Optimal. Leaf size=272 \[ -\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d} \]
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Rubi [A]
time = 0.25, antiderivative size = 272, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3647, 3711,
3609, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 210
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 1182
Rule 3609
Rule 3615
Rule 3647
Rule 3711
Rubi steps
\begin {align*} \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^3 \, dx &=\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}+\frac {2}{5} \int \sqrt {\tan (c+d x)} \left (\frac {1}{2} a \left (5 a^2-3 b^2\right )+\frac {5}{2} b \left (3 a^2-b^2\right ) \tan (c+d x)+6 a b^2 \tan ^2(c+d x)\right ) \, dx\\ &=\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}+\frac {2}{5} \int \sqrt {\tan (c+d x)} \left (\frac {5}{2} a \left (a^2-3 b^2\right )+\frac {5}{2} b \left (3 a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=\frac {2 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}+\frac {2}{5} \int \frac {-\frac {5}{2} b \left (3 a^2-b^2\right )+\frac {5}{2} a \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {2 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}+\frac {4 \text {Subst}\left (\int \frac {-\frac {5}{2} b \left (3 a^2-b^2\right )+\frac {5}{2} a \left (a^2-3 b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{5 d}\\ &=\frac {2 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}+\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}-\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {2 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}+\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}+\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}+\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\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} d}+\frac {\left ((a-b) \left (a^2+4 a b+b^2\right )\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} d}\\ &=\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}+\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left ((a+b) \left (a^2-4 a b+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}\\ &=-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {\tan (c+d x)}}{d}+\frac {8 a b^2 \tan ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 0.83, size = 120, normalized size = 0.44 \begin {gather*} \frac {-5 \sqrt [4]{-1} (i a+b)^3 \text {ArcTan}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-5 (-1)^{3/4} (a+i b)^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 b \sqrt {\tan (c+d x)} \left (15 a^2-5 b^2+5 a b \tan (c+d x)+b^2 \tan ^2(c+d x)\right )}{5 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.06, size = 250, normalized size = 0.92
method | result | size |
derivativedivides | \(\frac {\frac {2 b^{3} \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+2 a \,b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+6 a^{2} b \left (\sqrt {\tan }\left (d x +c \right )\right )-2 b^{3} \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (a^{3}-3 b^{2} a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(250\) |
default | \(\frac {\frac {2 b^{3} \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+2 a \,b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+6 a^{2} b \left (\sqrt {\tan }\left (d x +c \right )\right )-2 b^{3} \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (a^{3}-3 b^{2} a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(250\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.51, size = 238, normalized size = 0.88 \begin {gather*} \frac {8 \, b^{3} \tan \left (d x + c\right )^{\frac {5}{2}} + 40 \, a b^{2} \tan \left (d x + c\right )^{\frac {3}{2}} + 10 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 10 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 5 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 5 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 40 \, {\left (3 \, a^{2} b - b^{3}\right )} \sqrt {\tan \left (d x + c\right )}}{20 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 7179 vs.
\(2 (234) = 468\).
time = 3.74, size = 7179, normalized size = 26.39 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \sqrt {\tan {\left (c + d x \right )}}\, dx \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 = 6.09, size = 1742, normalized size = 6.40 \begin {gather*} \frac {2\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{5\,d}-\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {2\,b^3}{d}-\frac {6\,a^2\,b}{d}\right )+\frac {2\,a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{d}-\mathrm {atan}\left (\frac {\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\right )\,\sqrt {\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}-\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}\,1{}\mathrm {i}-\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\right )\,\sqrt {\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}+\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}\,1{}\mathrm {i}}{-\frac {16\,\left (-a^9+6\,a^5\,b^4+8\,a^3\,b^6+3\,a\,b^8\right )}{d^3}+\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\right )\,\sqrt {\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}-\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}+\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\right )\,\sqrt {\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}+\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}}\right )\,\sqrt {\frac {-a^6\,1{}\mathrm {i}+6\,a^5\,b+a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3-a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5+b^6\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\right )\,\sqrt {\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}-\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}\,1{}\mathrm {i}-\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\right )\,\sqrt {\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}+\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}\,1{}\mathrm {i}}{-\frac {16\,\left (-a^9+6\,a^5\,b^4+8\,a^3\,b^6+3\,a\,b^8\right )}{d^3}+\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\right )\,\sqrt {\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}-\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}+\left (\frac {8\,\left (4\,b^3\,d^2-12\,a^2\,b\,d^2\right )\,\sqrt {\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}}{d^3}+\frac {16\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (a^6-15\,a^4\,b^2+15\,a^2\,b^4-b^6\right )}{d^2}\right )\,\sqrt {\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}}\right )\,\sqrt {\frac {a^6\,1{}\mathrm {i}+6\,a^5\,b-a^4\,b^2\,15{}\mathrm {i}-20\,a^3\,b^3+a^2\,b^4\,15{}\mathrm {i}+6\,a\,b^5-b^6\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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