Optimal. Leaf size=120 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{5/2} f}+\frac {(a-3 b) \tan (e+f x)}{3 a^2 b f \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {(a+b) \tan (e+f x)}{3 a b f \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]
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Rubi [A] time = 0.27, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4141, 1975, 470, 527, 12, 377, 203} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{5/2} f}+\frac {(a-3 b) \tan (e+f x)}{3 a^2 b f \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {(a+b) \tan (e+f x)}{3 a b f \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]
Antiderivative was successfully verified.
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Rule 12
Rule 203
Rule 377
Rule 470
Rule 527
Rule 1975
Rule 4141
Rubi steps
\begin {align*} \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(a+b) \tan (e+f x)}{3 a b f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {a+b+(a-2 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 a b f}\\ &=-\frac {(a+b) \tan (e+f x)}{3 a b f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(a-3 b) \tan (e+f x)}{3 a^2 b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {3 b (a+b)}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a^2 b (a+b) f}\\ &=-\frac {(a+b) \tan (e+f x)}{3 a b f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(a-3 b) \tan (e+f x)}{3 a^2 b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{a^2 f}\\ &=-\frac {(a+b) \tan (e+f x)}{3 a b f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(a-3 b) \tan (e+f x)}{3 a^2 b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^2 f}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{5/2} f}-\frac {(a+b) \tan (e+f x)}{3 a b f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(a-3 b) \tan (e+f x)}{3 a^2 b f \sqrt {a+b+b \tan ^2(e+f x)}}\\ \end {align*}
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Mathematica [B] time = 6.02, size = 409, normalized size = 3.41 \[ \frac {\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b)^{5/2} \left (\frac {\sqrt {2} \csc (e+f x) \sec (e+f x) \left (\frac {16 \left (-a \sin ^2(e+f x)+a+b\right ) \left (1-\frac {a \sin ^2(e+f x)}{a+b}\right ) \left (\frac {a^2 (a+b) \sin ^4(e+f x)}{\left (-a \sin ^2(e+f x)+a+b\right )^2}+\frac {3 \sqrt {a} \sqrt {a+b} \sin (e+f x) \sin ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {\frac {-a \sin ^2(e+f x)+a+b}{a+b}}}-\frac {6 a (a+b) \sin ^2(e+f x)}{a \cos (2 (e+f x))+a+2 b}\right )}{a^3}-\frac {12 \sin ^4(e+f x)}{a+b}+\frac {\sin ^2(e+f x)}{a+b}+\frac {\sin ^2(e+f x) (a \cos (2 (e+f x))+a+2 b)}{(a+b)^2}\right )}{\left (-a \sin ^2(e+f x)+a+b\right )^{3/2}}+\frac {8 \tan (e+f x) (a \cos (2 (e+f x))+2 a+3 b)}{(a+b)^2 (a \cos (2 (e+f x))+a+2 b)^{3/2}}-\frac {12 \tan (e+f x) ((3 a+2 b) \cos (2 (e+f x))+b)}{(a+b)^2 (a \cos (2 (e+f x))+a+2 b)^{3/2}}\right )}{384 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.61, size = 661, normalized size = 5.51 \[ \left [-\frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a^{4} - a^{3} b\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 14 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 28 \, a^{3} b + 70 \, a^{2} b^{2} - 28 \, a b^{3} + b^{4} - 32 \, {\left (a^{4} - 7 \, a^{3} b + 7 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 14 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right ) + 8 \, {\left (4 \, a^{2} \cos \left (f x + e\right )^{3} - {\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{24 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}}, -\frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} - a^{2} b + a b^{2} - {\left (a^{3} - 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, {\left (4 \, a^{2} \cos \left (f x + e\right )^{3} - {\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{12 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}}\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 \[ \int \frac {\tan \left (f x + e\right )^{4}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 1.65, size = 1142, normalized size = 9.52 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^4}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \]
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 \[ \int \frac {\tan ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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