Optimal. Leaf size=205 \[ \frac{\sin (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{3 a^3 d}+\frac{2 b^2 \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{x \left (a^2 b (A+2 C)+a^3 (-B)-2 a b^2 B+2 A b^3\right )}{2 a^4}-\frac{(A b-a B) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{A \sin (c+d x) \cos ^2(c+d x)}{3 a d} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.740061, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {4104, 3919, 3831, 2659, 208} \[ \frac{\sin (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{3 a^3 d}+\frac{2 b^2 \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{x \left (a^2 b (A+2 C)+a^3 (-B)-2 a b^2 B+2 A b^3\right )}{2 a^4}-\frac{(A b-a B) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{A \sin (c+d x) \cos ^2(c+d x)}{3 a d} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 4104
Rule 3919
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 a d}-\frac{\int \frac{\cos ^2(c+d x) \left (3 (A b-a B)-a (2 A+3 C) \sec (c+d x)-2 A b \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 a}\\ &=-\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{\int \frac{\cos (c+d x) \left (2 \left (3 A b^2-3 a b B+\frac{1}{2} a^2 (4 A+6 C)\right )+a (A b+3 a B) \sec (c+d x)-3 b (A b-a B) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 a^2}\\ &=\frac{\left (3 A b^2-3 a b B+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 a d}-\frac{\int \frac{3 \left (2 A b^3-a^3 B-2 a b^2 B+a^2 b (A+2 C)\right )+3 a b (A b-a B) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 a^3}\\ &=-\frac{\left (2 A b^3-a^3 B-2 a b^2 B+a^2 b (A+2 C)\right ) x}{2 a^4}+\frac{\left (3 A b^2-3 a b B+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{\left (b^2 \left (A b^2-a (b B-a C)\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^4}\\ &=-\frac{\left (2 A b^3-a^3 B-2 a b^2 B+a^2 b (A+2 C)\right ) x}{2 a^4}+\frac{\left (3 A b^2-3 a b B+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{\left (b \left (A b^2-a (b B-a C)\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^4}\\ &=-\frac{\left (2 A b^3-a^3 B-2 a b^2 B+a^2 b (A+2 C)\right ) x}{2 a^4}+\frac{\left (3 A b^2-3 a b B+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{\left (2 b \left (A b^2-a (b B-a C)\right )\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 )}{a^4 d}\\ &=-\frac{\left (2 A b^3-a^3 B-2 a b^2 B+a^2 b (A+2 C)\right ) x}{2 a^4}+\frac{2 b^2 \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 \sqrt{a-b} \sqrt{a+b} d}+\frac{\left (3 A b^2-3 a b B+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 a^3 d}-\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 a d}\\ \end{align*}
Mathematica [A] time = 0.602023, size = 178, normalized size = 0.87 \[ \frac{6 (c+d x) \left (-a^2 b (A+2 C)+a^3 B+2 a b^2 B-2 A b^3\right )+3 a \sin (c+d x) \left (a^2 (3 A+4 C)-4 a b B+4 A b^2\right )-\frac{24 b^2 \left (a (a C-b B)+A b^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+3 a^2 (a B-A b) \sin (2 (c+d x))+a^3 A \sin (3 (c+d x))}{12 a^4 d} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] time = 0.137, size = 814, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Fricas [A] time = 0.624831, size = 1301, normalized size = 6.35 \begin{align*} \left [\frac{3 \,{\left (B a^{5} -{\left (A + 2 \, C\right )} a^{4} b + B a^{3} b^{2} -{\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} d x + 3 \,{\left (C a^{2} b^{2} - B a b^{3} + A b^{4}\right )} \sqrt{a^{2} - b^{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 \,{\left (2 \, A + 3 \, C\right )} a^{5} - 6 \, B a^{4} b + 2 \,{\left (A - 3 \, C\right )} a^{3} b^{2} + 6 \, B a^{2} b^{3} - 6 \, A a b^{4} + 2 \,{\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{6} - a^{4} b^{2}\right )} d}, \frac{3 \,{\left (B a^{5} -{\left (A + 2 \, C\right )} a^{4} b + B a^{3} b^{2} -{\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} d x + 6 \,{\left (C a^{2} b^{2} - B a b^{3} + A b^{4}\right )} \sqrt{-a^{2} + b^{2}} \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 ) +{\left (2 \,{\left (2 \, A + 3 \, C\right )} a^{5} - 6 \, B a^{4} b + 2 \,{\left (A - 3 \, C\right )} a^{3} b^{2} + 6 \, B a^{2} b^{3} - 6 \, A a b^{4} + 2 \,{\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{6} - a^{4} b^{2}\right )} d}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [B] time = 1.29894, size = 572, normalized size = 2.79 \begin{align*} \frac{\frac{3 \,{\left (B a^{3} - A a^{2} b - 2 \, C a^{2} b + 2 \, B a b^{2} - 2 \, A b^{3}\right )}{\left (d x + c\right )}}{a^{4}} + \frac{12 \,{\left (C a^{2} b^{2} - B a b^{3} + A b^{4}\right )}{\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 )}}{\sqrt{-a^{2} + b^{2}} a^{4}} + \frac{2 \,{\left (6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, A b^{2} \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 )}^{3} a^{3}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]