Optimal. Leaf size=534 \[ \frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\sqrt {b} \left (35 a^4 A b+6 a^2 A b^3+3 A b^5-15 a^5 B+18 a^3 b^2 B+a b^4 B\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{5/2} \left (a^2+b^2\right )^3 d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {b (A b-a B) \sqrt {\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \]
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Rubi [A]
time = 0.83, antiderivative size = 534, normalized size of antiderivative = 1.00, number of steps
used = 16, number of rules used = 13, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3690, 3730,
3734, 3615, 1182, 1176, 631, 210, 1179, 642, 3715, 65, 211} \begin {gather*} \frac {b (A b-a B) \sqrt {\tan (c+d x)}}{2 a d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {\left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}-\frac {\left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}+\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B+3 A b^3\right ) \sqrt {\tan (c+d x)}}{4 a^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {\left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}+\frac {\left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}+\frac {\sqrt {b} \left (-15 a^5 B+35 a^4 A b+18 a^3 b^2 B+6 a^2 A b^3+a b^4 B+3 A b^5\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{5/2} d \left (a^2+b^2\right )^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 210
Rule 211
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 1182
Rule 3615
Rule 3690
Rule 3715
Rule 3730
Rule 3734
Rubi steps
\begin {align*} \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3} \, dx &=\frac {b (A b-a B) \sqrt {\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\int \frac {\frac {1}{2} \left (4 a^2 A+3 A b^2+a b B\right )-2 a (A b-a B) \tan (c+d x)+\frac {3}{2} b (A b-a B) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx}{2 a \left (a^2+b^2\right )}\\ &=\frac {b (A b-a B) \sqrt {\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {\frac {1}{4} \left (8 a^4 A+3 a^2 A b^2+3 A b^4+9 a^3 b B+a b^3 B\right )-2 a^2 \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)+\frac {1}{4} b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{2 a^2 \left (a^2+b^2\right )^2}\\ &=\frac {b (A b-a B) \sqrt {\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {2 a^2 \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right )-2 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)}} \, dx}{2 a^2 \left (a^2+b^2\right )^3}+\frac {\left (b \left (35 a^4 A b+6 a^2 A b^3+3 A b^5-15 a^5 B+18 a^3 b^2 B+a b^4 B\right )\right ) \int \frac {1+\tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{8 a^2 \left (a^2+b^2\right )^3}\\ &=\frac {b (A b-a B) \sqrt {\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\text {Subst}\left (\int \frac {2 a^2 \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right )-2 a^2 \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 \left (a^2+b^2\right )^3 d}+\frac {\left (b \left (35 a^4 A b+6 a^2 A b^3+3 A b^5-15 a^5 B+18 a^3 b^2 B+a b^4 B\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{8 a^2 \left (a^2+b^2\right )^3 d}\\ &=\frac {b (A b-a B) \sqrt {\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\left (b \left (35 a^4 A b+6 a^2 A b^3+3 A b^5-15 a^5 B+18 a^3 b^2 B+a b^4 B\right )\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a^2 \left (a^2+b^2\right )^3 d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\left (a^2+b^2\right )^3 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\left (a^2+b^2\right )^3 d}\\ &=\frac {\sqrt {b} \left (35 a^4 A b+6 a^2 A b^3+3 A b^5-15 a^5 B+18 a^3 b^2 B+a b^4 B\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{5/2} \left (a^2+b^2\right )^3 d}+\frac {b (A b-a B) \sqrt {\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}\\ &=\frac {\sqrt {b} \left (35 a^4 A b+6 a^2 A b^3+3 A b^5-15 a^5 B+18 a^3 b^2 B+a b^4 B\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{5/2} \left (a^2+b^2\right )^3 d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {b (A b-a B) \sqrt {\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}\\ &=\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\sqrt {b} \left (35 a^4 A b+6 a^2 A b^3+3 A b^5-15 a^5 B+18 a^3 b^2 