Optimal. Leaf size=222 \[ -\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {a^2 \sec (c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {a^2 \left (2 a^2-7 b^2\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \]
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Rubi [A]
time = 0.30, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3930, 4165,
4088, 12, 3916, 2738, 214} \begin {gather*} -\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}-\frac {a^2 \left (2 a^2-7 b^2\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}-\frac {a^2 \tan (c+d x) \sec (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3}+\frac {a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 214
Rule 2738
Rule 3916
Rule 3930
Rule 4088
Rule 4165
Rubi steps
\begin {align*} \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^4} \, dx &=-\frac {a^2 \sec (c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {\int \frac {\sec (c+d x) \left (a^2-3 a b \sec (c+d x)-\left (2 a^2-3 b^2\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac {a^2 \sec (c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {a^2 \left (2 a^2-7 b^2\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {\int \frac {\sec (c+d x) \left (2 a b \left (a^2-6 b^2\right )+\left (2 a^4-3 a^2 b^2+6 b^4\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac {a^2 \sec (c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {a^2 \left (2 a^2-7 b^2\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\int \frac {3 b^3 \left (3 a^2+2 b^2\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 b^2 \left (a^2-b^2\right )^3}\\ &=-\frac {a^2 \sec (c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {a^2 \left (2 a^2-7 b^2\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\left (b \left (3 a^2+2 b^2\right )\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=-\frac {a^2 \sec (c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {a^2 \left (2 a^2-7 b^2\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\left (3 a^2+2 b^2\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=-\frac {a^2 \sec (c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {a^2 \left (2 a^2-7 b^2\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac {\left (3 a^2+2 b^2\right ) \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {a^2 \sec (c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}-\frac {a^2 \left (2 a^2-7 b^2\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac {a \left (2 a^4-5 a^2 b^2+18 b^4\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}\\ \end {align*}
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Mathematica [A]
time = 1.06, size = 158, normalized size = 0.71 \begin {gather*} \frac {\frac {6 b \left (3 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac {a \left (2 a^4-5 a^2 b^2+18 b^4+3 a b \left (a^2+9 b^2\right ) \cos (c+d x)+a^2 \left (4 a^2+11 b^2\right ) \cos ^2(c+d x)\right ) \sin (c+d x)}{(a-b)^3 (a+b)^3 (b+a \cos (c+d x))^3}}{6 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.40, size = 285, normalized size = 1.28
method | result | size |
derivativedivides | \(\frac {\frac {-\frac {\left (2 a^{2}+3 b a +6 b^{2}\right ) a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{3}+3 b \,a^{2}+3 b^{2} a +b^{3}\right )}+\frac {4 \left (a^{2}+9 b^{2}\right ) a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}+2 b a +b^{2}\right ) \left (a^{2}-2 b a +b^{2}\right )}-\frac {\left (2 a^{2}-3 b a +6 b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 b \,a^{2}+3 b^{2} a -b^{3}\right )}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )^{3}}-\frac {b \left (3 a^{2}+2 b^{2}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(285\) |
default | \(\frac {\frac {-\frac {\left (2 a^{2}+3 b a +6 b^{2}\right ) a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right ) \left (a^{3}+3 b \,a^{2}+3 b^{2} a +b^{3}\right )}+\frac {4 \left (a^{2}+9 b^{2}\right ) a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (a^{2}+2 b a +b^{2}\right ) \left (a^{2}-2 b a +b^{2}\right )}-\frac {\left (2 a^{2}-3 b a +6 b^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 b \,a^{2}+3 b^{2} a -b^{3}\right )}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )^{3}}-\frac {b \left (3 a^{2}+2 b^{2}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(285\) |
risch | \(-\frac {i \left (9 a^{4} b \,{\mathrm e}^{5 i \left (d x +c \right )}+6 a^{2} b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+45 a^{3} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+30 a \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}+24 a^{4} b \,{\mathrm e}^{3 i \left (d x +c \right )}+82 a^{2} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+44 b^{5} {\mathrm e}^{3 i \left (d x +c \right )}+12 a^{5} {\mathrm e}^{2 i \left (d x +c \right )}+36 a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+102 a \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+15 a^{4} b \,{\mathrm e}^{i \left (d x +c \right )}+60 a^{2} b^{3} {\mathrm e}^{i \left (d x +c \right )}+4 a^{5}+11 a^{3} b^{2}\right )}{3 \left (-a^{2}+b^{2}\right )^{3} d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )^{3}}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\) | \(585\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \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. 423 vs.
\(2 (207) = 414\).
time = 3.46, size = 903, normalized size = 4.07 \begin {gather*} \left [-\frac {3 \, {\left (3 \, a^{2} b^{4} + 2 \, b^{6} + {\left (3 \, a^{5} b + 2 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, a^{4} b^{2} + 2 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 2 \, {\left (2 \, a^{7} - 7 \, a^{5} b^{2} + 23 \, a^{3} b^{4} - 18 \, a b^{6} + {\left (4 \, a^{7} + 7 \, a^{5} b^{2} - 11 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{6} b + 8 \, a^{4} b^{3} - 9 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left ({\left (a^{11} - 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} - 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{10} b - 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} - 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} d \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d\right )}}, -\frac {3 \, {\left (3 \, a^{2} b^{4} + 2 \, b^{6} + {\left (3 \, a^{5} b + 2 \, a^{3} b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, a^{4} b^{2} + 2 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, a^{3} b^{3} + 2 \, a b^{5}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (2 \, a^{7} - 7 \, a^{5} b^{2} + 23 \, a^{3} b^{4} - 18 \, a b^{6} + {\left (4 \, a^{7} + 7 \, a^{5} b^{2} - 11 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{6} b + 8 \, a^{4} b^{3} - 9 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left ({\left (a^{11} - 4 \, a^{9} b^{2} + 6 \, a^{7} b^{4} - 4 \, a^{5} b^{6} + a^{3} b^{8}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{10} b - 4 \, a^{8} b^{3} + 6 \, a^{6} b^{5} - 4 \, a^{4} b^{7} + a^{2} b^{9}\right )} d \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{9} b^{2} - 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} - 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b^{3} - 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - 4 \, a^{2} b^{9} + b^{11}\right )} d\right )}}\right ] \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 \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{4}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.52, size = 403, normalized size = 1.82 \begin {gather*} -\frac {\frac {3 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {6 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 32 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}^{3}}}{3 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.41, size = 378, normalized size = 1.70 \begin {gather*} \frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (2\,a^3+3\,a^2\,b+6\,a\,b^2\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^3+9\,a\,b^2\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^3-3\,a^2\,b+6\,a\,b^2\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )+3\,a\,b^2+3\,a^2\,b+a^3+b^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}-\frac {b\,\mathrm {atanh}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2+2\,b^2\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\left (3\,a^2\,b+2\,b^3\right )\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (3\,a^2+2\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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