Optimal. Leaf size=125 \[ \frac {3 (a+b)^2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{8 \sqrt {a} f}+\frac {\sin (e+f x) \cos ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 f}+\frac {3 (a+b) \sin (e+f x) \cos (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{8 f} \]
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Rubi [A] time = 0.12, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4146, 378, 377, 203} \[ \frac {3 (a+b)^2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{8 \sqrt {a} f}+\frac {\sin (e+f x) \cos ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 f}+\frac {3 (a+b) \sin (e+f x) \cos (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{8 f} \]
Antiderivative was successfully verified.
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Rule 203
Rule 377
Rule 378
Rule 4146
Rubi steps
\begin {align*} \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b+b x^2\right )^{3/2}}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\cos ^3(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 f}+\frac {(3 (a+b)) \operatorname {Subst}\left (\int \frac {\sqrt {a+b+b x^2}}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac {3 (a+b) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}+\frac {\cos ^3(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 f}+\frac {\left (3 (a+b)^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac {3 (a+b) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}+\frac {\cos ^3(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 f}+\frac {\left (3 (a+b)^2\right ) \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 )}{8 f}\\ &=\frac {3 (a+b)^2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 \sqrt {a} f}+\frac {3 (a+b) \cos (e+f x) \sin (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 f}+\frac {\cos ^3(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 f}\\ \end {align*}
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Mathematica [A] time = 1.02, size = 191, normalized size = 1.53 \[ \frac {\cos (e+f x) \sqrt {-a \sin ^2(e+f x)+a+b} \left (a \cos ^2(e+f x)+b\right ) \sqrt {a+b \sec ^2(e+f x)} \left (3 (a+b)^{3/2} \sin ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )+\sqrt {a} \sin (e+f x) \sqrt {\frac {-a \sin ^2(e+f x)+a+b}{a+b}} (a \cos (2 (e+f x))+4 a+5 b)\right )}{2 \sqrt {a} f (a \cos (2 (e+f x))+a+2 b)^{3/2} \sqrt {\frac {a \cos (2 (e+f x))+a+2 b}{a+b}}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.89, size = 563, normalized size = 4.50 \[ \left [-\frac {3 \, {\left (a^{2} + 2 \, a b + 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 (2 \, a^{2} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{2} + 5 \, 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 )}{64 \, a f}, -\frac {3 \, {\left (a^{2} + 2 \, a b + 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 (2 \, a^{2} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{2} + 5 \, 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 )}{32 \, a 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 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 1.72, size = 1713, normalized size = 13.70 \[ \text {result too large to display} \]
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 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (e+f\,x\right )}^4\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [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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