Optimal. Leaf size=180 \[ -\frac {\text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 b \sqrt {a+b \tan (c+d x)}}{3 a^2 d \sqrt {\tan (c+d x)}} \]
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Rubi [A]
time = 0.26, antiderivative size = 180, 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 = {3650, 3730,
12, 3656, 926, 95, 211, 214} \begin {gather*} \frac {4 b \sqrt {a+b \tan (c+d x)}}{3 a^2 d \sqrt {\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 \sqrt {-b+i a}}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(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 \sqrt {b+i a}} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 95
Rule 211
Rule 214
Rule 926
Rule 3650
Rule 3656
Rule 3730
Rubi steps
\begin {align*} \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \int \frac {b+\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}{3 a}\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 b \sqrt {a+b \tan (c+d x)}}{3 a^2 d \sqrt {\tan (c+d x)}}+\frac {4 \int -\frac {3 a^2}{4 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a^2}\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 b \sqrt {a+b \tan (c+d x)}}{3 a^2 d \sqrt {\tan (c+d x)}}-\int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 b \sqrt {a+b \tan (c+d x)}}{3 a^2 d \sqrt {\tan (c+d x)}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 b \sqrt {a+b \tan (c+d x)}}{3 a^2 d \sqrt {\tan (c+d x)}}-\frac {\text {Subst}\left (\int \left (\frac {i}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {i}{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 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 b \sqrt {a+b \tan (c+d x)}}{3 a^2 d \sqrt {\tan (c+d x)}}-\frac {i \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 \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 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 b \sqrt {a+b \tan (c+d x)}}{3 a^2 d \sqrt {\tan (c+d x)}}-\frac {i \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 \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 {\tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 b \sqrt {a+b \tan (c+d x)}}{3 a^2 d \sqrt {\tan (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 1.62, size = 172, normalized size = 0.96 \begin {gather*} \frac {\frac {3 (-1)^{3/4} \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 {3 (-1)^{3/4} \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 (a-2 b \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}{a^2 \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 = 0.57, size = 945613, normalized size = 5253.41 \[\text {output too large to display}\]
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}{\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 {1}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\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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