Optimal. Leaf size=377 \[ -\frac{\sqrt{3} \sqrt [3]{a-i b} (B+i A) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{2 d}+\frac{\sqrt{3} \sqrt [3]{a+i b} (-B+i A) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{2 d}+\frac{3 \sqrt [3]{a-i b} (B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d}-\frac{3 \sqrt [3]{a+i b} (-B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d}-\frac{\sqrt [3]{a+i b} (-B+i A) \log (\cos (c+d x))}{4 d}+\frac{\sqrt [3]{a-i b} (B+i A) \log (\cos (c+d x))}{4 d}-\frac{1}{4} x \sqrt [3]{a-i b} (A-i B)-\frac{1}{4} x \sqrt [3]{a+i b} (A+i B)+\frac{3 B \sqrt [3]{a+b \tan (c+d x)}}{d} \]
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Rubi [A] time = 0.406038, antiderivative size = 377, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3528, 3539, 3537, 57, 617, 204, 31} \[ -\frac{\sqrt{3} \sqrt [3]{a-i b} (B+i A) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{2 d}+\frac{\sqrt{3} \sqrt [3]{a+i b} (-B+i A) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{2 d}+\frac{3 \sqrt [3]{a-i b} (B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d}-\frac{3 \sqrt [3]{a+i b} (-B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d}-\frac{\sqrt [3]{a+i b} (-B+i A) \log (\cos (c+d x))}{4 d}+\frac{\sqrt [3]{a-i b} (B+i A) \log (\cos (c+d x))}{4 d}-\frac{1}{4} x \sqrt [3]{a-i b} (A-i B)-\frac{1}{4} x \sqrt [3]{a+i b} (A+i B)+\frac{3 B \sqrt [3]{a+b \tan (c+d x)}}{d} \]
Antiderivative was successfully verified.
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Rule 3528
Rule 3539
Rule 3537
Rule 57
Rule 617
Rule 204
Rule 31
Rubi steps
\begin{align*} \int \sqrt [3]{a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx &=\frac{3 B \sqrt [3]{a+b \tan (c+d x)}}{d}+\int \frac{a A-b B+(A b+a B) \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx\\ &=\frac{3 B \sqrt [3]{a+b \tan (c+d x)}}{d}+\frac{1}{2} ((a-i b) (A-i B)) \int \frac{1+i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx+\frac{1}{2} ((a+i b) (A+i B)) \int \frac{1-i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx\\ &=\frac{3 B \sqrt [3]{a+b \tan (c+d x)}}{d}+\frac{(i (a-i b) (A-i B)) \operatorname{Subst}\left (\int \frac{1}{(-1+x) (a-i b x)^{2/3}} \, dx,x,i \tan (c+d x)\right )}{2 d}-\frac{((i a-b) (A+i B)) \operatorname{Subst}\left (\int \frac{1}{(-1+x) (a+i b x)^{2/3}} \, dx,x,-i \tan (c+d x)\right )}{2 d}\\ &=-\frac{1}{4} \sqrt [3]{a-i b} (A-i B) x-\frac{1}{4} \sqrt [3]{a+i b} (A+i B) x-\frac{\sqrt [3]{a+i b} (i A-B) \log (\cos (c+d x))}{4 d}+\frac{\sqrt [3]{a-i b} (i A+B) \log (\cos (c+d x))}{4 d}+\frac{3 B \sqrt [3]{a+b \tan (c+d x)}}{d}+\frac{\left (3 \sqrt [3]{a+i b} (i A-B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a+i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac{\left (3 (a+i b)^{2/3} (i A-B)\right ) \operatorname{Subst}\left (\int \frac{1}{(a+i b)^{2/3}+\sqrt [3]{a+i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac{\left (3 \sqrt [3]{a-i b} (i A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a-i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac{\left (3 (a-i b)^{2/3} (i A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{(a-i b)^{2/3}+\sqrt [3]{a-i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}\\ &=-\frac{1}{4} \sqrt [3]{a-i b} (A-i B) x-\frac{1}{4} \sqrt [3]{a+i b} (A+i B) x-\frac{\sqrt [3]{a+i b} (i A-B) \log (\cos (c+d x))}{4 d}+\frac{\sqrt [3]{a-i b} (i A+B) \log (\cos (c+d x))}{4 d}+\frac{3 \sqrt [3]{a-i b} (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac{3 \sqrt [3]{a+i b} (i A-B) \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac{3 B \sqrt [3]{a+b \tan (c+d x)}}{d}-\frac{\left (3 \sqrt [3]{a+i b} (i A-B)\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}\right )}{2 d}+\frac{\left (3 \sqrt [3]{a-i b} (i A+B)\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}\right )}{2 d}\\ &=-\frac{1}{4} \sqrt [3]{a-i b} (A-i B) x-\frac{1}{4} \sqrt [3]{a+i b} (A+i B) x-\frac{\sqrt{3} \sqrt [3]{a-i b} (i A+B) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )}{2 d}+\frac{\sqrt{3} \sqrt [3]{a+i b} (i A-B) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )}{2 d}-\frac{\sqrt [3]{a+i b} (i A-B) \log (\cos (c+d x))}{4 d}+\frac{\sqrt [3]{a-i b} (i A+B) \log (\cos (c+d x))}{4 d}+\frac{3 \sqrt [3]{a-i b} (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}-\frac{3 \sqrt [3]{a+i b} (i A-B) \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 d}+\frac{3 B \sqrt [3]{a+b \tan (c+d x)}}{d}\\ \end{align*}
