Optimal. Leaf size=129 \[ -\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{p+1}}{2 b^2 f (p+1)}+\frac {\left (a+b \tan ^2(e+f x)\right )^{p+2}}{2 b^2 f (p+2)}-\frac {\left (a+b \tan ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \tan ^2(e+f x)+a}{a-b}\right )}{2 f (p+1) (a-b)} \]
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Rubi [A] time = 0.17, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3670, 446, 88, 68} \[ -\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{p+1}}{2 b^2 f (p+1)}+\frac {\left (a+b \tan ^2(e+f x)\right )^{p+2}}{2 b^2 f (p+2)}-\frac {\left (a+b \tan ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \tan ^2(e+f x)+a}{a-b}\right )}{2 f (p+1) (a-b)} \]
Antiderivative was successfully verified.
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Rule 68
Rule 88
Rule 446
Rule 3670
Rubi steps
\begin {align*} \int \tan ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^5 \left (a+b x^2\right )^p}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^2 (a+b x)^p}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {(-a-b) (a+b x)^p}{b}+\frac {(a+b x)^p}{1+x}+\frac {(a+b x)^{1+p}}{b}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{2 b^2 f (1+p)}+\frac {\left (a+b \tan ^2(e+f x)\right )^{2+p}}{2 b^2 f (2+p)}+\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^p}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{2 b^2 f (1+p)}-\frac {\, _2F_1\left (1,1+p;2+p;\frac {a+b \tan ^2(e+f x)}{a-b}\right ) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{2 (a-b) f (1+p)}+\frac {\left (a+b \tan ^2(e+f x)\right )^{2+p}}{2 b^2 f (2+p)}\\ \end {align*}
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Mathematica [A] time = 0.91, size = 106, normalized size = 0.82 \[ \frac {\left (a+b \tan ^2(e+f x)\right )^{p+1} \left (b^2 (p+2) \, _2F_1\left (1,p+1;p+2;\frac {b \tan ^2(e+f x)+a}{a-b}\right )+(a-b) \left (a-b (p+1) \tan ^2(e+f x)+b (p+2)\right )\right )}{2 b^2 f (p+1) (p+2) (b-a)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{5}, x\right ) \]
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 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.16, size = 0, normalized size = 0.00 \[ \int \left (\tan ^{5}\left (f x +e \right )\right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]
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 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{5}\,{d x} \]
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 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^5\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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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.
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