Optimal. Leaf size=460 \[ -\frac{b \left (a^2-b^2 \sqrt{-c^2}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m \text{Hypergeometric2F1}\left (1,\frac{1}{2} (2 m+3),\frac{1}{2} (2 m+5),-\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{c f (2 m+3) \left (a^4+b^4 c^2\right )}-\frac{b \left (a^2+b^2 \sqrt{-c^2}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m \text{Hypergeometric2F1}\left (1,\frac{1}{2} (2 m+3),\frac{1}{2} (2 m+5),\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{c f (2 m+3) \left (a^4+b^4 c^2\right )}+\frac{a \left (a^2-b^2 \sqrt{-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \text{Hypergeometric2F1}\left (1,m+1,m+2,-\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{2 f (m+1) \left (a^4+b^4 c^2\right )}+\frac{a \left (a^2+b^2 \sqrt{-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{2 f (m+1) \left (a^4+b^4 c^2\right )}+\frac{b^4 c^2 \tan (e+f x) (d \tan (e+f x))^m \text{Hypergeometric2F1}\left (1,2 (m+1),2 m+3,-\frac{b \sqrt{c \tan (e+f x)}}{a}\right )}{a f (m+1) \left (a^4+b^4 c^2\right )} \]
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Rubi [A] time = 1.28252, antiderivative size = 460, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {3670, 15, 6725, 64, 1831, 1286, 364} \[ -\frac{b \left (a^2-b^2 \sqrt{-c^2}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m \, _2F_1\left (1,\frac{1}{2} (2 m+3);\frac{1}{2} (2 m+5);-\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{c f (2 m+3) \left (a^4+b^4 c^2\right )}-\frac{b \left (a^2+b^2 \sqrt{-c^2}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m \, _2F_1\left (1,\frac{1}{2} (2 m+3);\frac{1}{2} (2 m+5);\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{c f (2 m+3) \left (a^4+b^4 c^2\right )}+\frac{a \left (a^2-b^2 \sqrt{-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \, _2F_1\left (1,m+1;m+2;-\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{2 f (m+1) \left (a^4+b^4 c^2\right )}+\frac{a \left (a^2+b^2 \sqrt{-c^2}\right ) \tan (e+f x) (d \tan (e+f x))^m \, _2F_1\left (1,m+1;m+2;\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right )}{2 f (m+1) \left (a^4+b^4 c^2\right )}+\frac{b^4 c^2 \tan (e+f x) (d \tan (e+f x))^m \, _2F_1\left (1,2 (m+1);2 m+3;-\frac{b \sqrt{c \tan (e+f x)}}{a}\right )}{a f (m+1) \left (a^4+b^4 c^2\right )} \]
Antiderivative was successfully verified.
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Rule 3670
Rule 15
Rule 6725
Rule 64
Rule 1831
Rule 1286
Rule 364
Rubi steps
\begin{align*} \int \frac{(d \tan (e+f x))^m}{a+b \sqrt{c \tan (e+f x)}} \, dx &=\frac{c \operatorname{Subst}\left (\int \frac{\left (\frac{d x}{c}\right )^m}{\left (a+b \sqrt{x}\right ) \left (c^2+x^2\right )} \, 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) \left (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}}{(a+b x) \left (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 \left (\frac{b^4 x^{1+2 m}}{\left (a^4+b^4 c^2\right ) (a+b x)}+\frac{x^{1+2 m} \left (a^3-a^2 b x+a b^2 x^2-b^3 x^3\right )}{\left (a^4+b^4 c^2\right ) \left (c^2+x^4\right )}\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^3-a^2 b x+a b^2 x^2-b^3 x^3\right )}{c^2+x^4} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}+\frac{\left (2 b^4 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} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}\\ &=\frac{b^4 c^2 \, _2F_1\left (1,2 (1+m);3+2 m;-\frac{b \sqrt{c \tan (e+f x)}}{a}\right ) \tan (e+f x) (d \tan (e+f x))^m}{a \left (a^4+b^4 c^2\right ) f (1+m)}+\frac{\left (2 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \left (\frac{x^{1+2 m} \left (a^3+a b^2 x^2\right )}{c^2+x^4}+\frac{x^{2+2 m} \left (-a^2 b-b^3 x^2\right )}{c^2+x^4}\right ) \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}\\ &=\frac{b^4 c^2 \, _2F_1\left (1,2 (1+m);3+2 m;-\frac{b \sqrt{c \tan (e+f x)}}{a}\right ) \tan (e+f x) (d \tan (e+f x))^m}{a \left (a^4+b^4 c^2\right ) f (1+m)}+\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^3+a b^2 x^2\right )}{c^2+x^4} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\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^{2+2 m} \left (-a^2 b-b^3 x^2\right )}{c^2+x^4} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}\\ &=\frac{b^4 c^2 \, _2F_1\left (1,2 (1+m);3+2 m;-\frac{b \sqrt{c \tan (e+f x)}}{a}\right ) \tan (e+f x) (d \tan (e+f x))^m}{a \left (a^4+b^4 c^2\right ) f (1+m)}+\frac{\left (a 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 )}{\left (a^4+b^4 c^2\right ) f}-\frac{\left (b 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^{2+2 m}}{\sqrt{-c^2}+x^2} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}-\frac{\left (a 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 )}{\left (a^4+b^4 c^2\right ) f}+\frac{\left (b 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^{2+2 m}}{\sqrt{-c^2}-x^2} \, dx,x,\sqrt{c \tan (e+f x)}\right )}{\left (a^4+b^4 c^2\right ) f}\\ &=\frac{a \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 \left (a^4+b^4 c^2\right ) f (1+m)}+\frac{a \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 \left (a^4+b^4 c^2\right ) f (1+m)}+\frac{b^4 c^2 \, _2F_1\left (1,2 (1+m);3+2 m;-\frac{b \sqrt{c \tan (e+f x)}}{a}\right ) \tan (e+f x) (d \tan (e+f x))^m}{a \left (a^4+b^4 c^2\right ) f (1+m)}-\frac{b \left (a^2-b^2 \sqrt{-c^2}\right ) \, _2F_1\left (1,\frac{1}{2} (3+2 m);\frac{1}{2} (5+2 m);-\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c \left (a^4+b^4 c^2\right ) f (3+2 m)}-\frac{b \left (a^2+b^2 \sqrt{-c^2}\right ) \, _2F_1\left (1,\frac{1}{2} (3+2 m);\frac{1}{2} (5+2 m);\frac{c \tan (e+f x)}{\sqrt{-c^2}}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c \left (a^4+b^4 c^2\right ) f (3+2 m)}\\ \end{align*}
Mathematica [A] time = 6.27851, size = 385, normalized size = 0.84 \[ \frac{2 c (c \tan (e+f x))^{-m} (d \tan (e+f x))^m \left (\frac{a^3 (c \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(e+f x)\right )}{2 c^2 (m+1) \left (a^4+b^4 c^2\right )}+\frac{a b^2 (c \tan (e+f x))^{m+2} \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(e+f x)\right )}{2 c^2 (m+2) \left (a^4+b^4 c^2\right )}-\frac{a^2 b (c \tan (e+f x))^{\frac{1}{2} (2 m+3)} \text{Hypergeometric2F1}\left (1,\frac{1}{4} (2 m+3),\frac{1}{4} (2 m+7),-\tan ^2(e+f x)\right )}{c^2 (2 m+3) \left (a^4+b^4 c^2\right )}-\frac{b^3 (c \tan (e+f x))^{\frac{1}{2} (2 m+5)} \text{Hypergeometric2F1}\left (1,\frac{1}{4} (2 m+5),\frac{1}{4} (2 m+9),-\tan ^2(e+f x)\right )}{c^2 (2 m+5) \left (a^4+b^4 c^2\right )}+\frac{b^4 (c \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,2 (m+1),2 m+3,-\frac{b \sqrt{c \tan (e+f x)}}{a}\right )}{2 a (m+1) \left (a^4+b^4 c^2\right )}\right )}{f} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.201, size = 0, normalized size = 0. \begin{align*} \int{ \left ( d\tan \left ( fx+e \right ) \right ) ^{m} \left ( a+b\sqrt{c\tan \left ( fx+e \right ) } \right ) ^{-1}}\, 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 \frac{\left (d \tan \left (f x + e\right )\right )^{m}}{\sqrt{c \tan \left (f x + e\right )} b + a}\,{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 (\frac{\sqrt{c \tan \left (f x + e\right )} \left (d \tan \left (f x + e\right )\right )^{m} b - \left (d \tan \left (f x + e\right )\right )^{m} a}{b^{2} c \tan \left (f x + e\right ) - a^{2}}, 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 \frac{\left (d \tan{\left (e + f x \right )}\right )^{m}}{a + b \sqrt{c \tan{\left (e + f x \right )}}}\, 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{\left (d \tan \left (f x + e\right )\right )^{m}}{\sqrt{c \tan \left (f x + e\right )} b + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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