Optimal. Leaf size=232 \[ \frac {\text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}+\frac {\left (3 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{5/2} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d}-\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b^2 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d} \]
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Rubi [A]
time = 0.72, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps
used = 14, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3647, 3728,
3737, 6857, 65, 223, 212, 926, 95, 211, 214} \begin {gather*} \frac {\left (3 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{5/2} d}+\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 {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b^2 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}+\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 65
Rule 95
Rule 211
Rule 212
Rule 214
Rule 223
Rule 926
Rule 3647
Rule 3728
Rule 3737
Rule 6857
Rubi steps
\begin {align*} \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx &=\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}+\frac {\int \frac {\sqrt {\tan (c+d x)} \left (-\frac {3 a}{2}-2 b \tan (c+d x)-\frac {3}{2} a \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}} \, dx}{2 b}\\ &=-\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b^2 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}+\frac {\int \frac {\frac {3 a^2}{4}+\frac {1}{4} \left (3 a^2-8 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 b^2}\\ &=-\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b^2 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}+\frac {\text {Subst}\left (\int \frac {\frac {3 a^2}{4}+\frac {1}{4} \left (3 a^2-8 b^2\right ) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{2 b^2 d}\\ &=-\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b^2 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}+\frac {\text {Subst}\left (\int \left (\frac {3 a^2-8 b^2}{4 \sqrt {x} \sqrt {a+b x}}+\frac {\frac {3 a^2}{4}+\frac {1}{4} \left (-3 a^2+8 b^2\right )}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{2 b^2 d}\\ &=-\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b^2 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}+\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 {\left (3 a^2-8 b^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{8 b^2 d}\\ &=-\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b^2 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}+\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 {\left (3 a^2-8 b^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 b^2 d}\\ &=-\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b^2 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}+\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 {\left (3 a^2-8 b^2\right ) \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 )}{4 b^2 d}\\ &=\frac {\left (3 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{5/2} d}-\frac {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b^2 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b 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 {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 {\left (3 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{5/2} 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 {3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b^2 d}+\frac {\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 b d}\\ \end {align*}
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Mathematica [A]
time = 4.18, size = 270, normalized size = 1.16 \begin {gather*} \frac {-\frac {4 (-1)^{3/4} b^2 \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 {4 (-1)^{3/4} b^2 \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}}-3 a \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+2 b \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}+\frac {\sqrt {a} \left (3 a^2-8 b^2\right ) \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)}}}{4 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 = 1.03, size = 945945, normalized size = 4077.35 \[\text {output too large to display}\]
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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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 = -0.00 \begin {gather*} \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}{\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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