Optimal. Leaf size=275 \[ \frac {a^4 \cos ^3(c+d x)}{3 d}-\frac {a^4 \cos (c+d x)}{d}-\frac {4 a^3 b \sin ^3(c+d x)}{3 d}-\frac {4 a^3 b \sin (c+d x)}{d}+\frac {4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a^2 b^2 \cos ^3(c+d x)}{d}+\frac {12 a^2 b^2 \cos (c+d x)}{d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}+\frac {10 a b^3 \sin ^3(c+d x)}{3 d}+\frac {10 a b^3 \sin (c+d x)}{d}+\frac {2 a b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{d}-\frac {10 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^4 \cos ^3(c+d x)}{3 d}-\frac {3 b^4 \cos (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}-\frac {3 b^4 \sec (c+d x)}{d} \]
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Rubi [A] time = 0.25, antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3517, 2633, 2592, 302, 206, 2590, 270, 288} \[ -\frac {2 a^2 b^2 \cos ^3(c+d x)}{d}+\frac {12 a^2 b^2 \cos (c+d x)}{d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {4 a^3 b \sin ^3(c+d x)}{3 d}-\frac {4 a^3 b \sin (c+d x)}{d}+\frac {4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^4 \cos ^3(c+d x)}{3 d}-\frac {a^4 \cos (c+d x)}{d}+\frac {10 a b^3 \sin ^3(c+d x)}{3 d}+\frac {10 a b^3 \sin (c+d x)}{d}+\frac {2 a b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{d}-\frac {10 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^4 \cos ^3(c+d x)}{3 d}-\frac {3 b^4 \cos (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}-\frac {3 b^4 \sec (c+d x)}{d} \]
Antiderivative was successfully verified.
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Rule 206
Rule 270
Rule 288
Rule 302
Rule 2590
Rule 2592
Rule 2633
Rule 3517
Rubi steps
\begin {align*} \int \sin ^3(c+d x) (a+b \tan (c+d x))^4 \, dx &=\int \left (a^4 \sin ^3(c+d x)+4 a^3 b \sin ^3(c+d x) \tan (c+d x)+6 a^2 b^2 \sin ^3(c+d x) \tan ^2(c+d x)+4 a b^3 \sin ^3(c+d x) \tan ^3(c+d x)+b^4 \sin ^3(c+d x) \tan ^4(c+d x)\right ) \, dx\\ &=a^4 \int \sin ^3(c+d x) \, dx+\left (4 a^3 b\right ) \int \sin ^3(c+d x) \tan (c+d x) \, dx+\left (6 a^2 b^2\right ) \int \sin ^3(c+d x) \tan ^2(c+d x) \, dx+\left (4 a b^3\right ) \int \sin ^3(c+d x) \tan ^3(c+d x) \, dx+b^4 \int \sin ^3(c+d x) \tan ^4(c+d x) \, dx\\ &=-\frac {a^4 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (4 a^3 b\right ) \operatorname {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {\left (6 a^2 b^2\right ) \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (4 a b^3\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {b^4 \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^3}{x^4} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a^4 \cos (c+d x)}{d}+\frac {a^4 \cos ^3(c+d x)}{3 d}+\frac {2 a b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{d}+\frac {\left (4 a^3 b\right ) \operatorname {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac {\left (6 a^2 b^2\right ) \operatorname {Subst}\left (\int \left (-2+\frac {1}{x^2}+x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {\left (10 a b^3\right ) \operatorname {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {b^4 \operatorname {Subst}\left (\int \left (3+\frac {1}{x^4}-\frac {3}{x^2}-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a^4 \cos (c+d x)}{d}+\frac {12 a^2 b^2 \cos (c+d x)}{d}-\frac {3 b^4 \cos (c+d x)}{d}+\frac {a^4 \cos ^3(c+d x)}{3 d}-\frac {2 a^2 b^2 \cos ^3(c+d x)}{d}+\frac {b^4 \cos ^3(c+d x)}{3 d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {3 b^4 \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}-\frac {4 a^3 b \sin (c+d x)}{d}-\frac {4 a^3 b \sin ^3(c+d x)}{3 d}+\frac {2 a b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{d}+\frac {\left (4 a^3 b\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac {\left (10 a b^3\right ) \operatorname {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a^4 \cos (c+d x)}{d}+\frac {12 a^2 b^2 \cos (c+d x)}{d}-\frac {3 b^4 \cos (c+d x)}{d}+\frac {a^4 \cos ^3(c+d x)}{3 d}-\frac {2 a^2 b^2 \cos ^3(c+d x)}{d}+\frac {b^4 \cos ^3(c+d x)}{3 d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {3 b^4 \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}-\frac {4 a^3 b \sin (c+d x)}{d}+\frac {10 a b^3 \sin (c+d x)}{d}-\frac {4 a^3 b \sin ^3(c+d x)}{3 d}+\frac {10 a b^3 \sin ^3(c+d x)}{3 d}+\frac {2 a b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{d}-\frac {\left (10 a b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {10 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a^4 \cos (c+d x)}{d}+\frac {12 a^2 b^2 \cos (c+d x)}{d}-\frac {3 b^4 \cos (c+d x)}{d}+\frac {a^4 \cos ^3(c+d x)}{3 d}-\frac {2 a^2 b^2 \cos ^3(c+d x)}{d}+\frac {b^4 \cos ^3(c+d x)}{3 d}+\frac {6 a^2 b^2 \sec (c+d x)}{d}-\frac {3 b^4 \sec (c+d x)}{d}+\frac {b^4 \sec ^3(c+d x)}{3 d}-\frac {4 a^3 b \sin (c+d x)}{d}+\frac {10 a b^3 \sin (c+d x)}{d}-\frac {4 a^3 b \sin ^3(c+d x)}{3 d}+\frac {10 a b^3 \sin ^3(c+d x)}{3 d}+\frac {2 a b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{d}\\ \end {align*}
