Optimal. Leaf size=531 \[ -\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {b} \left (a^2+b^2\right )^3 d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \]
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Rubi [A]
time = 0.84, antiderivative size = 531, 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 = {3689, 3730,
3734, 3615, 1182, 1176, 631, 210, 1179, 642, 3715, 65, 211} \begin {gather*} -\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\left (a^3 (A-B)+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 (a^3 (A-B)+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 {\left (-3 a^3 B+7 a^2 A b+5 a b^2 B-A b^3\right ) \sqrt {\tan (c+d x)}}{4 a d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\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 ) \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 (-\left (a^3 (A+B)\right )+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 (-3 a^5 B+15 a^4 A b+26 a^3 b^2 B-18 a^2 A b^3-3 a b^4 B-A b^5\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {b} 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 3689
Rule 3715
Rule 3730
Rule 3734
Rubi steps
\begin {align*} \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx &=-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\int \frac {-\frac {1}{2} b (A b-a B)-2 b (a A+b 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 b \left (a^2+b^2\right )}\\ &=-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\int \frac {-\frac {1}{4} b \left (9 a^2 A b+A b^3-5 a^3 B+3 a b^2 B\right )-2 a b \left (a^2 A-A b^2+2 a b B\right ) \tan (c+d x)+\frac {1}{4} b \left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{2 a b \left (a^2+b^2\right )^2}\\ &=-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\int \frac {-2 a b \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right )-2 a b \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{2 a b \left (a^2+b^2\right )^3}-\frac {\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \int \frac {1+\tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx}{8 a \left (a^2+b^2\right )^3}\\ &=-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\text {Subst}\left (\int \frac {-2 a b \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right )-2 a b \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a b \left (a^2+b^2\right )^3 d}-\frac {\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} (a+b x)} \, dx,x,\tan (c+d x)\right )}{8 a \left (a^2+b^2\right )^3 d}\\ &=-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a \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 {\left (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {b} \left (a^2+b^2\right )^3 d}-\frac {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \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 {\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 (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 {\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 {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 {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 (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {b} \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 ) \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 (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (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 {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\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}{-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 (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}{-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 (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (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 (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (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 (15 a^4 A b-18 a^2 A b^3-A b^5-3 a^5 B+26 a^3 b^2 B-3 a b^4 B\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {b} \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 ) \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 (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (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 {(A b-a B) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B+5 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 a \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 = 4.01, size = 344, normalized size = 0.65 \begin {gather*} \frac {b (A b-a B) \tan ^{\frac {3}{2}}(c+d x)-(A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))+\frac {2 (a+b \tan (c+d x)) \left (\frac {1}{4} a^{3/2} b^{3/2} \left (a^2+b^2\right ) \left (-7 a^2 A b+A b^3+3 a^3 B-5 a b^2 B\right ) \sqrt {\tan (c+d x)}-\left (-\frac {1}{4} a b \left (-15 a^4 A b+18 a^2 A b^3+A b^5+3 a^5 B-26 a^3 b^2 B+3 a b^4 B\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )+\sqrt [4]{-1} a^{5/2} b^{3/2} \left ((i a-b)^3 (A-i B) \text {ArcTan}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-(i a+b)^3 (A+i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )\right )\right ) (a+b \tan (c+d x))\right )}{a^{3/2} b^{3/2} \left (a^2+b^2\right )^2}}{2 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.12, size = 451, normalized size = 0.85
method | result | size |
derivativedivides | \(\frac {\frac {\frac {\left (3 A \,a^{2} b -A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A \,a^{3}-3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}-\frac {2 \left (\frac {\frac {b \left (7 A \,a^{4} b +6 A \,a^{2} b^{3}-A \,b^{5}-3 B \,a^{5}+2 B \,a^{3} b^{2}+5 B a \,b^{4}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{8 a}+\left (\frac {9}{8} A \,a^{4} b +\frac {5}{4} A \,a^{2} b^{3}-\frac {5}{8} B \,a^{5}+\frac {3}{8} B a \,b^{4}+\frac {1}{8} A \,b^{5}-\frac {1}{4} B \,a^{3} b^{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (15 A \,a^{4} b -18 A \,a^{2} b^{3}-A \,b^{5}-3 B \,a^{5}+26 B \,a^{3} b^{2}-3 B a \,b^{4}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{8 a \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\) | \(451\) |
default | \(\frac {\frac {\frac {\left (3 A \,a^{2} b -A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A \,a^{3}-3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}-\frac {2 \left (\frac {\frac {b \left (7 A \,a^{4} b +6 A \,a^{2} b^{3}-A \,b^{5}-3 B \,a^{5}+2 B \,a^{3} b^{2}+5 B a \,b^{4}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{8 a}+\left (\frac {9}{8} A \,a^{4} b +\frac {5}{4} A \,a^{2} b^{3}-\frac {5}{8} B \,a^{5}+\frac {3}{8} B a \,b^{4}+\frac {1}{8} A \,b^{5}-\frac {1}{4} B \,a^{3} b^{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (15 A \,a^{4} b -18 A \,a^{2} b^{3}-A \,b^{5}-3 B \,a^{5}+26 B \,a^{3} b^{2}-3 B a \,b^{4}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{8 a \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{3}}}{d}\) | \(451\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.53, size = 537, normalized size = 1.01 \begin {gather*} \frac {\frac {{\left (3 \, B a^{5} - 15 \, A a^{4} b - 26 \, B a^{3} b^{2} + 18 \, A a^{2} b^{3} + 3 \, B a b^{4} + A b^{5}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a 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 (3 \, B a^{3} b - 7 \, A a^{2} b^{2} - 5 \, B a b^{3} + A b^{4}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + {\left (5 \, B a^{4} - 9 \, A a^{3} b - 3 \, B a^{2} b^{2} - A a b^{3}\right )} \sqrt {\tan \left (d x + c\right )}}{a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4} + {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{6} b + 2 \, a^{4} b^{3} + a^{2} 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 = 52.40, size = 2500, normalized size = 4.71 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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