Optimal. Leaf size=173 \[ \frac {F_1\left (1+m;\frac {1}{2},1;2+m;-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{2 d (1+m) \sqrt {a+b \tan (c+d x)}}+\frac {F_1\left (1+m;\frac {1}{2},1;2+m;-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{2 d (1+m) \sqrt {a+b \tan (c+d x)}} \]
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Rubi [A]
time = 0.13, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3656, 926, 140,
138} \begin {gather*} \frac {\tan ^{m+1}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac {1}{2},1;m+2;-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1) \sqrt {a+b \tan (c+d x)}}+\frac {\tan ^{m+1}(c+d x) \sqrt {\frac {b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac {1}{2},1;m+2;-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1) \sqrt {a+b \tan (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 138
Rule 140
Rule 926
Rule 3656
Rubi steps
\begin {align*} \int \frac {\tan ^m(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx &=\frac {\text {Subst}\left (\int \frac {x^m}{\sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (\frac {i x^m}{2 (i-x) \sqrt {a+b x}}+\frac {i x^m}{2 (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {i \text {Subst}\left (\int \frac {x^m}{(i-x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {i \text {Subst}\left (\int \frac {x^m}{(i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {\left (i \sqrt {1+\frac {b \tan (c+d x)}{a}}\right ) \text {Subst}\left (\int \frac {x^m}{(i-x) \sqrt {1+\frac {b x}{a}}} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt {a+b \tan (c+d x)}}+\frac {\left (i \sqrt {1+\frac {b \tan (c+d x)}{a}}\right ) \text {Subst}\left (\int \frac {x^m}{(i+x) \sqrt {1+\frac {b x}{a}}} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt {a+b \tan (c+d x)}}\\ &=\frac {F_1\left (1+m;\frac {1}{2},1;2+m;-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{2 d (1+m) \sqrt {a+b \tan (c+d x)}}+\frac {F_1\left (1+m;\frac {1}{2},1;2+m;-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{2 d (1+m) \sqrt {a+b \tan (c+d x)}}\\ \end {align*}
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Mathematica [F]
time = 8.26, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^m(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.40, size = 0, normalized size = 0.00 \[\int \frac {\tan ^{m}\left (d x +c \right )}{\sqrt {a +b \tan \left (d x +c \right )}}\, dx\]
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*} \int \frac {\tan ^{m}{\left (c + d x \right )}}{\sqrt {a + b \tan {\left (c + d x \right )}}}\, dx \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 = -0.01 \begin {gather*} \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^m}{\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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