Optimal. Leaf size=170 \[ \frac {2 C \tan ^{1+m}(c+d x)}{d (1+2 m) \sqrt {b \tan (c+d x)}}+\frac {2 (A-C) \, _2F_1\left (1,\frac {1}{4} (1+2 m);\frac {1}{4} (5+2 m);-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{d (1+2 m) \sqrt {b \tan (c+d x)}}+\frac {2 B \, _2F_1\left (1,\frac {1}{4} (3+2 m);\frac {1}{4} (7+2 m);-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{d (3+2 m) \sqrt {b \tan (c+d x)}} \]
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Rubi [A]
time = 0.10, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {20, 3711, 3619,
3557, 371} \begin {gather*} \frac {2 (A-C) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac {1}{4} (2 m+1);\frac {1}{4} (2 m+5);-\tan ^2(c+d x)\right )}{d (2 m+1) \sqrt {b \tan (c+d x)}}+\frac {2 B \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac {1}{4} (2 m+3);\frac {1}{4} (2 m+7);-\tan ^2(c+d x)\right )}{d (2 m+3) \sqrt {b \tan (c+d x)}}+\frac {2 C \tan ^{m+1}(c+d x)}{d (2 m+1) \sqrt {b \tan (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 20
Rule 371
Rule 3557
Rule 3619
Rule 3711
Rubi steps
\begin {align*} \int \frac {\tan ^m(c+d x) \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right )}{\sqrt {b \tan (c+d x)}} \, dx &=\frac {\sqrt {\tan (c+d x)} \int \tan ^{-\frac {1}{2}+m}(c+d x) \left (A+B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx}{\sqrt {b \tan (c+d x)}}\\ &=\frac {2 C \tan ^{1+m}(c+d x)}{d (1+2 m) \sqrt {b \tan (c+d x)}}+\frac {\sqrt {\tan (c+d x)} \int \tan ^{-\frac {1}{2}+m}(c+d x) (A-C+B \tan (c+d x)) \, dx}{\sqrt {b \tan (c+d x)}}\\ &=\frac {2 C \tan ^{1+m}(c+d x)}{d (1+2 m) \sqrt {b \tan (c+d x)}}+\frac {\left (B \sqrt {\tan (c+d x)}\right ) \int \tan ^{\frac {1}{2}+m}(c+d x) \, dx}{\sqrt {b \tan (c+d x)}}+\frac {\left ((A-C) \sqrt {\tan (c+d x)}\right ) \int \tan ^{-\frac {1}{2}+m}(c+d x) \, dx}{\sqrt {b \tan (c+d x)}}\\ &=\frac {2 C \tan ^{1+m}(c+d x)}{d (1+2 m) \sqrt {b \tan (c+d x)}}+\frac {\left (B \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {x^{\frac {1}{2}+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d \sqrt {b \tan (c+d x)}}+\frac {\left ((A-C) \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {x^{-\frac {1}{2}+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d \sqrt {b \tan (c+d x)}}\\ &=\frac {2 C \tan ^{1+m}(c+d x)}{d (1+2 m) \sqrt {b \tan (c+d x)}}+\frac {2 (A-C) \, _2F_1\left (1,\frac {1}{4} (1+2 m);\frac {1}{4} (5+2 m);-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{d (1+2 m) \sqrt {b \tan (c+d x)}}+\frac {2 B \, _2F_1\left (1,\frac {1}{4} (3+2 m);\frac {1}{4} (7+2 m);-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{d (3+2 m) \sqrt {b \tan (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 0.38, size = 133, normalized size = 0.78 \begin {gather*} \frac {2 \tan ^{1+m}(c+d x) \left (C (3+2 m)+(A-C) (3+2 m) \, _2F_1\left (1,\frac {1}{4} (1+2 m);\frac {1}{4} (5+2 m);-\tan ^2(c+d x)\right )+B (1+2 m) \, _2F_1\left (1,\frac {1}{4} (3+2 m);\frac {1}{4} (7+2 m);-\tan ^2(c+d x)\right ) \tan (c+d x)\right )}{d (1+2 m) (3+2 m) \sqrt {b \tan (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.62, size = 0, normalized size = 0.00 \[\int \frac {\left (\tan ^{m}\left (d x +c \right )\right ) \left (A +B \tan \left (d x +c \right )+C \left (\tan ^{2}\left (d x +c \right )\right )\right )}{\sqrt {b \tan \left (d x +c \right )}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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 {\left (A + B \tan {\left (c + d x \right )} + C \tan ^{2}{\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}}{\sqrt {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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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\,\left (C\,{\mathrm {tan}\left (c+d\,x\right )}^2+B\,\mathrm {tan}\left (c+d\,x\right )+A\right )}{\sqrt {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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