Optimal. Leaf size=126 \[ \frac {\left (a^2-3 a b+3 b^2\right ) \sin (c+d x)}{d (a-b)^3}-\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} d (a-b)^{7/2}}+\frac {\sin ^5(c+d x)}{5 d (a-b)}-\frac {(2 a-3 b) \sin ^3(c+d x)}{3 d (a-b)^2} \]
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Rubi [A] time = 0.15, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3676, 390, 208} \[ \frac {\left (a^2-3 a b+3 b^2\right ) \sin (c+d x)}{d (a-b)^3}-\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} d (a-b)^{7/2}}+\frac {\sin ^5(c+d x)}{5 d (a-b)}-\frac {(2 a-3 b) \sin ^3(c+d x)}{3 d (a-b)^2} \]
Antiderivative was successfully verified.
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Rule 208
Rule 390
Rule 3676
Rubi steps
\begin {align*} \int \frac {\cos ^5(c+d x)}{a+b \tan ^2(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^3}{a-(a-b) x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a^2-3 a b+3 b^2}{(a-b)^3}-\frac {(2 a-3 b) x^2}{(a-b)^2}+\frac {x^4}{a-b}-\frac {b^3}{(a-b)^3 \left (a-(a-b) x^2\right )}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {\left (a^2-3 a b+3 b^2\right ) \sin (c+d x)}{(a-b)^3 d}-\frac {(2 a-3 b) \sin ^3(c+d x)}{3 (a-b)^2 d}+\frac {\sin ^5(c+d x)}{5 (a-b) d}-\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{a-(a-b) x^2} \, dx,x,\sin (c+d x)\right )}{(a-b)^3 d}\\ &=-\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)^{7/2} d}+\frac {\left (a^2-3 a b+3 b^2\right ) \sin (c+d x)}{(a-b)^3 d}-\frac {(2 a-3 b) \sin ^3(c+d x)}{3 (a-b)^2 d}+\frac {\sin ^5(c+d x)}{5 (a-b) d}\\ \end {align*}
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Mathematica [A] time = 1.82, size = 148, normalized size = 1.17 \[ \frac {\frac {30 \left (5 a^2-16 a b+19 b^2\right ) \sin (c+d x)}{(a-b)^3}+\frac {120 b^3 \left (\log \left (\sqrt {a}-\sqrt {a-b} \sin (c+d x)\right )-\log \left (\sqrt {a-b} \sin (c+d x)+\sqrt {a}\right )\right )}{\sqrt {a} (a-b)^{7/2}}+\frac {5 (5 a-9 b) \sin (3 (c+d x))}{(a-b)^2}+\frac {3 \sin (5 (c+d x))}{a-b}}{240 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.55, size = 395, normalized size = 3.13 \[ \left [-\frac {15 \, \sqrt {a^{2} - a b} b^{3} \log \left (-\frac {{\left (a - b\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - a b} \sin \left (d x + c\right ) - 2 \, a + b}{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + b}\right ) - 2 \, {\left (3 \, {\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} - 34 \, a^{3} b + 59 \, a^{2} b^{2} - 33 \, a b^{3} + {\left (4 \, a^{4} - 17 \, a^{3} b + 22 \, a^{2} b^{2} - 9 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} d}, \frac {15 \, \sqrt {-a^{2} + a b} b^{3} \arctan \left (\frac {\sqrt {-a^{2} + a b} \sin \left (d x + c\right )}{a}\right ) + {\left (3 \, {\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} - 34 \, a^{3} b + 59 \, a^{2} b^{2} - 33 \, a b^{3} + {\left (4 \, a^{4} - 17 \, a^{3} b + 22 \, a^{2} b^{2} - 9 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{5} - 4 \, a^{4} b + 6 \, a^{3} b^{2} - 4 \, a^{2} b^{3} + a b^{4}\right )} d}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.50, size = 319, normalized size = 2.53 \[ -\frac {\frac {15 \, b^{3} \arctan \left (-\frac {a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{\sqrt {-a^{2} + a b}}\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {-a^{2} + a b}} - \frac {3 \, a^{4} \sin \left (d x + c\right )^{5} - 12 \, a^{3} b \sin \left (d x + c\right )^{5} + 18 \, a^{2} b^{2} \sin \left (d x + c\right )^{5} - 12 \, a b^{3} \sin \left (d x + c\right )^{5} + 3 \, b^{4} \sin \left (d x + c\right )^{5} - 10 \, a^{4} \sin \left (d x + c\right )^{3} + 45 \, a^{3} b \sin \left (d x + c\right )^{3} - 75 \, a^{2} b^{2} \sin \left (d x + c\right )^{3} + 55 \, a b^{3} \sin \left (d x + c\right )^{3} - 15 \, b^{4} \sin \left (d x + c\right )^{3} + 15 \, a^{4} \sin \left (d x + c\right ) - 75 \, a^{3} b \sin \left (d x + c\right ) + 150 \, a^{2} b^{2} \sin \left (d x + c\right ) - 135 \, a b^{3} \sin \left (d x + c\right ) + 45 \, b^{4} \sin \left (d x + c\right )}{a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}}}{15 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.83, size = 165, normalized size = 1.31 \[ \frac {\frac {\frac {\left (\sin ^{5}\left (d x +c \right )\right ) a^{2}}{5}-\frac {2 \left (\sin ^{5}\left (d x +c \right )\right ) a b}{5}+\frac {b^{2} \left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {2 \left (\sin ^{3}\left (d x +c \right )\right ) a^{2}}{3}+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right ) a b}{3}-\left (\sin ^{3}\left (d x +c \right )\right ) b^{2}+a^{2} \sin \left (d x +c \right )-3 \sin \left (d x +c \right ) a b +3 b^{2} \sin \left (d x +c \right )}{\left (a -b \right )^{3}}-\frac {b^{3} \arctanh \left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{\left (a -b \right )^{3} \sqrt {a \left (a -b \right )}}}{d} \]
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 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 15.08, size = 1493, normalized size = 11.85 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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