Optimal. Leaf size=154 \[ \frac{\tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e (a-b+c)^{3/2}}-\frac{c (2 a-b) \tan ^2(d+e x)+a (b-2 c)}{e (a-b+c) \left (b^2-4 a c\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \]
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Rubi [A] time = 0.284776, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3700, 1251, 822, 12, 724, 206} \[ \frac{\tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e (a-b+c)^{3/2}}-\frac{c (2 a-b) \tan ^2(d+e x)+a (b-2 c)}{e (a-b+c) \left (b^2-4 a c\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \]
Antiderivative was successfully verified.
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Rule 3700
Rule 1251
Rule 822
Rule 12
Rule 724
Rule 206
Rubi steps
\begin{align*} \int \frac{\tan ^3(d+e x)}{\left (a+b \tan ^2(d+e x)+c \tan ^4(d+e x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^3}{\left (1+x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}} \, dx,x,\tan (d+e x)\right )}{e}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x}{(1+x) \left (a+b x+c x^2\right )^{3/2}} \, dx,x,\tan ^2(d+e x)\right )}{2 e}\\ &=-\frac{a (b-2 c)+(2 a-b) c \tan ^2(d+e x)}{(a-b+c) \left (b^2-4 a c\right ) e \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac{\operatorname{Subst}\left (\int \frac{b^2-4 a c}{2 (1+x) \sqrt{a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{(a-b+c) \left (b^2-4 a c\right ) e}\\ &=-\frac{a (b-2 c)+(2 a-b) c \tan ^2(d+e x)}{(a-b+c) \left (b^2-4 a c\right ) e \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{2 (a-b+c) e}\\ &=-\frac{a (b-2 c)+(2 a-b) c \tan ^2(d+e x)}{(a-b+c) \left (b^2-4 a c\right ) e \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{4 a-4 b+4 c-x^2} \, dx,x,\frac{2 a-b-(-b+2 c) \tan ^2(d+e x)}{\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{(a-b+c) e}\\ &=\frac{\tanh ^{-1}\left (\frac{2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 (a-b+c)^{3/2} e}-\frac{a (b-2 c)+(2 a-b) c \tan ^2(d+e x)}{(a-b+c) \left (b^2-4 a c\right ) e \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\\ \end{align*}
Mathematica [A] time = 2.83612, size = 155, normalized size = 1.01 \[ \frac{\frac{2 c (2 a-b) \tan ^2(d+e x)+2 a (b-2 c)}{(a-b+c) \left (4 a c-b^2\right ) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}+\frac{\tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{(a-b+c)^{3/2}}}{2 e} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.159, size = 508, normalized size = 3.3 \begin{align*} 2\,{\frac{c \left ( \tan \left ( ex+d \right ) \right ) ^{2}}{e\sqrt{a+b \left ( \tan \left ( ex+d \right ) \right ) ^{2}+c \left ( \tan \left ( ex+d \right ) \right ) ^{4}} \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{b}{e \left ( 4\,ac-{b}^{2} \right ) }{\frac{1}{\sqrt{a+b \left ( \tan \left ( ex+d \right ) \right ) ^{2}+c \left ( \tan \left ( ex+d \right ) \right ) ^{4}}}}}-2\,{\frac{c}{e \left ( \sqrt{-4\,ac+{b}^{2}}-b+2\,c \right ) \left ( b-2\,c+\sqrt{-4\,ac+{b}^{2}} \right ) \sqrt{a-b+c}}\ln \left ({\frac{2\,a-2\,b+2\,c+ \left ( b-2\,c \right ) \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ) +2\,\sqrt{a-b+c}\sqrt{ \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ) ^{2}c+ \left ( b-2\,c \right ) \left ( 1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2} \right ) +a-b+c}}{1+ \left ( \tan \left ( ex+d \right ) \right ) ^{2}}} \right ) }+2\,{\frac{c}{e \left ( \sqrt{-4\,ac+{b}^{2}}-b+2\,c \right ) \left ( -4\,ac+{b}^{2} \right ) }\sqrt{ \left ( \left ( \tan \left ( ex+d \right ) \right ) ^{2}-1/2\,{\frac{-b+\sqrt{-4\,ac+{b}^{2}}}{c}} \right ) ^{2}c+\sqrt{-4\,ac+{b}^{2}} \left ( \left ( \tan \left ( ex+d \right ) \right ) ^{2}-1/2\,{\frac{-b+\sqrt{-4\,ac+{b}^{2}}}{c}} \right ) } \left ( \left ( \tan \left ( ex+d \right ) \right ) ^{2}-1/2\,{\frac{\sqrt{-4\,ac+{b}^{2}}}{c}}+1/2\,{\frac{b}{c}} \right ) ^{-1}}-2\,{\frac{c}{e \left ( b-2\,c+\sqrt{-4\,ac+{b}^{2}} \right ) \left ( -4\,ac+{b}^{2} \right ) }\sqrt{ \left ( \left ( \tan \left ( ex+d \right ) \right ) ^{2}+1/2\,{\frac{b+\sqrt{-4\,ac+{b}^{2}}}{c}} \right ) ^{2}c-\sqrt{-4\,ac+{b}^{2}} \left ( \left ( \tan \left ( ex+d \right ) \right ) ^{2}+1/2\,{\frac{b+\sqrt{-4\,ac+{b}^{2}}}{c}} \right ) } \left ( \left ( \tan \left ( ex+d \right ) \right ) ^{2}+1/2\,{\frac{\sqrt{-4\,ac+{b}^{2}}}{c}}+1/2\,{\frac{b}{c}} \right ) ^{-1}} \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 \frac{\tan \left (e x + d\right )^{3}}{{\left (c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 5.53981, size = 2384, normalized size = 15.48 \begin{align*} \text{result too large to display} \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 \frac{\tan ^{3}{\left (d + e x \right )}}{\left (a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (e x + d\right )^{3}}{{\left (c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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