Optimal. Leaf size=252 \[ \frac {3 e^5 (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}+\frac {e^5 \, _2F_1\left (1,\frac {1}{2} (-5+m);\frac {1}{2} (-3+m);-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {3 e^5 \cos ^2(c+d x)^{\frac {1}{2} (-4+m)} \, _2F_1\left (\frac {1}{2} (-5+m),\frac {1}{2} (-4+m);\frac {1}{2} (-3+m);\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {e^5 \cos ^2(c+d x)^{\frac {1}{2} (-2+m)} \, _2F_1\left (\frac {1}{2} (-5+m),\frac {1}{2} (-2+m);\frac {1}{2} (-3+m);\sin ^2(c+d x)\right ) \sec ^3(c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)} \]
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Rubi [A]
time = 0.25, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3973, 3971,
3557, 371, 2697, 2687, 32} \begin {gather*} \frac {e^5 (e \tan (c+d x))^{m-5} \, _2F_1\left (1,\frac {m-5}{2};\frac {m-3}{2};-\tan ^2(c+d x)\right )}{a^3 d (5-m)}-\frac {e^5 \sec ^3(c+d x) \cos ^2(c+d x)^{\frac {m-2}{2}} (e \tan (c+d x))^{m-5} \, _2F_1\left (\frac {m-5}{2},\frac {m-2}{2};\frac {m-3}{2};\sin ^2(c+d x)\right )}{a^3 d (5-m)}-\frac {3 e^5 \sec (c+d x) \cos ^2(c+d x)^{\frac {m-4}{2}} (e \tan (c+d x))^{m-5} \, _2F_1\left (\frac {m-5}{2},\frac {m-4}{2};\frac {m-3}{2};\sin ^2(c+d x)\right )}{a^3 d (5-m)}+\frac {3 e^5 (e \tan (c+d x))^{m-5}}{a^3 d (5-m)} \end {gather*}
Antiderivative was successfully verified.
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Rule 32
Rule 371
Rule 2687
Rule 2697
Rule 3557
Rule 3971
Rule 3973
Rubi steps
\begin {align*} \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx &=\frac {e^6 \int (-a+a \sec (c+d x))^3 (e \tan (c+d x))^{-6+m} \, dx}{a^6}\\ &=\frac {e^6 \int \left (-a^3 (e \tan (c+d x))^{-6+m}+3 a^3 \sec (c+d x) (e \tan (c+d x))^{-6+m}-3 a^3 \sec ^2(c+d x) (e \tan (c+d x))^{-6+m}+a^3 \sec ^3(c+d x) (e \tan (c+d x))^{-6+m}\right ) \, dx}{a^6}\\ &=-\frac {e^6 \int (e \tan (c+d x))^{-6+m} \, dx}{a^3}+\frac {e^6 \int \sec ^3(c+d x) (e \tan (c+d x))^{-6+m} \, dx}{a^3}+\frac {\left (3 e^6\right ) \int \sec (c+d x) (e \tan (c+d x))^{-6+m} \, dx}{a^3}-\frac {\left (3 e^6\right ) \int \sec ^2(c+d x) (e \tan (c+d x))^{-6+m} \, dx}{a^3}\\ &=-\frac {3 e^5 \cos ^2(c+d x)^{\frac {1}{2} (-4+m)} \, _2F_1\left (\frac {1}{2} (-5+m),\frac {1}{2} (-4+m);\frac {1}{2} (-3+m);\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {e^5 \cos ^2(c+d x)^{\frac {1}{2} (-2+m)} \, _2F_1\left (\frac {1}{2} (-5+m),\frac {1}{2} (-2+m);\frac {1}{2} (-3+m);\sin ^2(c+d x)\right ) \sec ^3(c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {\left (3 e^6\right ) \text {Subst}\left (\int (e x)^{-6+m} \, dx,x,\tan (c+d x)\right )}{a^3 d}-\frac {e^7 \text {Subst}\left (\int \frac {x^{-6+m}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a^3 d}\\ &=\frac {3 e^5 (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}+\frac {e^5 \, _2F_1\left (1,\frac {1}{2} (-5+m);\frac {1}{2} (-3+m);-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {3 e^5 \cos ^2(c+d x)^{\frac {1}{2} (-4+m)} \, _2F_1\left (\frac {1}{2} (-5+m),\frac {1}{2} (-4+m);\frac {1}{2} (-3+m);\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {e^5 \cos ^2(c+d x)^{\frac {1}{2} (-2+m)} \, _2F_1\left (\frac {1}{2} (-5+m),\frac {1}{2} (-2+m);\frac {1}{2} (-3+m);\sin ^2(c+d x)\right ) \sec ^3(c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}\\ \end {align*}
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Mathematica [F]
time = 10.78, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.15, size = 0, normalized size = 0.00 \[\int \frac {\left (e \tan \left (d x +c \right )\right )^{m}}{\left (a +a \sec \left (d x +c \right )\right )^{3}}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{m}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^3\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m}{a^3\,{\left (\cos \left (c+d\,x\right )+1\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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