Optimal. Leaf size=121 \[ -\frac {\sqrt {2} a \tanh ^{-1}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{d^{7/2} f}-\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac {2 a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {2 a}{d^3 f \sqrt {d \tan (e+f x)}} \]
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Rubi [A]
time = 0.12, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3610, 3613,
214} \begin {gather*} -\frac {\sqrt {2} a \tanh ^{-1}\left (\frac {\sqrt {d} \tan (e+f x)+\sqrt {d}}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{d^{7/2} f}+\frac {2 a}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {2 a}{3 d^2 f (d \tan (e+f x))^{3/2}}-\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 214
Rule 3610
Rule 3613
Rubi steps
\begin {align*} \int \frac {a+a \tan (e+f x)}{(d \tan (e+f x))^{7/2}} \, dx &=-\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}+\frac {\int \frac {a d-a d \tan (e+f x)}{(d \tan (e+f x))^{5/2}} \, dx}{d^2}\\ &=-\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac {2 a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {\int \frac {-a d^2-a d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2}} \, dx}{d^4}\\ &=-\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac {2 a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {2 a}{d^3 f \sqrt {d \tan (e+f x)}}+\frac {\int \frac {-a d^3+a d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{d^6}\\ &=-\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac {2 a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {2 a}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{-2 a^2 d^6+d x^2} \, dx,x,\frac {-a d^3-a d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{f}\\ &=-\frac {\sqrt {2} a \tanh ^{-1}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{d^{7/2} f}-\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac {2 a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {2 a}{d^3 f \sqrt {d \tan (e+f x)}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in
optimal.
time = 0.20, size = 68, normalized size = 0.56 \begin {gather*} -\frac {\left (\frac {1}{5}+\frac {i}{5}\right ) a \left (\, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};-i \tan (e+f x)\right )-i \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};i \tan (e+f x)\right )\right )}{d f (d \tan (e+f x))^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(322\) vs.
\(2(100)=200\).
time = 0.20, size = 323, normalized size = 2.67
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {2}{3 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {2}{5 d \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {2}{d^{3} \sqrt {d \tan \left (f x +e \right )}}+\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{3}}\right )}{f}\) | \(323\) |
default | \(\frac {a \left (-\frac {2}{3 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {2}{5 d \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {2}{d^{3} \sqrt {d \tan \left (f x +e \right )}}+\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{3}}\right )}{f}\) | \(323\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.53, size = 143, normalized size = 1.18 \begin {gather*} -\frac {\frac {15 \, a {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{d^{2}} - \frac {4 \, {\left (15 \, a d^{2} \tan \left (f x + e\right )^{2} - 5 \, a d^{2} \tan \left (f x + e\right ) - 3 \, a d^{2}\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} d^{2}}}{30 \, d f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.88, size = 261, normalized size = 2.16 \begin {gather*} \left [\frac {15 \, \sqrt {2} a \sqrt {d} \log \left (\frac {\tan \left (f x + e\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) + 1\right )}}{\sqrt {d}} + 4 \, \tan \left (f x + e\right ) + 1}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{3} + 4 \, {\left (15 \, a \tan \left (f x + e\right )^{2} - 5 \, a \tan \left (f x + e\right ) - 3 \, a\right )} \sqrt {d \tan \left (f x + e\right )}}{30 \, d^{4} f \tan \left (f x + e\right )^{3}}, \frac {15 \, \sqrt {2} a d \sqrt {-\frac {1}{d}} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-\frac {1}{d}} {\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{3} + 2 \, {\left (15 \, a \tan \left (f x + e\right )^{2} - 5 \, a \tan \left (f x + e\right ) - 3 \, a\right )} \sqrt {d \tan \left (f x + e\right )}}{15 \, d^{4} f \tan \left (f x + e\right )^{3}}\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*} a \left (\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx + \int \frac {\tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 295 vs.
\(2 (105) = 210\).
time = 0.77, size = 295, normalized size = 2.44 \begin {gather*} -\frac {\sqrt {2} {\left (a d \sqrt {{\left | d \right |}} - a {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{2 \, d^{5} f} - \frac {\sqrt {2} {\left (a d \sqrt {{\left | d \right |}} - a {\left | d \right |}^{\frac {3}{2}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{2 \, d^{5} f} - \frac {\sqrt {2} {\left (a d \sqrt {{\left | d \right |}} + a {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{4 \, d^{5} f} + \frac {\sqrt {2} {\left (a d \sqrt {{\left | d \right |}} + a {\left | d \right |}^{\frac {3}{2}}\right )} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{4 \, d^{5} f} + \frac {2 \, {\left (15 \, a d^{2} \tan \left (f x + e\right )^{2} - 5 \, a d^{2} \tan \left (f x + e\right ) - 3 \, a d^{2}\right )}}{15 \, \sqrt {d \tan \left (f x + e\right )} d^{5} f \tan \left (f x + e\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.80, size = 120, normalized size = 0.99 \begin {gather*} -\frac {\frac {2\,a}{5\,d}-\frac {2\,a\,{\mathrm {tan}\left (e+f\,x\right )}^2}{d}}{f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}-\frac {2\,a}{3\,d^2\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,\left (1+1{}\mathrm {i}\right )}{d^{7/2}\,f}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,\left (-1+1{}\mathrm {i}\right )}{d^{7/2}\,f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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