Optimal. Leaf size=119 \[ \frac {\left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac {2 a^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^3 \sqrt {a+b} d}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\sec (c+d x) \tan (c+d x)}{2 b d} \]
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Rubi [A]
time = 0.19, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3936, 4167,
4083, 3855, 3916, 2738, 214} \begin {gather*} -\frac {2 a^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d \sqrt {a-b} \sqrt {a+b}}+\frac {\left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 214
Rule 2738
Rule 3855
Rule 3916
Rule 3936
Rule 4083
Rule 4167
Rubi steps
\begin {align*} \int \frac {\sec ^4(c+d x)}{a+b \sec (c+d x)} \, dx &=\frac {\sec (c+d x) \tan (c+d x)}{2 b d}+\frac {\int \frac {\sec (c+d x) \left (a+b \sec (c+d x)-2 a \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b}\\ &=-\frac {a \tan (c+d x)}{b^2 d}+\frac {\sec (c+d x) \tan (c+d x)}{2 b d}+\frac {\int \frac {\sec (c+d x) \left (a b+\left (2 a^2+b^2\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^2}\\ &=-\frac {a \tan (c+d x)}{b^2 d}+\frac {\sec (c+d x) \tan (c+d x)}{2 b d}-\frac {a^3 \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{b^3}+\frac {\left (2 a^2+b^2\right ) \int \sec (c+d x) \, dx}{2 b^3}\\ &=\frac {\left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\sec (c+d x) \tan (c+d x)}{2 b d}-\frac {a^3 \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{b^4}\\ &=\frac {\left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\sec (c+d x) \tan (c+d x)}{2 b d}-\frac {\left (2 a^3\right ) \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^4 d}\\ &=\frac {\left (2 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac {2 a^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^3 \sqrt {a+b} d}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\sec (c+d x) \tan (c+d x)}{2 b d}\\ \end {align*}
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Mathematica [A]
time = 1.19, size = 238, normalized size = 2.00 \begin {gather*} \frac {\frac {8 a^3 \tanh ^{-1}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-4 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {b^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-4 a b \tan (c+d x)}{4 b^3 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.21, size = 192, normalized size = 1.61
method | result | size |
derivativedivides | \(\frac {-\frac {1}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-2 a -b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (2 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{3}}-\frac {2 a^{3} \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {1}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-2 a -b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-2 a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{3}}}{d}\) | \(192\) |
default | \(\frac {-\frac {1}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-2 a -b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (2 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{3}}-\frac {2 a^{3} \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {1}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-2 a -b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-2 a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{3}}}{d}\) | \(192\) |
risch | \(-\frac {i \left (b \,{\mathrm e}^{3 i \left (d x +c \right )}+2 a \,{\mathrm e}^{2 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}+2 a \right )}{d \,b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{b^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 b d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d \,b^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{b^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 b d}\) | \(299\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \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. 214 vs.
\(2 (106) = 212\).
time = 3.13, size = 485, normalized size = 4.08 \begin {gather*} \left [\frac {2 \, \sqrt {a^{2} - b^{2}} a^{3} \cos \left (d x + c\right )^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{2} b^{2} - b^{4} - 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} b^{3} - b^{5}\right )} d \cos \left (d x + c\right )^{2}}, -\frac {4 \, \sqrt {-a^{2} + b^{2}} a^{3} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{2} - {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{2} b^{2} - b^{4} - 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} b^{3} - b^{5}\right )} d \cos \left (d x + c\right )^{2}}\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 \frac {\sec ^{4}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.53, size = 211, normalized size = 1.77 \begin {gather*} -\frac {\frac {4 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a^{3}}{\sqrt {-a^{2} + b^{2}} b^{3}} - \frac {{\left (2 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} + \frac {{\left (2 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}} - \frac {2 \, {\left (2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} b^{2}}}{2 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.65, size = 1002, normalized size = 8.42 \begin {gather*} \frac {\sin \left (c+d\,x\right )}{2\,b\,d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )}+\frac {\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2\,b\,d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )}+\frac {a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^3\,d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )}+\frac {\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )}{2\,b\,d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )}-\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{2\,b^2\,d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )}-\frac {a^3\,\mathrm {atan}\left (\frac {\left (8\,a^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-8\,a^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+8\,a^7\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,a\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-8\,a^5\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+5\,a^2\,b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-8\,a^3\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+8\,a^4\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )\right )\,1{}\mathrm {i}}{b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,\left (4\,a^4\,\left (a^2-b^2\right )+b^4\,\left (a^2-b^2\right )+2\,a^5\,b-4\,a^6+2\,a^3\,b^3+4\,a^2\,b^2\,\left (a^2-b^2\right )-a\,b^3\,\left (a^2-b^2\right )-2\,a^3\,b\,\left (a^2-b^2\right )\right )}\right )\,1{}\mathrm {i}}{b^3\,d\,\sqrt {a^2-b^2}\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )}+\frac {a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )}{b^3\,d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )}-\frac {a^3\,\cos \left (2\,c+2\,d\,x\right )\,\mathrm {atan}\left (\frac {\left (8\,a^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-8\,a^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+8\,a^7\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,a\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-8\,a^5\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+5\,a^2\,b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-8\,a^3\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+8\,a^4\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )\right )\,1{}\mathrm {i}}{b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,\left (4\,a^4\,\left (a^2-b^2\right )+b^4\,\left (a^2-b^2\right )+2\,a^5\,b-4\,a^6+2\,a^3\,b^3+4\,a^2\,b^2\,\left (a^2-b^2\right )-a\,b^3\,\left (a^2-b^2\right )-2\,a^3\,b\,\left (a^2-b^2\right )\right )}\right )\,1{}\mathrm {i}}{b^3\,d\,\sqrt {a^2-b^2}\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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