Optimal. Leaf size=204 \[ \frac{(6 A-55 B+244 C) \tan (c+d x)}{105 a^4 d}+\frac{(3 A+25 B-88 C) \tan (c+d x) \sec ^2(c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac{(B-4 C) \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{(B-4 C) \tan (c+d x)}{a^4 d (\sec (c+d x)+1)}-\frac{(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac{(2 A+5 B-12 C) \tan (c+d x) \sec ^3(c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]
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Rubi [A] time = 0.629072, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4084, 4019, 4008, 3787, 3770, 3767, 8} \[ \frac{(6 A-55 B+244 C) \tan (c+d x)}{105 a^4 d}+\frac{(3 A+25 B-88 C) \tan (c+d x) \sec ^2(c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac{(B-4 C) \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{(B-4 C) \tan (c+d x)}{a^4 d (\sec (c+d x)+1)}-\frac{(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac{(2 A+5 B-12 C) \tan (c+d x) \sec ^3(c+d x)}{35 a d (a \sec (c+d x)+a)^3} \]
Antiderivative was successfully verified.
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Rule 4084
Rule 4019
Rule 4008
Rule 3787
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \frac{\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx &=-\frac{(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{\int \frac{\sec ^4(c+d x) (a (3 A+4 B-4 C)+a (A-B+8 C) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(2 A+5 B-12 C) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec ^3(c+d x) \left (3 a^2 (2 A+5 B-12 C)+a^2 (3 A-10 B+52 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=\frac{(3 A+25 B-88 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(2 A+5 B-12 C) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec ^2(c+d x) \left (2 a^3 (3 A+25 B-88 C)+a^3 (6 A-55 B+244 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=\frac{(3 A+25 B-88 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(2 A+5 B-12 C) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{(B-4 C) \tan (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}-\frac{\int \sec (c+d x) \left (-105 a^4 (B-4 C)-a^4 (6 A-55 B+244 C) \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=\frac{(3 A+25 B-88 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(2 A+5 B-12 C) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{(B-4 C) \tan (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}+\frac{(B-4 C) \int \sec (c+d x) \, dx}{a^4}+\frac{(6 A-55 B+244 C) \int \sec ^2(c+d x) \, dx}{105 a^4}\\ &=\frac{(B-4 C) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{(3 A+25 B-88 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(2 A+5 B-12 C) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{(B-4 C) \tan (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}-\frac{(6 A-55 B+244 C) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d}\\ &=\frac{(B-4 C) \tanh ^{-1}(\sin (c+d x))}{a^4 d}+\frac{(6 A-55 B+244 C) \tan (c+d x)}{105 a^4 d}+\frac{(3 A+25 B-88 C) \sec ^2(c+d x) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac{(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac{(2 A+5 B-12 C) \sec ^3(c+d x) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac{(B-4 C) \tan (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end{align*}
Mathematica [B] time = 6.39385, size = 1208, normalized size = 5.92 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.076, size = 363, normalized size = 1.8 \begin{align*}{\frac{A}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{B}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{C}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{3\,A}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{B}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{7\,C}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{A}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{11\,B}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{23\,C}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{A}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{15\,B}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{49\,C}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{B}{d{a}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-4\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) C}{d{a}^{4}}}-{\frac{C}{d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{B}{d{a}^{4}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+4\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) C}{d{a}^{4}}}-{\frac{C}{d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 0.994374, size = 555, normalized size = 2.72 \begin{align*} \frac{C{\left (\frac{1680 \, \sin \left (d x + c\right )}{{\left (a^{4} - \frac{a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} + \frac{\frac{5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac{3360 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac{3360 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - 5 \, B{\left (\frac{\frac{315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac{168 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac{168 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} + \frac{3 \, A{\left (\frac{35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.530904, size = 872, normalized size = 4.27 \begin{align*} \frac{105 \,{\left ({\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{5} + 4 \,{\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{4} + 6 \,{\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (B - 4 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \,{\left ({\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{5} + 4 \,{\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{4} + 6 \,{\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (B - 4 \, C\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (3 \, A - 80 \, B + 332 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (24 \, A - 535 \, B + 2236 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (39 \, A - 620 \, B + 2636 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (9 \, A - 65 \, B + 296 \, C\right )} \cos \left (d x + c\right ) + 105 \, C\right )} \sin \left (d x + c\right )}{210 \,{\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sec ^{4}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{B \sec ^{5}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \sec ^{6}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec{\left (c + d x \right )} + 1}\, dx}{a^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.36198, size = 385, normalized size = 1.89 \begin{align*} \frac{\frac{840 \,{\left (B - 4 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac{840 \,{\left (B - 4 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac{1680 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} a^{4}} + \frac{15 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 63 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 105 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 147 \, C a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 105 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 385 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 805 \, C a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 105 \, A a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1575 \, B a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5145 \, C a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{28}}}{840 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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