Optimal. Leaf size=83 \[ -\frac{\sin ^3(c+d x) \cos (c+d x) (a+b \tan (c+d x))}{4 d}-\frac{\sin (c+d x) \cos (c+d x) (3 a+4 b \tan (c+d x))}{8 d}+\frac{3 a x}{8}-\frac{b \log (\cos (c+d x))}{d} \]
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Rubi [A] time = 0.168075, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {819, 635, 203, 260} \[ -\frac{\sin ^3(c+d x) \cos (c+d x) (a+b \tan (c+d x))}{4 d}-\frac{\sin (c+d x) \cos (c+d x) (3 a+4 b \tan (c+d x))}{8 d}+\frac{3 a x}{8}-\frac{b \log (\cos (c+d x))}{d} \]
Antiderivative was successfully verified.
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Rule 819
Rule 635
Rule 203
Rule 260
Rubi steps
\begin{align*} \int \sin ^4(c+d x) (a+b \tan (c+d x)) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 (a+b x)}{\left (1+x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{\cos (c+d x) \sin ^3(c+d x) (a+b \tan (c+d x))}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{x^2 (3 a+4 b x)}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{4 d}\\ &=-\frac{\cos (c+d x) \sin ^3(c+d x) (a+b \tan (c+d x))}{4 d}-\frac{\cos (c+d x) \sin (c+d x) (3 a+4 b \tan (c+d x))}{8 d}+\frac{\operatorname{Subst}\left (\int \frac{3 a+8 b x}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=-\frac{\cos (c+d x) \sin ^3(c+d x) (a+b \tan (c+d x))}{4 d}-\frac{\cos (c+d x) \sin (c+d x) (3 a+4 b \tan (c+d x))}{8 d}+\frac{(3 a) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{8 d}+\frac{b \operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{3 a x}{8}-\frac{b \log (\cos (c+d x))}{d}-\frac{\cos (c+d x) \sin ^3(c+d x) (a+b \tan (c+d x))}{4 d}-\frac{\cos (c+d x) \sin (c+d x) (3 a+4 b \tan (c+d x))}{8 d}\\ \end{align*}
Mathematica [A] time = 0.078587, size = 82, normalized size = 0.99 \[ \frac{3 a (c+d x)}{8 d}-\frac{a \sin (2 (c+d x))}{4 d}+\frac{a \sin (4 (c+d x))}{32 d}-\frac{b \left (\frac{1}{4} \cos ^4(c+d x)-\cos ^2(c+d x)+\log (\cos (c+d x))\right )}{d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.033, size = 92, normalized size = 1.1 \begin{align*} -{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{b\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-{\frac{a\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{3\,a\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{8\,d}}+{\frac{3\,ax}{8}}+{\frac{3\,ac}{8\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 2.15915, size = 117, normalized size = 1.41 \begin{align*} \frac{3 \,{\left (d x + c\right )} a + 4 \, b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - \frac{5 \, a \tan \left (d x + c\right )^{3} - 8 \, b \tan \left (d x + c\right )^{2} + 3 \, a \tan \left (d x + c\right ) - 6 \, b}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1}}{8 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.35933, size = 189, normalized size = 2.28 \begin{align*} -\frac{2 \, b \cos \left (d x + c\right )^{4} - 3 \, a d x - 8 \, b \cos \left (d x + c\right )^{2} + 8 \, b \log \left (-\cos \left (d x + c\right )\right ) -{\left (2 \, a \cos \left (d x + c\right )^{3} - 5 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \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 \left (a + b \tan{\left (c + d x \right )}\right ) \sin ^{4}{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.69996, size = 1439, normalized size = 17.34 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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