Mathematica [A] time = 0.875988, size = 347, normalized size = 0.92 \[ \frac{i \left ((A-i B) \left (3 \sqrt [3]{a+b \tan (c+d x)}-\frac{1}{2} \sqrt [3]{a-i b} \left (2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt{3}}\right )-2 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )+\log \left (\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}+(a-i b)^{2/3}\right )\right )\right )-(A+i B) \left (3 \sqrt [3]{a+b \tan (c+d x)}-\frac{1}{2} \sqrt [3]{a+i b} \left (2 \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt{3}}\right )-2 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )+\log \left (\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}+(a+i b)^{2/3}\right )\right )\right )\right )}{2 d} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.079, size = 99, normalized size = 0.3 \begin{align*} 3\,{\frac{B\sqrt [3]{a+b\tan \left ( dx+c \right ) }}{d}}+{\frac{1}{2\,d}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{6}-2\,{{\it \_Z}}^{3}a+{a}^{2}+{b}^{2} \right ) }{\frac{ \left ( Ab+aB \right ){{\it \_R}}^{3}-{a}^{2}B-{b}^{2}B}{{{\it \_R}}^{5}-{{\it \_R}}^{2}a}\ln \left ( \sqrt [3]{a+b\tan \left ( dx+c \right ) }-{\it \_R} \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [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.
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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 ) \sqrt [3]{a + b \tan{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 13.5897, size = 666, normalized size = 1.77 \begin{align*} -\frac{1}{8} \,{\left (i \, \sqrt{3} + 1\right )} \left (\frac{8 i \, A^{3} a - 24 \, A^{2} B a - 24 i \, A B^{2} a + 8 \, B^{3} a - 8 \, A^{3} b - 24 i \, A^{2} B b + 24 \, A B^{2} b + 8 i \, B^{3} b}{d^{3}}\right )^{\frac{1}{3}} \log \left (d^{2}\right ) - \frac{1}{8} \,{\left (-i \, \sqrt{3} + 1\right )} \left (\frac{8 i \, A^{3} a - 24 \, A^{2} B a - 24 i \, A B^{2} a + 8 \, B^{3} a - 8 \, A^{3} b - 24 i \, A^{2} B b + 24 \, A B^{2} b + 8 i \, B^{3} b}{d^{3}}\right )^{\frac{1}{3}} \log \left (d^{2}\right ) - \frac{1}{8} \,{\left (i \, \sqrt{3} + 1\right )} \left (\frac{-8 i \, A^{3} a - 24 \, A^{2} B a + 24 i \, A B^{2} a + 8 \, B^{3} a - 8 \, A^{3} b + 24 i \, A^{2} B b + 24 \, A B^{2} b - 8 i \, B^{3} b}{d^{3}}\right )^{\frac{1}{3}} \log \left (d^{2}\right ) - \frac{1}{8} \,{\left (-i \, \sqrt{3} + 1\right )} \left (\frac{-8 i \, A^{3} a - 24 \, A^{2} B a + 24 i \, A B^{2} a + 8 \, B^{3} a - 8 \, A^{3} b + 24 i \, A^{2} B b + 24 \, A B^{2} b - 8 i \, B^{3} b}{d^{3}}\right )^{\frac{1}{3}} \log \left (d^{2}\right ) + \frac{1}{4} \, \left (\frac{8 i \, A^{3} a - 24 \, A^{2} B a - 24 i \, A B^{2} a + 8 \, B^{3} a - 8 \, A^{3} b - 24 i \, A^{2} B b + 24 \, A B^{2} b + 8 i \, B^{3} b}{d^{3}}\right )^{\frac{1}{3}} \log \left (i \,{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}} d^{2} +{\left (i \, a - b\right )}^{\frac{1}{3}} d^{2}\right ) + \frac{1}{4} \, \left (\frac{-8 i \, A^{3} a - 24 \, A^{2} B a + 24 i \, A B^{2} a + 8 \, B^{3} a - 8 \, A^{3} b + 24 i \, A^{2} B b + 24 \, A B^{2} b - 8 i \, B^{3} b}{d^{3}}\right )^{\frac{1}{3}} \log \left (-i \,{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}} d^{2} +{\left (-i \, a - b\right )}^{\frac{1}{3}} d^{2}\right ) + \frac{3 \,{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}} B}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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