Optimal. Leaf size=246 \[ \frac {\sqrt {2} a^2 d^{3/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {\sqrt {2} a^2 d^{3/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {a^2 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} f}+\frac {a^2 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} f}+\frac {4 a^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f} \]
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Rubi [A]
time = 0.16, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps
used = 15, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3624, 12, 16,
3554, 3557, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {\sqrt {2} a^2 d^{3/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {\sqrt {2} a^2 d^{3/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{f}-\frac {a^2 d^{3/2} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} f}+\frac {a^2 d^{3/2} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} f}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\frac {4 a^2 (d \tan (e+f x))^{3/2}}{3 f} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 16
Rule 210
Rule 303
Rule 335
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 3554
Rule 3557
Rule 3624
Rubi steps
\begin {align*} \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2 \, dx &=\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\int 2 a^2 \tan (e+f x) (d \tan (e+f x))^{3/2} \, dx\\ &=\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\left (2 a^2\right ) \int \tan (e+f x) (d \tan (e+f x))^{3/2} \, dx\\ &=\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\frac {\left (2 a^2\right ) \int (d \tan (e+f x))^{5/2} \, dx}{d}\\ &=\frac {4 a^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}-\left (2 a^2 d\right ) \int \sqrt {d \tan (e+f x)} \, dx\\ &=\frac {4 a^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}-\frac {\left (2 a^2 d^2\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}\\ &=\frac {4 a^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}-\frac {\left (4 a^2 d^2\right ) \text {Subst}\left (\int \frac {x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=\frac {4 a^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}+\frac {\left (2 a^2 d^2\right ) \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}-\frac {\left (2 a^2 d^2\right ) \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=\frac {4 a^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}-\frac {\left (a^2 d^{3/2}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} f}-\frac {\left (a^2 d^{3/2}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} f}-\frac {\left (a^2 d^2\right ) \text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}-\frac {\left (a^2 d^2\right ) \text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=-\frac {a^2 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} f}+\frac {a^2 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} f}+\frac {4 a^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}-\frac {\left (\sqrt {2} a^2 d^{3/2}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {\left (\sqrt {2} a^2 d^{3/2}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}\\ &=\frac {\sqrt {2} a^2 d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {\sqrt {2} a^2 d^{3/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {a^2 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} f}+\frac {a^2 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} f}+\frac {4 a^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in
optimal.
time = 0.40, size = 52, normalized size = 0.21 \begin {gather*} \frac {2 a^2 (d \tan (e+f x))^{3/2} \left (10-10 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(e+f x)\right )+3 \tan (e+f x)\right )}{15 f} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.26, size = 172, normalized size = 0.70
method | result | size |
derivativedivides | \(\frac {2 a^{2} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{3} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f d}\) | \(172\) |
default | \(\frac {2 a^{2} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{3} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f d}\) | \(172\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.52, size = 203, normalized size = 0.83 \begin {gather*} -\frac {15 \, a^{2} d^{3} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {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}} + \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 )} - 12 \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{2} - 40 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{2} d}{30 \, d f} \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. 774 vs.
\(2 (203) = 406\).
time = 1.18, size = 774, normalized size = 3.15 \begin {gather*} \frac {60 \, \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \arctan \left (-\frac {a^{8} d^{6} + \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} a^{6} d^{4} f \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} - \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \sqrt {\frac {a^{12} d^{9} \sin \left (f x + e\right ) + \sqrt {\frac {a^{8} d^{6}}{f^{4}}} a^{8} d^{6} f^{2} \cos \left (f x + e\right ) + \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {3}{4}} a^{6} d^{4} f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\cos \left (f x + e\right )}}}{a^{8} d^{6}}\right ) \cos \left (f x + e\right )^{2} + 60 \, \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \arctan \left (\frac {a^{8} d^{6} - \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} a^{6} d^{4} f \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} + \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \sqrt {\frac {a^{12} d^{9} \sin \left (f x + e\right ) + \sqrt {\frac {a^{8} d^{6}}{f^{4}}} a^{8} d^{6} f^{2} \cos \left (f x + e\right ) - \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {3}{4}} a^{6} d^{4} f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\cos \left (f x + e\right )}}}{a^{8} d^{6}}\right ) \cos \left (f x + e\right )^{2} + 15 \, \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \cos \left (f x + e\right )^{2} \log \left (\frac {a^{12} d^{9} \sin \left (f x + e\right ) + \sqrt {\frac {a^{8} d^{6}}{f^{4}}} a^{8} d^{6} f^{2} \cos \left (f x + e\right ) + \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {3}{4}} a^{6} d^{4} f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) - 15 \, \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \cos \left (f x + e\right )^{2} \log \left (\frac {a^{12} d^{9} \sin \left (f x + e\right ) + \sqrt {\frac {a^{8} d^{6}}{f^{4}}} a^{8} d^{6} f^{2} \cos \left (f x + e\right ) - \sqrt {2} \left (\frac {a^{8} d^{6}}{f^{4}}\right )^{\frac {3}{4}} a^{6} d^{4} f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) - 4 \, {\left (3 \, a^{2} d \cos \left (f x + e\right )^{2} - 10 \, a^{2} d \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \, a^{2} d\right )} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{30 \, f \cos \left (f x + e\right )^{2}} \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^{2} \left (\int \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx + \int 2 \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )}\, dx + \int \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (e + f x \right )}\, dx\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.68, size = 274, normalized size = 1.11 \begin {gather*} -\frac {1}{30} \, {\left (\frac {30 \, \sqrt {2} a^{2} {\left | d \right |}^{\frac {3}{2}} \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 )}{d f} + \frac {30 \, \sqrt {2} a^{2} {\left | d \right |}^{\frac {3}{2}} \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 )}{d f} - \frac {15 \, \sqrt {2} a^{2} {\left | d \right |}^{\frac {3}{2}} \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 )}{d f} + \frac {15 \, \sqrt {2} a^{2} {\left | d \right |}^{\frac {3}{2}} \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 )}{d f} - \frac {4 \, {\left (3 \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{10} f^{4} \tan \left (f x + e\right )^{2} + 10 \, \sqrt {d \tan \left (f x + e\right )} a^{2} d^{10} f^{4} \tan \left (f x + e\right )\right )}}{d^{10} f^{5}}\right )} d \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.64, size = 104, normalized size = 0.42 \begin {gather*} \frac {4\,a^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,f}+\frac {2\,a^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{5\,d\,f}-\frac {2\,{\left (-1\right )}^{1/4}\,a^2\,d^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{f}-\frac {{\left (-1\right )}^{1/4}\,a^2\,d^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{\sqrt {d}}\right )\,2{}\mathrm {i}}{f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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