Optimal. Leaf size=175 \[ \frac {a F_1\left (1+m;-\frac {3}{2},1;2+m;-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}}+\frac {a F_1\left (1+m;-\frac {3}{2},1;2+m;-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}} \]
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Rubi [A]
time = 0.15, antiderivative size = 175, 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 {a \tan ^{m+1}(c+d x) \sqrt {a+b \tan (c+d x)} F_1\left (m+1;-\frac {3}{2},1;m+2;-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1) \sqrt {\frac {b \tan (c+d x)}{a}+1}}+\frac {a \tan ^{m+1}(c+d x) \sqrt {a+b \tan (c+d x)} F_1\left (m+1;-\frac {3}{2},1;m+2;-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1) \sqrt {\frac {b \tan (c+d x)}{a}+1}} \end {gather*}
Antiderivative was successfully verified.
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Rule 138
Rule 140
Rule 926
Rule 3656
Rubi steps
\begin {align*} \int \tan ^m(c+d x) (a+b \tan (c+d x))^{3/2} \, dx &=\frac {\text {Subst}\left (\int \frac {x^m (a+b x)^{3/2}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\text {Subst}\left (\int \left (\frac {i x^m (a+b x)^{3/2}}{2 (i-x)}+\frac {i x^m (a+b x)^{3/2}}{2 (i+x)}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {i \text {Subst}\left (\int \frac {x^m (a+b x)^{3/2}}{i-x} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {i \text {Subst}\left (\int \frac {x^m (a+b x)^{3/2}}{i+x} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {\left (i a \sqrt {a+b \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {x^m \left (1+\frac {b x}{a}\right )^{3/2}}{i-x} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt {1+\frac {b \tan (c+d x)}{a}}}+\frac {\left (i a \sqrt {a+b \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {x^m \left (1+\frac {b x}{a}\right )^{3/2}}{i+x} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt {1+\frac {b \tan (c+d x)}{a}}}\\ &=\frac {a F_1\left (1+m;-\frac {3}{2},1;2+m;-\frac {b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}}+\frac {a F_1\left (1+m;-\frac {3}{2},1;2+m;-\frac {b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d (1+m) \sqrt {1+\frac {b \tan (c+d x)}{a}}}\\ \end {align*}
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Mathematica [F]
time = 18.93, size = 0, normalized size = 0.00 \begin {gather*} \int \tan ^m(c+d x) (a+b \tan (c+d x))^{3/2} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.39, size = 0, normalized size = 0.00 \[\int \left (\tan ^{m}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}\, dx\]
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 \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{m}{\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 {\mathrm {tan}\left (c+d\,x\right )}^m\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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