Optimal. Leaf size=115 \[ \frac {a^2}{3 b^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b)}{b^2 f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f (a-b)^{5/2}} \]
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Rubi [A] time = 0.20, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3670, 446, 87, 63, 208} \[ \frac {a^2}{3 b^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b)}{b^2 f (a-b)^2 \sqrt {a+b \tan ^2(e+f x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f (a-b)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 63
Rule 87
Rule 208
Rule 446
Rule 3670
Rubi steps
\begin {align*} \int \frac {\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^5}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{(1+x) (a+b x)^{5/2}} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {\operatorname {Subst}\left (\int \left (-\frac {a^2}{(a-b) b (a+b x)^{5/2}}+\frac {a (a-2 b)}{(a-b)^2 b (a+b x)^{3/2}}+\frac {1}{(a-b)^2 (1+x) \sqrt {a+b x}}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {a^2}{3 (a-b) b^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b)}{(a-b)^2 b^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 (a-b)^2 f}\\ &=\frac {a^2}{3 (a-b) b^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b)}{(a-b)^2 b^2 f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \tan ^2(e+f x)}\right )}{(a-b)^2 b f}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{(a-b)^{5/2} f}+\frac {a^2}{3 (a-b) b^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {a (a-2 b)}{(a-b)^2 b^2 f \sqrt {a+b \tan ^2(e+f x)}}\\ \end {align*}
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Mathematica [C] time = 0.46, size = 91, normalized size = 0.79 \[ \frac {b^2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {b \tan ^2(e+f x)+a}{a-b}\right )-(a-b) \left (2 a+3 b \tan ^2(e+f x)-b\right )}{3 b^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.55, size = 608, normalized size = 5.29 \[ \left [\frac {3 \, {\left (b^{4} \tan \left (f x + e\right )^{4} + 2 \, a b^{3} \tan \left (f x + e\right )^{2} + a^{2} b^{2}\right )} \sqrt {a - b} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) - 4 \, {\left (2 \, a^{4} - 7 \, a^{3} b + 5 \, a^{2} b^{2} + 3 \, {\left (a^{3} b - 3 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{12 \, {\left ({\left (a^{3} b^{4} - 3 \, a^{2} b^{5} + 3 \, a b^{6} - b^{7}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{3} - 3 \, a^{3} b^{4} + 3 \, a^{2} b^{5} - a b^{6}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} f\right )}}, \frac {3 \, {\left (b^{4} \tan \left (f x + e\right )^{4} + 2 \, a b^{3} \tan \left (f x + e\right )^{2} + a^{2} b^{2}\right )} \sqrt {-a + b} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) - 2 \, {\left (2 \, a^{4} - 7 \, a^{3} b + 5 \, a^{2} b^{2} + 3 \, {\left (a^{3} b - 3 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{6 \, {\left ({\left (a^{3} b^{4} - 3 \, a^{2} b^{5} + 3 \, a b^{6} - b^{7}\right )} f \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{3} - 3 \, a^{3} b^{4} + 3 \, a^{2} b^{5} - a b^{6}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} f\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 4.22, size = 137, normalized size = 1.19 \[ \frac {\arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {-a + b}}\right )}{{\left (a^{2} f - 2 \, a b f + b^{2} f\right )} \sqrt {-a + b}} - \frac {3 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )} a^{2} - a^{3} - 6 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )} a b + a^{2} b}{3 \, {\left (a^{2} b^{2} f - 2 \, a b^{3} f + b^{4} f\right )} {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.34, size = 169, normalized size = 1.47 \[ -\frac {\tan ^{2}\left (f x +e \right )}{f b \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}-\frac {2 a}{3 f \,b^{2} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}+\frac {1}{3 f b \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}+\frac {1}{3 \left (a -b \right ) f \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}+\frac {1}{\left (a -b \right )^{2} f \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}+\frac {\arctan \left (\frac {\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{\sqrt {-a +b}}\right )}{f \left (a -b \right )^{2} \sqrt {-a +b}} \]
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 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 16.03, size = 148, normalized size = 1.29 \[ \frac {\frac {a^2}{3\,\left (a-b\right )}+\frac {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )\,\left (2\,a\,b-a^2\right )}{{\left (a-b\right )}^2}}{b^2\,f\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}}+\frac {\mathrm {atan}\left (\frac {a^2\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}+b^2\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,1{}\mathrm {i}-a\,b\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,2{}\mathrm {i}}{{\left (a-b\right )}^{5/2}}\right )\,1{}\mathrm {i}}{f\,{\left (a-b\right )}^{5/2}} \]
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 \[ \int \frac {\tan ^{5}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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