = 0.77, size = 170, normalized size = 1.81 \[ -\frac {2 b^2 \left (a^2+b^2\right ) \cos ^2(c+d x)+2 a b \left (a^2+b^2\right ) \tan ^{-1}(\tan (c+d x))+a \left (b \left (a^2+b^2\right ) \sin (2 (c+d x))+2 a \left (-2 b^2 \log (a+b \tan (c+d x))+\left (a \sqrt {-b^2}+b^2\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+\left (b^2-a \sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )\right )\right )}{4 b d \left (a^2+b^2\right )^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 122, normalized size = 1.30 \[ \frac {a^{2} b \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) + {\left (a^{3} - a b^{2}\right )} d x - {\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.62, size = 184, normalized size = 1.96 \[ \frac {\frac {2 \, a^{2} b^{2} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac {a^{2} b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{3} - a b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {a^{2} b \tan \left (d x + c\right )^{2} - a^{3} \tan \left (d x + c\right ) - a b^{2} \tan \left (d x + c\right ) - b^{3}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (d x + c\right )^{2} + 1\right )}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.36, size = 238, normalized size = 2.53 \[ \frac {b \,a^{2} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )^{2}}-\frac {\tan \left (d x +c \right ) a^{3}}{2 d \left (a^{2}+b^{2}\right )^{2} \left (1+\tan ^{2}\left (d x +c \right )\right )}-\frac {\tan \left (d x +c \right ) b^{2} a}{2 d \left (a^{2}+b^{2}\right )^{2} \left (1+\tan ^{2}\left (d x +c \right )\right )}-\frac {a^{2} b}{2 d \left (a^{2}+b^{2}\right )^{2} \left (1+\tan ^{2}\left (d x +c \right )\right )}-\frac {b^{3}}{2 d \left (a^{2}+b^{2}\right )^{2} \left (1+\tan ^{2}\left (d x +c \right )\right )}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{2} b}{2 d \left (a^{2}+b^{2}\right )^{2}}+\frac {\arctan \left (\tan \left (d x +c \right )\right ) a^{3}}{2 d \left (a^{2}+b^{2}\right )^{2}}-\frac {\arctan \left (\tan \left (d x +c \right )\right ) a \,b^{2}}{2 d \left (a^{2}+b^{2}\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.58, size = 144, normalized size = 1.53 \[ \frac {\frac {2 \, a^{2} b \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {a^{2} b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{3} - a b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {a \tan \left (d x + c\right ) + b}{{\left (a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{2} + a^{2} + b^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.87, size = 147, normalized size = 1.56 \[ \frac {a^2\,b\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d\,{\left (a^2+b^2\right )}^2}-\frac {a\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{4\,d\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )}-\frac {{\cos \left (c+d\,x\right )}^2\,\left (\frac {b}{2\,\left (a^2+b^2\right )}+\frac {a\,\mathrm {tan}\left (c+d\,x\right )}{2\,\left (a^2+b^2\right )}\right )}{d}-\frac {a\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,d\,\left (-a^2+a\,b\,2{}\mathrm {i}+b^2\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{2}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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