B+a b^4 B\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{5/2} \left (a^2+b^2\right )^3 d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {b (A b-a B) \sqrt {\tan (c+d x)}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 2.94, size = 288, normalized size = 0.54 \begin {gather*} \frac {\frac {2 \left (\frac {1}{2} \sqrt {b} \left (35 a^4 A b+6 a^2 A b^3+3 A b^5-15 a^5 B+18 a^3 b^2 B+a b^4 B\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )-2 \sqrt [4]{-1} a^{5/2} \left ((a+i b)^3 (A-i B) \text {ArcTan}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+(a-i b)^3 (A+i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )\right )\right )}{a^{3/2} \left (a^2+b^2\right )^2}+\frac {2 b (A b-a B) \sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\tan (c+d x)}}{a \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 a \left (a^2+b^2\right ) d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.12, size = 452, normalized size = 0.85
method | result | size |
derivativedivides | \(\frac {\frac {\frac {\left (A \,a^{3}-3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-3 A \,a^{2} b +A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}+\frac {2 b \left (\frac {\frac {b \left (11 A \,a^{4} b +14 A \,a^{2} b^{3}+3 A \,b^{5}-7 B \,a^{5}-6 B \,a^{3} b^{2}+B a \,b^{4}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{8 a^{2}}+\frac {\left (13 A \,a^{4} b +18 A \,a^{2} b^{3}+5 A \,b^{5}-9 B \,a^{5}-10 B \,a^{3} b^{2}-B a \,b^{4}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{8 a}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (35 A \,a^{4} b +6 A \,a^{2} b^{3}+3 A \,b^{5}-15 B \,a^{5}+18 B \,a^{3} b^{2}+B a \,b^{4}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{8 a^{2} \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\) | \(452\) |
default | \(\frac {\frac {\frac {\left (A \,a^{3}-3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-3 A \,a^{2} b +A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}+\frac {2 b \left (\frac {\frac {b \left (11 A \,a^{4} b +14 A \,a^{2} b^{3}+3 A \,b^{5}-7 B \,a^{5}-6 B \,a^{3} b^{2}+B a \,b^{4}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{8 a^{2}}+\frac {\left (13 A \,a^{4} b +18 A \,a^{2} b^{3}+5 A \,b^{5}-9 B \,a^{5}-10 B \,a^{3} b^{2}-B a \,b^{4}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{8 a}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (35 A \,a^{4} b +6 A \,a^{2} b^{3}+3 A \,b^{5}-15 B \,a^{5}+18 B \,a^{3} b^{2}+B a \,b^{4}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{8 a^{2} \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\) | \(452\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.52, size = 551, normalized size = 1.03 \begin {gather*} -\frac {\frac {{\left (15 \, B a^{5} b - 35 \, A a^{4} b^{2} - 18 \, B a^{3} b^{3} - 6 \, A a^{2} b^{4} - B a b^{5} - 3 \, A b^{6}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} \sqrt {a b}} - \frac {2 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} - 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} + {\left (A - B\right )} b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} - 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} + {\left (A - B\right )} b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left ({\left (A - B\right )} a^{3} + 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} - {\left (A + B\right )} b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left ({\left (A - B\right )} a^{3} + 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} - {\left (A + B\right )} b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (7 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3} - B a b^{4} - 3 \, A b^{5}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + {\left (9 \, B a^{4} b - 13 \, A a^{3} b^{2} + B a^{2} b^{3} - 5 \, A a b^{4}\right )} \sqrt {\tan \left (d x + c\right )}}{a^{8} + 2 \, a^{6} b^{2} + a^{4} b^{4} + {\left (a^{6} b^{2} + 2 \, a^{4} b^{4} + a^{2} b^{6}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \tan \left (d x + c\right )}}{4 \, d} \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 [B]
time = 53.70, size = 2500, normalized size = 4.68 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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