Optimal. Leaf size=212 \[ \frac{\left (a^2-b^2 \sqrt{-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \text{Hypergeometric2F1}\left (1,m+1,m+2,-\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{2 f (m+1)}+\frac{\left (a^2+b^2 \sqrt{-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{2 f (m+1)}+\frac{4 a b (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m \text{Hypergeometric2F1}\left (1,\frac{1}{4} (2 m+3),\frac{1}{4} (2 m+7),-\tan ^2(e+f x)\right )}{c f (2 m+3)} \]
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Rubi [A] time = 0.712759, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {3670, 15, 1831, 364, 1286} \[ \frac{\left (a^2-b^2 \sqrt{-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \, _2F_1\left (1,m+1;m+2;-\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{2 f (m+1)}+\frac{\left (a^2+b^2 \sqrt{-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \, _2F_1\left (1,m+1;m+2;\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{2 f (m+1)}+\frac{4 a b (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m \, _2F_1\left (1,\frac{1}{4} (2 m+3);\frac{1}{4} (2 m+7);-\tan ^2(e+f x)\right )}{c f (2 m+3)} \]
Antiderivative was successfully verified.
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Rule 3670
Rule 15
Rule 1831
Rule 364
Rule 1286
Rubi steps
\begin{align*} \int (d \tan (e+f x))^m \left (a+b \sqrt{c \tan (e+f x)}\right )^2 \, dx &=\frac{c \operatorname{Subst}\left (\int \frac{\left (a+b \sqrt{x}\right )^2 \left (\frac{d x}{c}\right )^m}{c^2+x^2} \, dx,x,c \tan (e+f x)\right )}{f}\\ &=\frac{(2 c) \operatorname{Subst}\left (\int \frac{x \left (\frac{d x^2}{c}\right )^m (a+b x)^2}{c^2+x^4} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{f}\\ &=\frac{\left (2 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+2 m} (a+b x)^2}{c^2+x^4} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{f}\\ &=\frac{\left (2 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \left (\frac{2 a b x^{2+2 m}}{c^2+x^4}+\frac{x^{1+2 m} \left (a^2+b^2 x^2\right )}{c^2+x^4}\right ) \, dx,x,\sqrt{c \tan (e+f x)}\right )}{f}\\ &=\frac{\left (2 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+2 m} \left (a^2+b^2 x^2\right )}{c^2+x^4} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{f}+\frac{\left (4 a b c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{x^{2+2 m}}{c^2+x^4} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{f}\\ &=\frac{4 a b \, _2F_1\left (1,\frac{1}{4} (3+2 m);\frac{1}{4} (7+2 m);-\tan ^2(e+f x)\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c f (3+2 m)}+\frac{\left (c \left (b^2-\frac{a^2}{\sqrt{-c^2}}\right ) (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+2 m}}{\sqrt{-c^2}+x^2} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{f}-\frac{\left (c \left (b^2+\frac{a^2}{\sqrt{-c^2}}\right ) (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{x^{1+2 m}}{\sqrt{-c^2}-x^2} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{f}\\ &=\frac{\left (a^2-b^2 \sqrt{-c^2}\right ) \, _2F_1\left (1,1+m;2+m;-\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 f (1+m)}+\frac{\left (a^2+b^2 \sqrt{-c^2}\right ) \, _2F_1\left (1,1+m;2+m;\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 f (1+m)}+\frac{4 a b \, _2F_1\left (1,\frac{1}{4} (3+2 m);\frac{1}{4} (7+2 m);-\tan ^2(e+f x)\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c f (3+2 m)}\\ \end{align*}
Mathematica [A] time = 1.28323, size = 151, normalized size = 0.71 \[ \frac{\tan (e+f x) (d \tan (e+f x))^m \left (\frac{a^2 \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(e+f x)\right )}{m+1}+b \left (\frac{4 a \sqrt{c \tan (e+f x)} \text{Hypergeometric2F1}\left (1,\frac{1}{4} (2 m+3),\frac{1}{4} (2 m+7),-\tan ^2(e+f x)\right )}{2 m+3}+\frac{b c \tan (e+f x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(e+f x)\right )}{m+2}\right )\right )}{f} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.257, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sqrt{c\tan \left ( fx+e \right ) } \right ) ^{2} \left ( d\tan \left ( fx+e \right ) \right ) ^{m}\, dx \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 (\sqrt{c \tan \left (f x + e\right )} b + a\right )}^{2} \left (d \tan \left (f x + e\right )\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (2 \, \sqrt{c \tan \left (f x + e\right )} \left (d \tan \left (f x + e\right )\right )^{m} a b +{\left (b^{2} c \tan \left (f x + e\right ) + a^{2}\right )} \left (d \tan \left (f x + e\right )\right )^{m}, x\right ) \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 (d \tan{\left (e + f x \right )}\right )^{m} \left (a + b \sqrt{c \tan{\left (e + f x \right )}}\right )^{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{\left (\sqrt{c \tan \left (f x + e\right )} b + a\right )}^{2} \left (d \tan \left (f x + e\right )\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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