Optimal. Leaf size=193 \[ \frac {\text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}-\frac {2}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 b \left (a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \]
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Rubi [A]
time = 0.45, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3650, 3730,
3697, 3696, 95, 209, 212} \begin {gather*} -\frac {2 b \left (a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{a^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}-\frac {2}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 95
Rule 209
Rule 212
Rule 3650
Rule 3696
Rule 3697
Rule 3730
Rubi steps
\begin {align*} \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx &=-\frac {2}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 \int \frac {b+\frac {1}{2} a \tan (c+d x)+b \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{a}\\ &=-\frac {2}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 b \left (a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}-\frac {4 \int \frac {\frac {a^2 b}{4}+\frac {1}{4} a^3 \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{a^2 \left (a^2+b^2\right )}\\ &=-\frac {2}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 b \left (a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {\int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (i a-b)}-\frac {\int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (i a+b)}\\ &=-\frac {2}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 b \left (a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {\text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (i a-b) d}-\frac {\text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (i a+b) d}\\ &=-\frac {2}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 b \left (a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {\text {Subst}\left (\int \frac {1}{1-(-i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b) d}-\frac {\text {Subst}\left (\int \frac {1}{1-(i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b) d}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}-\frac {2}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 b \left (a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 4.44, size = 202, normalized size = 1.05 \begin {gather*} -\frac {\frac {\sqrt [4]{-1} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{3/2}}+\frac {\frac {\sqrt [4]{-1} (a-i b) \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+\frac {2 \left (a \left (a^2+b^2\right )+b \left (a^2+2 b^2\right ) \tan (c+d x)\right )}{a^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}}{a^2+b^2}}{d} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 0.98, size = 799513, normalized size = 4142.55 \[\text {output too large to display}\]
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(-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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{\frac {3}{2}}{\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 {1}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,{\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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