Optimal. Leaf size=224 \[ \frac {\left (a^2-10 a b+8 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{16 b f}-\frac {\left (a^3+6 a^2 b-24 a b^2+16 b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{16 b^{3/2} f}+\frac {(a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}+\frac {b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}+\frac {(7 a-6 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{24 f} \]
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Rubi [A] time = 0.35, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3670, 477, 582, 523, 217, 206, 377, 203} \[ \frac {\left (a^2-10 a b+8 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{16 b f}-\frac {\left (6 a^2 b+a^3-24 a b^2+16 b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{16 b^{3/2} f}+\frac {b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}+\frac {(7 a-6 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{24 f}+\frac {(a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f} \]
Antiderivative was successfully verified.
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Rule 203
Rule 206
Rule 217
Rule 377
Rule 477
Rule 523
Rule 582
Rule 3670
Rubi steps
\begin {align*} \int \tan ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4 \left (a+b x^2\right )^{3/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}+\frac {\operatorname {Subst}\left (\int \frac {x^4 \left (a (6 a-5 b)+(7 a-6 b) b x^2\right )}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{6 f}\\ &=\frac {(7 a-6 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{24 f}+\frac {b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}-\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (3 a (7 a-6 b) b-3 b \left (a^2-10 a b+8 b^2\right ) x^2\right )}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{24 b f}\\ &=\frac {\left (a^2-10 a b+8 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{16 b f}+\frac {(7 a-6 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{24 f}+\frac {b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}+\frac {\operatorname {Subst}\left (\int \frac {-3 a b \left (a^2-10 a b+8 b^2\right )-3 b \left (a^3+6 a^2 b-24 a b^2+16 b^3\right ) x^2}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{48 b^2 f}\\ &=\frac {\left (a^2-10 a b+8 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{16 b f}+\frac {(7 a-6 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{24 f}+\frac {b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}+\frac {(a-b)^2 \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}-\frac {\left (a^3+6 a^2 b-24 a b^2+16 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{16 b f}\\ &=\frac {\left (a^2-10 a b+8 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{16 b f}+\frac {(7 a-6 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{24 f}+\frac {b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}+\frac {(a-b)^2 \operatorname {Subst}\left (\int \frac {1}{1-(-a+b) x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\left (a^3+6 a^2 b-24 a b^2+16 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{16 b f}\\ &=\frac {(a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}-\frac {\left (a^3+6 a^2 b-24 a b^2+16 b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{16 b^{3/2} f}+\frac {\left (a^2-10 a b+8 b^2\right ) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{16 b f}+\frac {(7 a-6 b) \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{24 f}+\frac {b \tan ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{6 f}\\ \end {align*}
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Mathematica [C] time = 6.44, size = 833, normalized size = 3.72 \[ \frac {\sqrt {\frac {\cos (2 (e+f x)) a+a+b-b \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \left (\frac {1}{6} b \tan (e+f x) \sec ^4(e+f x)+\frac {7}{24} (a \sin (e+f x)-2 b \sin (e+f x)) \sec ^3(e+f x)+\frac {\left (3 \sin (e+f x) a^2-44 b \sin (e+f x) a+44 b^2 \sin (e+f x)\right ) \sec (e+f x)}{48 b}\right )}{f}-\frac {-\frac {b \left (a^3-2 b a^2-8 b^2 a+8 b^3\right ) \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right ) \sin ^4(e+f x)}{a (a+b+(a-b) \cos (2 (e+f x)))}-\frac {4 b \left (-8 b^3+16 a b^2-8 a^2 b\right ) \sqrt {\cos (2 (e+f x))+1} \sqrt {\frac {a+b+(a-b) \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \left (\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right ) \sin ^4(e+f x)}{4 a \sqrt {\cos (2 (e+f x))+1} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}-\frac {\sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \Pi \left (-\frac {b}{a-b};\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right ) \sin ^4(e+f x)}{2 (a-b) \sqrt {\cos (2 (e+f x))+1} \sqrt {a+b+(a-b) \cos (2 (e+f x))}}\right )}{\sqrt {a+b+(a-b) \cos (2 (e+f x))}}}{8 b f} \]
Antiderivative was successfully verified.
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fricas [A] time = 2.99, size = 861, normalized size = 3.84 \[ \left [\frac {3 \, {\left (a^{3} + 6 \, a^{2} b - 24 \, a b^{2} + 16 \, b^{3}\right )} \sqrt {b} \log \left (2 \, b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {b} \tan \left (f x + e\right ) + a\right ) - 48 \, {\left (a b^{2} - b^{3}\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left (8 \, b^{3} \tan \left (f x + e\right )^{5} + 2 \, {\left (7 \, a b^{2} - 6 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (a^{2} b - 10 \, a b^{2} + 8 \, b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{96 \, b^{2} f}, \frac {3 \, {\left (a^{3} + 6 \, a^{2} b - 24 \, a b^{2} + 16 \, b^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-b}}{b \tan \left (f x + e\right )}\right ) - 24 \, {\left (a b^{2} - b^{3}\right )} \sqrt {-a + b} \log \left (-\frac {{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) + {\left (8 \, b^{3} \tan \left (f x + e\right )^{5} + 2 \, {\left (7 \, a b^{2} - 6 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (a^{2} b - 10 \, a b^{2} + 8 \, b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{48 \, b^{2} f}, \frac {96 \, {\left (a b^{2} - b^{3}\right )} \sqrt {a - b} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right ) + 3 \, {\left (a^{3} + 6 \, a^{2} b - 24 \, a b^{2} + 16 \, b^{3}\right )} \sqrt {b} \log \left (2 \, b \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {b} \tan \left (f x + e\right ) + a\right ) + 2 \, {\left (8 \, b^{3} \tan \left (f x + e\right )^{5} + 2 \, {\left (7 \, a b^{2} - 6 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (a^{2} b - 10 \, a b^{2} + 8 \, b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{96 \, b^{2} f}, \frac {48 \, {\left (a b^{2} - b^{3}\right )} \sqrt {a - b} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right ) + 3 \, {\left (a^{3} + 6 \, a^{2} b - 24 \, a b^{2} + 16 \, b^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-b}}{b \tan \left (f x + e\right )}\right ) + {\left (8 \, b^{3} \tan \left (f x + e\right )^{5} + 2 \, {\left (7 \, a b^{2} - 6 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (a^{2} b - 10 \, a b^{2} + 8 \, b^{3}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{48 \, b^{2} f}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.38, size = 510, normalized size = 2.28 \[ \frac {\tan \left (f x +e \right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {5}{2}}}{6 f b}-\frac {a \tan \left (f x +e \right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}{24 f b}-\frac {a^{2} \tan \left (f x +e \right ) \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{16 f b}-\frac {a^{3} \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{16 f \,b^{\frac {3}{2}}}-\frac {\tan \left (f x +e \right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}{4 f}-\frac {3 a \tan \left (f x +e \right ) \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{8 f}-\frac {3 a^{2} \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{8 f \sqrt {b}}+\frac {b \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )}{2 f}+\frac {3 \sqrt {b}\, a \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{2 f}-\frac {b^{\frac {3}{2}} \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{f}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f \left (a -b \right )}-\frac {2 a \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f b \left (a -b \right )}+\frac {a^{2} \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f \,b^{2} \left (a -b \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^4\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]
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 \[ \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{4}{\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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