Optimal. Leaf size=243 \[ \frac {a^3 (e \tan (c+d x))^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )}{d e (m+1)}+\frac {a^3 \sec ^3(c+d x) \cos ^2(c+d x)^{\frac {m+4}{2}} (e \tan (c+d x))^{m+1} \, _2F_1\left (\frac {m+1}{2},\frac {m+4}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {3 a^3 \sec (c+d x) \cos ^2(c+d x)^{\frac {m+2}{2}} (e \tan (c+d x))^{m+1} \, _2F_1\left (\frac {m+1}{2},\frac {m+2}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {3 a^3 (e \tan (c+d x))^{m+1}}{d e (m+1)} \]
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Rubi [A] time = 0.23, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3886, 3476, 364, 2617, 2607, 32} \[ \frac {a^3 (e \tan (c+d x))^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )}{d e (m+1)}+\frac {a^3 \sec ^3(c+d x) \cos ^2(c+d x)^{\frac {m+4}{2}} (e \tan (c+d x))^{m+1} \, _2F_1\left (\frac {m+1}{2},\frac {m+4}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {3 a^3 \sec (c+d x) \cos ^2(c+d x)^{\frac {m+2}{2}} (e \tan (c+d x))^{m+1} \, _2F_1\left (\frac {m+1}{2},\frac {m+2}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {3 a^3 (e \tan (c+d x))^{m+1}}{d e (m+1)} \]
Antiderivative was successfully verified.
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Rule 32
Rule 364
Rule 2607
Rule 2617
Rule 3476
Rule 3886
Rubi steps
\begin {align*} \int (a+a \sec (c+d x))^3 (e \tan (c+d x))^m \, dx &=\int \left (a^3 (e \tan (c+d x))^m+3 a^3 \sec (c+d x) (e \tan (c+d x))^m+3 a^3 \sec ^2(c+d x) (e \tan (c+d x))^m+a^3 \sec ^3(c+d x) (e \tan (c+d x))^m\right ) \, dx\\ &=a^3 \int (e \tan (c+d x))^m \, dx+a^3 \int \sec ^3(c+d x) (e \tan (c+d x))^m \, dx+\left (3 a^3\right ) \int \sec (c+d x) (e \tan (c+d x))^m \, dx+\left (3 a^3\right ) \int \sec ^2(c+d x) (e \tan (c+d x))^m \, dx\\ &=\frac {3 a^3 \cos ^2(c+d x)^{\frac {2+m}{2}} \, _2F_1\left (\frac {1+m}{2},\frac {2+m}{2};\frac {3+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{1+m}}{d e (1+m)}+\frac {a^3 \cos ^2(c+d x)^{\frac {4+m}{2}} \, _2F_1\left (\frac {1+m}{2},\frac {4+m}{2};\frac {3+m}{2};\sin ^2(c+d x)\right ) \sec ^3(c+d x) (e \tan (c+d x))^{1+m}}{d e (1+m)}+\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int (e x)^m \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (a^3 e\right ) \operatorname {Subst}\left (\int \frac {x^m}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{d}\\ &=\frac {3 a^3 (e \tan (c+d x))^{1+m}}{d e (1+m)}+\frac {a^3 \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{1+m}}{d e (1+m)}+\frac {3 a^3 \cos ^2(c+d x)^{\frac {2+m}{2}} \, _2F_1\left (\frac {1+m}{2},\frac {2+m}{2};\frac {3+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{1+m}}{d e (1+m)}+\frac {a^3 \cos ^2(c+d x)^{\frac {4+m}{2}} \, _2F_1\left (\frac {1+m}{2},\frac {4+m}{2};\frac {3+m}{2};\sin ^2(c+d x)\right ) \sec ^3(c+d x) (e \tan (c+d x))^{1+m}}{d e (1+m)}\\ \end {align*}
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Mathematica [C] time = 6.23, size = 391, normalized size = 1.61 \[ \frac {a^3 e \sec ^6\left (\frac {1}{2} (c+d x)\right ) (\sec (c+d x)+1)^3 \left (-\tan ^2(c+d x)\right )^{\frac {1-m}{2}} (e \tan (c+d x))^{m-1} \left (\, _2F_1\left (\frac {3}{2},\frac {1-m}{2};\frac {5}{2};\sec ^2(c+d x)\right )+9 \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3}{2};\sec ^2(c+d x)\right )\right )}{24 d}+\frac {a^3 2^{-m-4} (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \tan ^{-m}(c+d x) (e \tan (c+d x))^m \left (i 2^m (m+1) \left (-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^m \cos (c+d x) \, _2F_1\left (1,m;m+1;-\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )-i (m+1) \left (1+e^{2 i (c+d x)}\right )^m \left (-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^m \cos (c+d x) \, _2F_1\left (m,m;m+1;\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+3\ 2^{m+1} m \sin (c+d x) \tan ^m(c+d x)\right )}{d m (m+1)} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{3} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \sec \left (d x + c\right )^{2} + 3 \, a^{3} \sec \left (d x + c\right ) + a^{3}\right )} \left (e \tan \left (d x + c\right )\right )^{m}, x\right ) \]
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 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{3} \left (e \tan \left (d x + c\right )\right )^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.90, size = 0, normalized size = 0.00 \[ \int \left (a +a \sec \left (d x +c \right )\right )^{3} \left (e \tan \left (d x +c \right )\right )^{m}\, dx \]
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 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{3} \left (e \tan \left (d x + c\right )\right )^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3 \,d x \]
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 \[ a^{3} \left (\int \left (e \tan {\left (c + d x \right )}\right )^{m}\, dx + \int 3 \left (e \tan {\left (c + d x \right )}\right )^{m} \sec {\left (c + d x \right )}\, dx + \int 3 \left (e \tan {\left (c + d x \right )}\right )^{m} \sec ^{2}{\left (c + d x \right )}\, dx + \int \left (e \tan {\left (c + d x \right )}\right )^{m} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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