Optimal. Leaf size=181 \[ \frac {i \sqrt {i a-b} \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {i \sqrt {i a+b} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}} \]
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Rubi [A]
time = 0.24, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3649, 3730,
21, 3656, 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 {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}+\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 21
Rule 95
Rule 211
Rule 214
Rule 924
Rule 3649
Rule 3656
Rule 3730
Rubi steps
\begin {align*} \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2}{3} \int \frac {-\frac {b}{2}+\frac {3}{2} a \tan (c+d x)+b \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}+\frac {4 \int \frac {-\frac {3 a^2}{4}-\frac {3}{4} a b \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a}\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\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 {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\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 {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\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 {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\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 {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 {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 0.60, size = 161, normalized size = 0.89 \begin {gather*} \frac {-3 (-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 )+3 (-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 )-\frac {2 (a+b \tan (c+d x))^{3/2}}{a \tan ^{\frac {3}{2}}(c+d x)}}{3 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 = 1.08, size = 1090997, normalized size = 6027.61 \[\text {output too large to display}\]
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 {\sqrt {a + b \tan {\left (c + d x \right )}}}{\tan ^{\frac {5}{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 {\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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