Optimal. Leaf size=290 \[ \frac {2 \left (15 a^2 A-5 a b B+2 A b^2\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a^2 d}-\frac {2 (5 a B+A b) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}+\frac {\sqrt {-b+i a} (A+i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\sqrt {b+i a} (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d} \]
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Rubi [A] time = 1.19, antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4241, 3608, 3649, 3616, 3615, 93, 203, 206} \[ \frac {2 \left (15 a^2 A-5 a b B+2 A b^2\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a^2 d}-\frac {2 (5 a B+A b) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}+\frac {\sqrt {-b+i a} (A+i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\sqrt {b+i a} (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d} \]
Antiderivative was successfully verified.
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Rule 93
Rule 203
Rule 206
Rule 3608
Rule 3615
Rule 3616
Rule 3649
Rule 4241
Rubi steps
\begin {align*} \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\\ &=-\frac {2 A \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 {1}{2} (-A b-5 a B)+\frac {5}{2} (a A-b B) \tan (c+d x)+2 A b \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 (A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 A \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} \left (-15 a^2 A-2 A b^2+5 a b B\right )-\frac {15}{4} a (A b+a B) \tan (c+d x)-\frac {1}{2} b (A b+5 a B) \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 A+2 A b^2-5 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a^2 d}-\frac {2 (A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 A \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 (A b+a B)-\frac {15}{8} a^2 (a A-b B) \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 A+2 A b^2-5 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a^2 d}-\frac {2 (A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}-\frac {\left (4 \left (\frac {15}{8} a^2 (A b+a B)-\frac {15}{8} i a^2 (a A-b B)\right ) \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}{15 a^2}-\frac {\left (4 \left (\frac {15}{8} a^2 (A b+a B)+\frac {15}{8} i a^2 (a A-b B)\right ) \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}{15 a^2}\\ &=\frac {2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a^2 d}-\frac {2 (A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}-\frac {\left (4 \left (\frac {15}{8} a^2 (A b+a B)-\frac {15}{8} i a^2 (a A-b B)\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{15 a^2 d}-\frac {\left (4 \left (\frac {15}{8} a^2 (A b+a B)+\frac {15}{8} i a^2 (a A-b B)\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{15 a^2 d}\\ &=\frac {2 \left (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a^2 d}-\frac {2 (A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}-\frac {\left (8 \left (\frac {15}{8} a^2 (A b+a B)-\frac {15}{8} i a^2 (a A-b B)\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {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 )}{15 a^2 d}-\frac {\left (8 \left (\frac {15}{8} a^2 (A b+a B)+\frac {15}{8} i a^2 (a A-b B)\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {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 )}{15 a^2 d}\\ &=\frac {\sqrt {i a-b} (A+i 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 {\sqrt {i a+b} (A-i 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 (15 a^2 A+2 A b^2-5 a b B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{15 a^2 d}-\frac {2 (A b+5 a B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{15 a d}-\frac {2 A \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 = 2.20, size = 252, normalized size = 0.87 \[ \frac {\cot ^{\frac {5}{2}}(c+d x) \left (2 \sqrt {a+b \tan (c+d x)} \left (\left (15 a^2 A-5 a b B+2 A b^2\right ) \tan ^2(c+d x)-3 a^2 A-a (5 a B+A b) \tan (c+d x)\right )+15 \sqrt [4]{-1} a^2 \sqrt {-a+i b} (A-i B) \tan ^{\frac {5}{2}}(c+d x) \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+15 \sqrt [4]{-1} a^2 \sqrt {a+i b} (A+i B) \tan ^{\frac {5}{2}}(c+d x) \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )}{15 a^2 d} \]
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 2.75, size = 43089, normalized size = 148.58 \[ \text {output too large to display} \]
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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {b \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac {7}{2}}\,{d x} \]
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 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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