Optimal. Leaf size=261 \[ -\frac{b \left (a^2 (4 A+3 C)+b^2 (A+2 C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac{a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac{\left (a^2 b^2 (2 A+9 C)-4 a^4 C+3 A b^4\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac{a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]
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Rubi [A] time = 0.672904, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4091, 4080, 4003, 12, 3831, 2659, 208} \[ -\frac{b \left (a^2 (4 A+3 C)+b^2 (A+2 C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{7/2} (a+b)^{7/2}}+\frac{a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}+\frac{\left (a^2 b^2 (2 A+9 C)-4 a^4 C+3 A b^4\right ) \tan (c+d x)}{6 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^2}+\frac{a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^3} \]
Antiderivative was successfully verified.
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Rule 4091
Rule 4080
Rule 4003
Rule 12
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^4} \, dx &=\frac{a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\int \frac{\sec (c+d x) \left (-3 b \left (A b^2+a^2 C\right )+a \left (2 A b^2-\left (a^2-3 b^2\right ) C\right ) \sec (c+d x)+3 b \left (a^2-b^2\right ) C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=\frac{a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}-\frac{\int \frac{\sec (c+d x) \left (-2 a b^2 \left (a^2 C-b^2 (5 A+6 C)\right )-b \left (a^2 b^2 (2 A-3 C)+2 a^4 C+3 b^4 (A+2 C)\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{6 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}+\frac{\int -\frac{3 b^4 \left (b^2 (A+2 C)+a^2 (4 A+3 C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )^3}\\ &=\frac{a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (b \left (b^2 (A+2 C)+a^2 (4 A+3 C)\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=\frac{a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (b^2 (A+2 C)+a^2 (4 A+3 C)\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=\frac{a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}-\frac{\left (b^2 (A+2 C)+a^2 (4 A+3 C)\right ) \operatorname{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 )}{\left (a^2-b^2\right )^3 d}\\ &=-\frac{b \left (4 a^2 A+A b^2+3 a^2 C+2 b^2 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}+\frac{a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^3}+\frac{\left (3 A b^4-4 a^4 C+a^2 b^2 (2 A+9 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^2}+\frac{a \left (a^2 b^2 (2 A-5 C)+2 a^4 C+b^4 (13 A+18 C)\right ) \tan (c+d x)}{6 b^2 \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))}\\ \end{align*}
Mathematica [A] time = 1.27293, size = 221, normalized size = 0.85 \[ -\frac{\frac{2 \sin (c+d x) \left (6 b \left (9 a^2 b^2 (A+C)+a^4 (2 A+C)-A b^4\right ) \cos (c+d x)+a \left (\left (a^2 b^2 (10 A+11 C)+a^4 (6 A+4 C)-A b^4\right ) \cos (2 (c+d x))+a^2 b^2 (14 A+C)+a^4 (6 A+8 C)+b^4 (25 A+36 C)\right )\right )}{(a \cos (c+d x)+b)^3}+\frac{24 b \left (a^2 (4 A+3 C)+b^2 (A+2 C)\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{24 d \left (b^2-a^2\right )^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.095, size = 374, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ( 2\,{\frac{1}{ \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) ^{3}} \left ( -1/2\,{\frac{ \left ( 2\,A{a}^{3}+2\,A{a}^{2}b+6\,Aa{b}^{2}+A{b}^{3}+2\,{a}^{3}C+3\,{a}^{2}bC+6\,Ca{b}^{2} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{ \left ( a-b \right ) \left ({a}^{3}+3\,{a}^{2}b+3\,a{b}^{2}+{b}^{3} \right ) }}+2/3\,{\frac{ \left ( 3\,{a}^{2}A+7\,A{b}^{2}+{a}^{2}C+9\,{b}^{2}C \right ) a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{ \left ({a}^{2}+2\,ab+{b}^{2} \right ) \left ({a}^{2}-2\,ab+{b}^{2} \right ) }}-1/2\,{\frac{ \left ( 2\,A{a}^{3}-2\,A{a}^{2}b+6\,Aa{b}^{2}-A{b}^{3}+2\,{a}^{3}C-3\,{a}^{2}bC+6\,Ca{b}^{2} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{ \left ( a+b \right ) \left ({a}^{3}-3\,{a}^{2}b+3\,a{b}^{2}-{b}^{3} \right ) }} \right ) }-{\frac{b \left ( 4\,{a}^{2}A+A{b}^{2}+3\,{a}^{2}C+2\,{b}^{2}C \right ) }{{a}^{6}-3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}-{b}^{6}}{\it Artanh} \left ({(a-b)\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 0.755258, size = 2475, normalized size = 9.48 \begin{align*} \text{result too large to display} \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*} \int \frac{\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{4}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.29461, size = 936, normalized size = 3.59 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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