Optimal. Leaf size=329 \[ \frac {2 c (c \sec (a+b x))^{3/2}}{3 b d (d \csc (a+b x))^{3/2}}+\frac {c^2 \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {c \sec (a+b x)}}{\sqrt {2} b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {c^2 \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {c \sec (a+b x)}}{\sqrt {2} b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {c^2 \log \left (1-\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{2 \sqrt {2} b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {c^2 \log \left (1+\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{2 \sqrt {2} b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}} \]
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Rubi [A]
time = 0.16, antiderivative size = 329, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2704, 2709,
3557, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {c^2 \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {c \sec (a+b x)}}{\sqrt {2} b d^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {c^2 \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (a+b x)}+1\right ) \sqrt {c \sec (a+b x)}}{\sqrt {2} b d^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}-\frac {c^2 \sqrt {c \sec (a+b x)} \log \left (\tan (a+b x)-\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{2 \sqrt {2} b d^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}+\frac {c^2 \sqrt {c \sec (a+b x)} \log \left (\tan (a+b x)+\sqrt {2} \sqrt {\tan (a+b x)}+1\right )}{2 \sqrt {2} b d^2 \sqrt {\tan (a+b x)} \sqrt {d \csc (a+b x)}}+\frac {2 c (c \sec (a+b x))^{3/2}}{3 b d (d \csc (a+b x))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 210
Rule 303
Rule 335
Rule 631
Rule 642
Rule 1176
Rule 1179
Rule 2704
Rule 2709
Rule 3557
Rubi steps
\begin {align*} \int \frac {(c \sec (a+b x))^{5/2}}{(d \csc (a+b x))^{5/2}} \, dx &=\frac {2 c (c \sec (a+b x))^{3/2}}{3 b d (d \csc (a+b x))^{3/2}}-\frac {c^2 \int \frac {\sqrt {c \sec (a+b x)}}{\sqrt {d \csc (a+b x)}} \, dx}{d^2}\\ &=\frac {2 c (c \sec (a+b x))^{3/2}}{3 b d (d \csc (a+b x))^{3/2}}-\frac {\left (c^2 \sqrt {c \sec (a+b x)}\right ) \int \sqrt {\tan (a+b x)} \, dx}{d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=\frac {2 c (c \sec (a+b x))^{3/2}}{3 b d (d \csc (a+b x))^{3/2}}-\frac {\left (c^2 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{1+x^2} \, dx,x,\tan (a+b x)\right )}{b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=\frac {2 c (c \sec (a+b x))^{3/2}}{3 b d (d \csc (a+b x))^{3/2}}-\frac {\left (2 c^2 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\sqrt {\tan (a+b x)}\right )}{b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=\frac {2 c (c \sec (a+b x))^{3/2}}{3 b d (d \csc (a+b x))^{3/2}}+\frac {\left (c^2 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (a+b x)}\right )}{b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {\left (c^2 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (a+b x)}\right )}{b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=\frac {2 c (c \sec (a+b x))^{3/2}}{3 b d (d \csc (a+b x))^{3/2}}-\frac {\left (c^2 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (a+b x)}\right )}{2 b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {\left (c^2 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (a+b x)}\right )}{2 b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {\left (c^2 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (a+b x)}\right )}{2 \sqrt {2} b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {\left (c^2 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (a+b x)}\right )}{2 \sqrt {2} b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=\frac {2 c (c \sec (a+b x))^{3/2}}{3 b d (d \csc (a+b x))^{3/2}}-\frac {c^2 \log \left (1-\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{2 \sqrt {2} b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {c^2 \log \left (1+\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{2 \sqrt {2} b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {\left (c^2 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (a+b x)}\right )}{\sqrt {2} b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {\left (c^2 \sqrt {c \sec (a+b x)}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (a+b x)}\right )}{\sqrt {2} b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ &=\frac {2 c (c \sec (a+b x))^{3/2}}{3 b d (d \csc (a+b x))^{3/2}}+\frac {c^2 \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {c \sec (a+b x)}}{\sqrt {2} b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {c^2 \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (a+b x)}\right ) \sqrt {c \sec (a+b x)}}{\sqrt {2} b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}-\frac {c^2 \log \left (1-\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{2 \sqrt {2} b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}+\frac {c^2 \log \left (1+\sqrt {2} \sqrt {\tan (a+b x)}+\tan (a+b x)\right ) \sqrt {c \sec (a+b x)}}{2 \sqrt {2} b d^2 \sqrt {d \csc (a+b x)} \sqrt {\tan (a+b x)}}\\ \end {align*}
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Mathematica [A]
time = 1.31, size = 143, normalized size = 0.43 \begin {gather*} \frac {c \left (4+3 \sqrt {2} \text {ArcTan}\left (\frac {-1+\sqrt {\cot ^2(a+b x)}}{\sqrt {2} \sqrt [4]{\cot ^2(a+b x)}}\right ) \cot ^2(a+b x)^{3/4}+3 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{\cot ^2(a+b x)}}{1+\sqrt {\cot ^2(a+b x)}}\right ) \cot ^2(a+b x)^{3/4}\right ) (c \sec (a+b x))^{3/2}}{6 b d (d \csc (a+b x))^{3/2}} \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 = 58.86, size = 544, normalized size = 1.65
method | result | size |
default | \(\frac {\left (-3 i \cos \left (b x +a \right ) \sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+3 i \cos \left (b x +a \right ) \sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-3 \cos \left (b x +a \right ) \sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-3 \cos \left (b x +a \right ) \sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticPi \left (\sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+2 \sqrt {2}\, \cos \left (b x +a \right )-2 \sqrt {2}\right ) \cos \left (b x +a \right ) \left (\frac {c}{\cos \left (b x +a \right )}\right )^{\frac {5}{2}} \sqrt {2}}{6 b \left (-1+\cos \left (b x +a \right )\right ) \left (\frac {d}{\sin \left (b x +a \right )}\right )^{\frac {5}{2}} \sin \left (b x +a \right )}\) | \(544\) |
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 [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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 = -0.00 \begin {gather*} \int \frac {{\left (\frac {c}{\cos \left (a+b\,x\right )}\right )}^{5/2}}{{\left (\frac {d}{\sin \left (a+b\,x\right )}\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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