Optimal. Leaf size=121 \[ -\frac {\sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 f}+\frac {(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{f}-\frac {b \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 f} \]
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Rubi [A]
time = 0.09, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3269, 427, 537,
223, 212, 385} \begin {gather*} -\frac {b \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 f}+\frac {(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{f}-\frac {\sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 f} \end {gather*}
Antiderivative was successfully verified.
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Rule 212
Rule 223
Rule 385
Rule 427
Rule 537
Rule 3269
Rubi steps
\begin {align*} \int \sec (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^{3/2}}{1-x^2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac {b \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 f}-\frac {\text {Subst}\left (\int \frac {-a (2 a+b)-b (3 a+2 b) x^2}{\left (1-x^2\right ) \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{2 f}\\ &=-\frac {b \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 f}+\frac {(a+b)^2 \text {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{f}-\frac {(b (3 a+2 b)) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{2 f}\\ &=-\frac {b \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 f}+\frac {(a+b)^2 \text {Subst}\left (\int \frac {1}{1-(a+b) x^2} \, dx,x,\frac {\sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{f}-\frac {(b (3 a+2 b)) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 f}\\ &=-\frac {\sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 f}+\frac {(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{f}-\frac {b \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 f}\\ \end {align*}
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Mathematica [A]
time = 0.48, size = 233, normalized size = 1.93 \begin {gather*} \frac {\sqrt {2} b (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a+b} \sin (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )+\sqrt {2} \left (4 a^2+5 a b+2 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {2 a+2 b} \sin (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )-2 \sqrt {b} \sqrt {a+b} \left (\sqrt {2} (3 a+2 b) \log \left (\sqrt {2 a+b-b \cos (2 (e+f x))}+\sqrt {2} \sqrt {b} \sin (e+f x)\right )+\sqrt {b} \sqrt {2 a+b-b \cos (2 (e+f x))} \sin (e+f x)\right )}{4 \sqrt {2} \sqrt {a+b} f} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(416\) vs.
\(2(103)=206\).
time = 19.57, size = 417, normalized size = 3.45
method | result | size |
default | \(-\frac {2 b^{\frac {3}{2}} \ln \left (\sin \left (f x +e \right ) \sqrt {b}+\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\right ) \sqrt {a +b}+3 a \sqrt {b}\, \ln \left (\sin \left (f x +e \right ) \sqrt {b}+\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\right ) \sqrt {a +b}+b \sin \left (f x +e \right ) \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \sqrt {a +b}+\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2}+2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a b +\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{2}-\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2}-2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a b -\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{2}}{2 \sqrt {a +b}\, f}\) | \(417\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.53, size = 178, normalized size = 1.47 \begin {gather*} -\frac {3 \, a \sqrt {b} \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right ) + 2 \, b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right ) - {\left (a + b\right )}^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right ) - {\left (a + b\right )}^{\frac {3}{2}} \operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right ) + \sqrt {b \sin \left (f x + e\right )^{2} + a} b \sin \left (f x + e\right )}{2 \, 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. 280 vs.
\(2 (103) = 206\).
time = 0.79, size = 1381, normalized size = 11.41 \begin {gather*} \left [\frac {{\left (3 \, a + 2 \, b\right )} \sqrt {b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{2} b^{2} + 24 \, a b^{3} + 24 \, b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 32 \, a^{3} b + 160 \, a^{2} b^{2} + 256 \, a b^{3} + 128 \, b^{4} - 32 \, {\left (a^{3} b + 10 \, a^{2} b^{2} + 24 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{6} - 24 \, {\left (a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} - 10 \, a^{2} b - 24 \, a b^{2} - 16 \, b^{3} + 2 \, {\left (5 \, a^{2} b + 24 \, a b^{2} + 24 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b} \sin \left (f x + e\right )\right ) + 4 \, {\left (a + b\right )}^{\frac {3}{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 8 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \sin \left (f x + e\right )}{16 \, f}, -\frac {8 \, {\left (a + b\right )} \sqrt {-a - b} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) - {\left (3 \, a + 2 \, b\right )} \sqrt {b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{2} b^{2} + 24 \, a b^{3} + 24 \, b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 32 \, a^{3} b + 160 \, a^{2} b^{2} + 256 \, a b^{3} + 128 \, b^{4} - 32 \, {\left (a^{3} b + 10 \, a^{2} b^{2} + 24 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{6} - 24 \, {\left (a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} - 10 \, a^{2} b - 24 \, a b^{2} - 16 \, b^{3} + 2 \, {\left (5 \, a^{2} b + 24 \, a b^{2} + 24 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b} \sin \left (f x + e\right )\right ) + 8 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \sin \left (f x + e\right )}{16 \, f}, \frac {{\left (3 \, a + 2 \, b\right )} \sqrt {-b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 8 \, a b + 8 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{4} + a^{2} b + 3 \, a b^{2} + 2 \, b^{3} - {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) + 2 \, {\left (a + b\right )}^{\frac {3}{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \sin \left (f x + e\right )}{8 \, f}, -\frac {4 \, {\left (a + b\right )} \sqrt {-a - b} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) - {\left (3 \, a + 2 \, b\right )} \sqrt {-b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 8 \, a b + 8 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{4} + a^{2} b + 3 \, a b^{2} + 2 \, b^{3} - {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \sin \left (f x + e\right )}{8 \, f}\right ] \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*} \int \left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \sec {\left (e + f x \right )}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}}{\cos \left (e+f\,x\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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