Optimal. Leaf size=169 \[ \frac {3 b^{5/2} d \text {ArcTan}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{4 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}-\frac {3 b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{4 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}+\frac {b \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 f} \]
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Rubi [A]
time = 0.11, antiderivative size = 169, 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 = {2691, 2696,
2644, 335, 304, 209, 212} \begin {gather*} \frac {3 b^{5/2} d \sqrt {b \tan (e+f x)} \text {ArcTan}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{4 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {3 b^{5/2} d \sqrt {b \tan (e+f x)} \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{4 f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {b (b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}{2 f} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 212
Rule 304
Rule 335
Rule 2644
Rule 2691
Rule 2696
Rubi steps
\begin {align*} \int \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2} \, dx &=\frac {b \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 f}-\frac {1}{4} \left (3 b^2\right ) \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)} \, dx\\ &=\frac {b \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 f}-\frac {\left (3 b^2 d \sqrt {b \tan (e+f x)}\right ) \int \sec (e+f x) \sqrt {b \sin (e+f x)} \, dx}{4 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}\\ &=\frac {b \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 f}-\frac {\left (3 b d \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{b^2}} \, dx,x,b \sin (e+f x)\right )}{4 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}\\ &=\frac {b \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 f}-\frac {\left (3 b d \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{b^2}} \, dx,x,\sqrt {b \sin (e+f x)}\right )}{2 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}\\ &=\frac {b \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 f}-\frac {\left (3 b^3 d \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \sin (e+f x)}\right )}{4 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}+\frac {\left (3 b^3 d \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \sin (e+f x)}\right )}{4 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}\\ &=\frac {3 b^{5/2} d \tan ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{4 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}-\frac {3 b^{5/2} d \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{4 f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}+\frac {b \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}}{2 f}\\ \end {align*}
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Mathematica [A]
time = 1.70, size = 182, normalized size = 1.08 \begin {gather*} \frac {\csc ^3(e+f x) \sqrt {d \sec (e+f x)} (b \tan (e+f x))^{5/2} \left (-4 \sqrt {\sec (e+f x)}+4 \sec ^{\frac {5}{2}}(e+f x)-6 \text {ArcTan}\left (\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right ) \sqrt [4]{\tan ^2(e+f x)}+3 \left (\log \left (1-\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )-\log \left (1+\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )\right ) \sqrt [4]{\tan ^2(e+f x)}\right )}{8 f \sec ^{\frac {7}{2}}(e+f x)} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 0.35, size = 602, normalized size = 3.56
method | result | size |
default | \(\frac {\left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sqrt {\frac {d}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \left (3 i \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (\cos ^{2}\left (f x +e \right )\right )-3 i \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+3 \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+3 \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right ) \sqrt {2}-2 \sqrt {2}\right ) \sqrt {2}}{8 f \left (\cos \left (f x +e \right )-1\right ) \sin \left (f x +e \right )}\) | \(602\) |
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 \begin {gather*} \text {Failed to integrate} \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. 425 vs.
\(2 (145) = 290\).
time = 0.53, size = 858, normalized size = 5.08 \begin {gather*} \left [\frac {6 \, \sqrt {-b d} b^{2} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {-b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b d \cos \left (f x + e\right )^{2} - b d - {\left (b d \cos \left (f x + e\right ) + b d\right )} \sin \left (f x + e\right )\right )}}\right ) \cos \left (f x + e\right ) + 3 \, \sqrt {-b d} b^{2} \cos \left (f x + e\right ) \log \left (\frac {b d \cos \left (f x + e\right )^{4} - 72 \, b d \cos \left (f x + e\right )^{2} + 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} - {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \cos \left (f x + e\right )\right )} \sqrt {-b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 72 \, b d + 28 \, {\left (b d \cos \left (f x + e\right )^{2} - 2 \, b d\right )} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right ) + 16 \, b^{2} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{32 \, f \cos \left (f x + e\right )}, \frac {6 \, \sqrt {b d} b^{2} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b d \cos \left (f x + e\right )^{2} - b d + {\left (b d \cos \left (f x + e\right ) + b d\right )} \sin \left (f x + e\right )\right )}}\right ) \cos \left (f x + e\right ) + 3 \, \sqrt {b d} b^{2} \cos \left (f x + e\right ) \log \left (\frac {b d \cos \left (f x + e\right )^{4} - 72 \, b d \cos \left (f x + e\right )^{2} + 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} + {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \cos \left (f x + e\right )\right )} \sqrt {b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 72 \, b d - 28 \, {\left (b d \cos \left (f x + e\right )^{2} - 2 \, b d\right )} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right ) + 16 \, b^{2} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{32 \, f \cos \left (f x + e\right )}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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 {\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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