Integrand size = 21, antiderivative size = 276 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^n \, dx=-\frac {\left (a b^2 n+\sqrt {-b^2} \left (a^2+b^2 (1+n)\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{4 b \left (a^2+b^2\right ) \left (a-\sqrt {-b^2}\right ) d (1+n)}-\frac {\left (a b^2 n-\sqrt {-b^2} \left (a^2+b^2 (1+n)\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{4 b \left (a^2+b^2\right ) \left (a+\sqrt {-b^2}\right ) d (1+n)}-\frac {\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d} \]
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Time = 0.49 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3597, 1663, 845, 70} \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^n \, dx=-\frac {\left (\sqrt {-b^2} \left (a^2+b^2 (n+1)\right )+a b^2 n\right ) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )}{4 b d (n+1) \left (a^2+b^2\right ) \left (a-\sqrt {-b^2}\right )}-\frac {\left (a b^2 n-\sqrt {-b^2} \left (a^2+b^2 (n+1)\right )\right ) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{4 b d (n+1) \left (a^2+b^2\right ) \left (a+\sqrt {-b^2}\right )}-\frac {\cos ^2(c+d x) (a \tan (c+d x)+b) (a+b \tan (c+d x))^{n+1}}{2 d \left (a^2+b^2\right )} \]
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Rule 70
Rule 845
Rule 1663
Rule 3597
Rubi steps \begin{align*} \text {integral}& = \frac {b \text {Subst}\left (\int \frac {x^2 (a+x)^n}{\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)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d}-\frac {\text {Subst}\left (\int \frac {(a+x)^n \left (-b^2 \left (a^2+b^2 (1+n)\right )-a b^2 n x\right )}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 b \left (a^2+b^2\right ) d} \\ & = -\frac {\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d}-\frac {\text {Subst}\left (\int \left (\frac {\left (a b^4 n-b^2 \sqrt {-b^2} \left (a^2+b^2 (1+n)\right )\right ) (a+x)^n}{2 b^2 \left (\sqrt {-b^2}-x\right )}+\frac {\left (-a b^4 n-b^2 \sqrt {-b^2} \left (a^2+b^2 (1+n)\right )\right ) (a+x)^n}{2 b^2 \left (\sqrt {-b^2}+x\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{2 b \left (a^2+b^2\right ) d} \\ & = -\frac {\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d}-\frac {\left (a b^2 n-\sqrt {-b^2} \left (a^2+b^2 (1+n)\right )\right ) \text {Subst}\left (\int \frac {(a+x)^n}{\sqrt {-b^2}-x} \, dx,x,b \tan (c+d x)\right )}{4 b \left (a^2+b^2\right ) d}+\frac {\left (a b^2 n+\sqrt {-b^2} \left (a^2+b^2 (1+n)\right )\right ) \text {Subst}\left (\int \frac {(a+x)^n}{\sqrt {-b^2}+x} \, dx,x,b \tan (c+d x)\right )}{4 b \left (a^2+b^2\right ) d} \\ & = -\frac {\left (a b^2 n+\sqrt {-b^2} \left (a^2+b^2 (1+n)\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{4 b \left (a^2+b^2\right ) \left (a-\sqrt {-b^2}\right ) d (1+n)}-\frac {\left (a b^2 n-\sqrt {-b^2} \left (a^2+b^2 (1+n)\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{4 b \left (a^2+b^2\right ) \left (a+\sqrt {-b^2}\right ) d (1+n)}-\frac {\cos ^2(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{2 \left (a^2+b^2\right ) d} \\ \end{align*}
Time = 1.33 (sec) , antiderivative size = 270, normalized size of antiderivative = 0.98 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {\left (\left (a^3 \sqrt {-b^2}+a^2 b^2 (-1+n)-b^4 (1+n)-a \left (-b^2\right )^{3/2} (1+2 n)\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )-\left (a^3 \sqrt {-b^2}-a^2 b^2 (-1+n)+b^4 (1+n)-a \left (-b^2\right )^{3/2} (1+2 n)\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )+2 b \left (a^2+b^2\right ) (1+n) \cos (c+d x) (b \cos (c+d x)+a \sin (c+d x))\right ) (a+b \tan (c+d x))^{1+n}}{4 b \left (a^2+b^2\right ) \left (-a+\sqrt {-b^2}\right ) \left (a+\sqrt {-b^2}\right ) d (1+n)} \]
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\[\int \left (\sin ^{2}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{n}d x\]
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\[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{2} \,d x } \]
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\[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^n \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{n} \sin ^{2}{\left (c + d x \right )}\, dx \]
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\[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{2} \,d x } \]
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\[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{2} \,d x } \]
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Timed out. \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^n \, dx=\int {\sin \left (c+d\,x\right )}^2\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \]
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