Optimal. Leaf size=259 \[ -\frac {(i a-b)^{5/2} \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {(i a+b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {2 \left (15 a^2-23 b^2\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {22 a b \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d} \]
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Rubi [A]
time = 0.73, antiderivative size = 259, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4326, 3646,
3730, 3697, 3696, 95, 209, 212} \begin {gather*} \frac {2 \left (15 a^2-23 b^2\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}-\frac {(-b+i a)^{5/2} \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 {22 a b \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}+\frac {(b+i a)^{5/2} \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} \end {gather*}
Antiderivative was successfully verified.
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Rule 95
Rule 209
Rule 212
Rule 3646
Rule 3696
Rule 3697
Rule 3730
Rule 4326
Rubi steps
\begin {align*} \int \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}+\frac {1}{5} \left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {11 a^2 b}{2}-\frac {5}{2} a \left (a^2-3 b^2\right ) \tan (c+d x)-\frac {1}{2} b \left (4 a^2-5 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {22 a b \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}-\frac {\left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {1}{4} a^2 \left (15 a^2-23 b^2\right )+\frac {15}{4} a b \left (3 a^2-b^2\right ) \tan (c+d x)+\frac {11}{2} a^2 b^2 \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{15 a}\\ &=\frac {2 \left (15 a^2-23 b^2\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {22 a b \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}+\frac {\left (8 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {-\frac {15}{8} a^2 b \left (3 a^2-b^2\right )+\frac {15}{8} a^3 \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^2}\\ &=\frac {2 \left (15 a^2-23 b^2\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {22 a b \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}-\frac {1}{2} \left ((i a-b)^3 \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 ((i a+b)^3 \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 (15 a^2-23 b^2\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {22 a b \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}-\frac {\left ((i a-b)^3 \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 ((i a+b)^3 \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 (15 a^2-23 b^2\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {22 a b \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}-\frac {\left ((i a-b)^3 \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 ((i a+b)^3 \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} \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} \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 (15 a^2-23 b^2\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {22 a b \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 d}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}\\ \end {align*}
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Mathematica [A]
time = 1.80, size = 214, normalized size = 0.83 \begin {gather*} \frac {\cot ^{\frac {5}{2}}(c+d x) \left (15 \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 ) \tan ^{\frac {5}{2}}(c+d x)+15 \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 ) \tan ^{\frac {5}{2}}(c+d x)+2 \sqrt {a+b \tan (c+d x)} \left (-3 a^2-11 a b \tan (c+d x)+\left (15 a^2-23 b^2\right ) \tan ^2(c+d x)\right )\right )}{15 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 = 35.14, size = 33357, normalized size = 128.79
method | result | size |
default | \(\text {Expression too large to display}\) | \(33357\) |
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 )}^{7/2}\,{\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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