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Mathematica [B] time = 6.27, size = 1017, normalized size = 3.70 \[ -\frac {\left (3 a^4-42 b^2 a^2+11 b^4\right ) (a+b \tan (c+d x))^4 \cos ^5(c+d x)}{4 d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {a b \left (a^2-b^2\right ) \sin (3 (c+d x)) (a+b \tan (c+d x))^4 \cos ^4(c+d x)}{3 d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {\left (a^4-6 b^2 a^2+b^4\right ) \cos (3 (c+d x)) (a+b \tan (c+d x))^4 \cos ^4(c+d x)}{12 d (a \cos (c+d x)+b \sin (c+d x))^4}-\frac {2 \left (2 a^3 b-5 a b^3\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^4 \cos ^4(c+d x)}{d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {2 \left (2 a^3 b-5 a b^3\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^4 \cos ^4(c+d x)}{d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {b^4 \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^4 \cos ^4(c+d x)}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {\left (36 a^2 b^2 \sin \left (\frac {1}{2} (c+d x)\right )-17 b^4 \sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^4 \cos ^4(c+d x)}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {\left (17 b^4 \sin \left (\frac {1}{2} (c+d x)\right )-36 a^2 b^2 \sin \left (\frac {1}{2} (c+d x)\right )\right ) (a+b \tan (c+d x))^4 \cos ^4(c+d x)}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^4}-\frac {a b \left (5 a^2-9 b^2\right ) \sin (c+d x) (a+b \tan (c+d x))^4 \cos ^4(c+d x)}{d (a \cos (c+d x)+b \sin (c+d x))^4}-\frac {b^2 \left (17 b^2-36 a^2\right ) (a+b \tan (c+d x))^4 \cos ^4(c+d x)}{6 d (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {\left (b^4+12 a b^3\right ) (a+b \tan (c+d x))^4 \cos ^4(c+d x)}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^4}+\frac {\left (b^4-12 a b^3\right ) (a+b \tan (c+d x))^4 \cos ^4(c+d x)}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^4}-\frac {b^4 \sin \left (\frac {1}{2} (c+d x)\right ) (a+b \tan (c+d x))^4 \cos ^4(c+d x)}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (a \cos (c+d x)+b \sin (c+d x))^4} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.49, size = 224, normalized size = 0.81 \[ \frac {{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{4} - 12 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (2 \, a^{3} b - 5 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, a^{3} b - 5 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + b^{4} + 9 \, {\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{5} + 3 \, a b^{3} \cos \left (d x + c\right ) - 2 \, {\left (4 \, a^{3} b - 7 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [A] time = 0.53, size = 412, normalized size = 1.50 \[ -\frac {\cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{4}}{3 d}-\frac {2 a^{4} \cos \left (d x +c \right )}{3 d}-\frac {4 a^{3} b \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}-\frac {4 a^{3} b \sin \left (d x +c \right )}{d}+\frac {4 a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {6 a^{2} b^{2} \left (\sin ^{6}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {16 a^{2} b^{2} \cos \left (d x +c \right )}{d}+\frac {6 a^{2} b^{2} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{d}+\frac {8 a^{2} b^{2} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{d}+\frac {2 a \,b^{3} \left (\sin ^{7}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}}+\frac {2 a \,b^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{d}+\frac {10 a \,b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d}+\frac {10 a \,b^{3} \sin \left (d x +c \right )}{d}-\frac {10 a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{4} \left (\sin ^{8}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3}}-\frac {5 b^{4} \left (\sin ^{8}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )}-\frac {16 b^{4} \cos \left (d x +c \right )}{3 d}-\frac {5 b^{4} \cos \left (d x +c \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{3 d}-\frac {2 b^{4} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{d}-\frac {8 b^{4} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{3 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.54, size = 218, normalized size = 0.79 \[ \frac {{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{4} - 2 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{3} b - 6 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a^{2} b^{2} + {\left (4 \, \sin \left (d x + c\right )^{3} - \frac {6 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 24 \, \sin \left (d x + c\right )\right )} a b^{3} + {\left (\cos \left (d x + c\right )^{3} - \frac {9 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} - 9 \, \cos \left (d x + c\right )\right )} b^{4}}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.21, size = 319, normalized size = 1.16 \[ -\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (8\,a^4-96\,a^2\,b^2+32\,b^4\right )+4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (20\,a\,b^3-8\,a^3\,b\right )-\frac {4\,a^4}{3}-\frac {32\,b^4}{3}+32\,a^2\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {32\,a^4}{3}-64\,a^2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {20\,a\,b^3}{3}-\frac {8\,a^3\,b}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (20\,a\,b^3-8\,a^3\,b\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {20\,a\,b^3}{3}-\frac {8\,a^3\,b}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (56\,a\,b^3-48\,a^3\,b\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (56\,a\,b^3-48\,a^3\,b\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (20\,a\,b^3-8\,a^3\,b\right )}{d} \]
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 \[ \int \left (a + b \tan {\left (c + d x \right )}\right )^{4} \sin ^{3}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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