Optimal. Leaf size=184 \[ \frac {i \sqrt {i a-b} \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b} d}+\frac {i \sqrt {i a+b} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d} \]
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Rubi [A]
time = 0.49, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3651, 3736,
6857, 65, 223, 212, 924, 95, 211, 214} \begin {gather*} \frac {i \sqrt {-b+i a} \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b} d}+\frac {i \sqrt {b+i a} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 95
Rule 211
Rule 212
Rule 214
Rule 223
Rule 924
Rule 3651
Rule 3736
Rule 6857
Rubi steps
\begin {align*} \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} \, dx &=\frac {\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\int \frac {-\frac {a}{2}-b \tan (c+d x)+\frac {1}{2} a \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=\frac {\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {\text {Subst}\left (\int \frac {-\frac {a}{2}-b x+\frac {a x^2}{2}}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {\text {Subst}\left (\int \left (\frac {a}{2 \sqrt {x} \sqrt {a+b x}}-\frac {\sqrt {a+b x}}{\sqrt {x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\text {Subst}\left (\int \frac {\sqrt {a+b x}}{\sqrt {x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}+\frac {a \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\text {Subst}\left (\int \left (\frac {i a-b}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {i a+b}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac {a \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {a \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a-b) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {(i a+b) \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b} d}+\frac {\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {(i a-b) \text {Subst}\left (\int \frac {1}{i-(a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b) \text {Subst}\left (\int \frac {1}{i-(-a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\\ &=\frac {i \sqrt {i a-b} \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b} d}+\frac {i \sqrt {i a+b} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\\ \end {align*}
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Mathematica [A]
time = 1.56, size = 217, normalized size = 1.18 \begin {gather*} \frac {-(-1)^{3/4} \sqrt {-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 )+(-1)^{3/4} \sqrt {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 {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+\frac {a^{3/2} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}}{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 = 0.56, size = 1091177, normalized size = 5930.31 \[\text {output too large to display}\]
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 \sqrt {a + b \tan {\left (c + d x \right )}} \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]
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 )}^{3/2}\,\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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