Optimal. Leaf size=231 \[ -\frac {(i a-b)^{5/2} \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {b} \left (15 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 d}+\frac {(i a+b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d} \]
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Rubi [A]
time = 1.51, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps
used = 14, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3647, 3728,
3736, 6857, 65, 223, 212, 95, 211, 214} \begin {gather*} \frac {\sqrt {b} \left (15 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 d}-\frac {(-b+i a)^{5/2} \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {(b+i a)^{5/2} \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 3647
Rule 3728
Rule 3736
Rule 6857
Rubi steps
\begin {align*} \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx &=\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{2} \int \frac {\sqrt {\tan (c+d x)} \left (\frac {1}{2} a \left (4 a^2-3 b^2\right )+2 b \left (3 a^2-b^2\right ) \tan (c+d x)+\frac {9}{2} a b^2 \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}} \, dx\\ &=\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {\int \frac {-\frac {9}{4} a^2 b^2+2 a b \left (a^2-3 b^2\right ) \tan (c+d x)+\frac {1}{4} b^2 \left (15 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}\\ &=\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {\text {Subst}\left (\int \frac {-\frac {9}{4} a^2 b^2+2 a b \left (a^2-3 b^2\right ) x+\frac {1}{4} b^2 \left (15 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 d}\\ &=\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {\text {Subst}\left (\int \left (\frac {b^2 \left (15 a^2-8 b^2\right )}{4 \sqrt {x} \sqrt {a+b x}}-\frac {2 \left (3 a^2 b^2-b^4-a b \left (a^2-3 b^2\right ) x\right )}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{2 b d}\\ &=\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {\text {Subst}\left (\int \frac {3 a^2 b^2-b^4-a b \left (a^2-3 b^2\right ) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (b \left (15 a^2-8 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}-\frac {\text {Subst}\left (\int \left (\frac {a b \left (a^2-3 b^2\right )+i \left (3 a^2 b^2-b^4\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {-a b \left (a^2-3 b^2\right )+i \left (3 a^2 b^2-b^4\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b d}+\frac {\left (b \left (15 a^2-8 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 d}\\ &=\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {(a-i b)^3 \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+i b)^3 \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (b \left (15 a^2-8 b^2\right )\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 d}\\ &=\frac {\sqrt {b} \left (15 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 d}+\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {(a-i b)^3 \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 {(a+i b)^3 \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)^{5/2} \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {b} \left (15 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 d}+\frac {(i a+b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\\ \end {align*}
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Mathematica [A]
time = 2.02, size = 264, normalized size = 1.14 \begin {gather*} \frac {4 \sqrt [4]{-1} (-a+i b)^{5/2} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+4 \sqrt [4]{-1} (a+i b)^{5/2} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+2 b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}+\frac {\sqrt {a} \sqrt {b} \left (15 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 {a+b \tan (c+d x)}}}{4 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 = 0.31, size = 1345303, normalized size = 5823.82 \[\text {output too large to display}\]
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 \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}} \sqrt {\tan {\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 = -0.00 \begin {gather*} \int \sqrt {\mathrm {tan}\left (c+d\,x\right )}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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