Optimal. Leaf size=167 \[ \frac {b (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a-\sqrt {-b^2}}\right )}{2 \sqrt {-b^2} f (m+1) \left (a-\sqrt {-b^2}\right )}-\frac {b (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a+\sqrt {-b^2}}\right )}{2 \sqrt {-b^2} f (m+1) \left (a+\sqrt {-b^2}\right )} \]
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Rubi [A] time = 0.25, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3485, 712, 68} \[ \frac {b (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a-\sqrt {-b^2}}\right )}{2 \sqrt {-b^2} f (m+1) \left (a-\sqrt {-b^2}\right )}-\frac {b (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a+\sqrt {-b^2}}\right )}{2 \sqrt {-b^2} f (m+1) \left (a+\sqrt {-b^2}\right )} \]
Antiderivative was successfully verified.
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Rule 68
Rule 712
Rule 3485
Rubi steps
\begin {align*} \int (a+b \tan (e+f x))^m \, dx &=\frac {b \operatorname {Subst}\left (\int \frac {(a+x)^m}{b^2+x^2} \, dx,x,b \tan (e+f x)\right )}{f}\\ &=\frac {b \operatorname {Subst}\left (\int \left (\frac {\sqrt {-b^2} (a+x)^m}{2 b^2 \left (\sqrt {-b^2}-x\right )}+\frac {\sqrt {-b^2} (a+x)^m}{2 b^2 \left (\sqrt {-b^2}+x\right )}\right ) \, dx,x,b \tan (e+f x)\right )}{f}\\ &=-\frac {b \operatorname {Subst}\left (\int \frac {(a+x)^m}{\sqrt {-b^2}-x} \, dx,x,b \tan (e+f x)\right )}{2 \sqrt {-b^2} f}-\frac {b \operatorname {Subst}\left (\int \frac {(a+x)^m}{\sqrt {-b^2}+x} \, dx,x,b \tan (e+f x)\right )}{2 \sqrt {-b^2} f}\\ &=\frac {b \, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a-\sqrt {-b^2}}\right ) (a+b \tan (e+f x))^{1+m}}{2 \sqrt {-b^2} \left (a-\sqrt {-b^2}\right ) f (1+m)}-\frac {b \, _2F_1\left (1,1+m;2+m;\frac {a+b \tan (e+f x)}{a+\sqrt {-b^2}}\right ) (a+b \tan (e+f x))^{1+m}}{2 \sqrt {-b^2} \left (a+\sqrt {-b^2}\right ) f (1+m)}\\ \end {align*}
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Mathematica [C] time = 0.16, size = 118, normalized size = 0.71 \[ \frac {(a+b \tan (e+f x))^{m+1} \left ((a+i b) \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a-i b}\right )-(a-i b) \, _2F_1\left (1,m+1;m+2;\frac {a+b \tan (e+f x)}{a+i b}\right )\right )}{2 f (m+1) (a+i b) (b+i a)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \tan \left (f x + e\right ) + a\right )}^{m}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.70, size = 0, normalized size = 0.00 \[ \int \left (a +b \tan \left (f x +e \right )\right )^{m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m \,d x \]
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 \[ \int \left (a + b \tan {\left (e + f x \right )}\right )^{m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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