Optimal. Leaf size=349 \[ \frac {(i a-b)^{5/2} (i A-B) \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {(i a+b)^{5/2} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{105 a d}+\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{105 d}-\frac {2 a (10 A b+7 a B) \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{35 d}-\frac {2 a A \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d} \]
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Rubi [A]
time = 1.09, antiderivative size = 349, normalized size of antiderivative = 1.00, number of steps
used = 12, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {4326, 3686,
3726, 3730, 3697, 3696, 95, 209, 212} \begin {gather*} \frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{105 d}+\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{105 a d}+\frac {(-b+i a)^{5/2} (-B+i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (7 a B+10 A b) \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{35 d}+\frac {(b+i a)^{5/2} (B+i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a A \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 95
Rule 209
Rule 212
Rule 3686
Rule 3696
Rule 3697
Rule 3726
Rule 3730
Rule 4326
Rubi steps
\begin {align*} \int \cot ^{\frac {9}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx\\ &=-\frac {2 a A \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}+\frac {1}{7} \left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+b \tan (c+d x)} \left (\frac {1}{2} a (10 A b+7 a B)-\frac {7}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-\frac {1}{2} b (4 a A-7 b B) \tan ^2(c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\\ &=-\frac {2 a (10 A b+7 a B) \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{35 d}-\frac {2 a A \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}+\frac {1}{35} \left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {-\frac {1}{4} a \left (35 a^2 A-45 A b^2-77 a b B\right )-\frac {35}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)-\frac {1}{4} b \left (60 a A b+28 a^2 B-35 b^2 B\right ) \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\\ &=\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{105 d}-\frac {2 a (10 A b+7 a B) \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{35 d}-\frac {2 a A \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}-\frac {\left (8 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {1}{8} a \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right )-\frac {105}{8} a \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)-\frac {1}{4} a b \left (35 a^2 A-45 A b^2-77 a b B\right ) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{105 a}\\ &=\frac {2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{105 a d}+\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{105 d}-\frac {2 a (10 A b+7 a B) \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{35 d}-\frac {2 a A \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}+\frac {\left (16 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {105}{16} a^2 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )+\frac {105}{16} a^2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{105 a^2}\\ &=\frac {2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{105 a d}+\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{105 d}-\frac {2 a (10 A b+7 a B) \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{35 d}-\frac {2 a A \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}+\frac {1}{2} \left ((a-i b)^3 (A-i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx+\frac {1}{2} \left ((a+i b)^3 (A+i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=\frac {2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{105 a d}+\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{105 d}-\frac {2 a (10 A b+7 a B) \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{35 d}-\frac {2 a A \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}+\frac {\left ((a-i b)^3 (A-i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left ((a+i b)^3 (A+i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{105 a d}+\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{105 d}-\frac {2 a (10 A b+7 a B) \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{35 d}-\frac {2 a A \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}+\frac {\left ((a-i b)^3 (A-i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-(i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left ((a+i b)^3 (A+i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-(-i a+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} (i A-B) \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {(i a+b)^{5/2} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{105 a d}+\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{105 d}-\frac {2 a (10 A b+7 a B) \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{35 d}-\frac {2 a A \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d}\\ \end {align*}
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Mathematica [A]
time = 3.41, size = 382, normalized size = 1.09 \begin {gather*} -\frac {\cot ^{\frac {7}{2}}(c+d x) \left (35 a b (4 A b+a B) \sqrt {a+b \tan (c+d x)}+5 a \left (24 a^2 A-28 A b^2-49 a b B\right ) \sqrt {a+b \tan (c+d x)}+6 a \left (60 a A b+28 a^2 B-35 b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}+210 a b B (a+b \tan (c+d x))^{3/2}-4 \tan ^2(c+d x) \left (105 \sqrt [4]{-1} a \left ((-a+i b)^{5/2} (i A+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 )+(a+i b)^{5/2} (-i A+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 )\right ) \tan ^{\frac {3}{2}}(c+d x)+2 a \left (35 a^2 A-45 A b^2-77 a b B\right ) \sqrt {a+b \tan (c+d x)}+2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}\right )\right )}{420 a d} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 31.67, size = 67595, normalized size = 193.68
method | result | size |
default | \(\text {Expression too large to display}\) | \(67595\) |
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(-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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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 {\mathrm {cot}\left (c+d\,x\right )}^{9/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\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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