Optimal. Leaf size=337 \[ \frac{\sqrt{b} \left (15 a^2 d^2+10 a b c d+b^2 \left (-\left (c^2+8 d^2\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{4 d^{3/2} f}+\frac{b^2 \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac{b (b c-9 a d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}-\frac{i (a-i b)^{5/2} \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{i (a+i b)^{5/2} \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f} \]
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Rubi [A] time = 3.94853, antiderivative size = 337, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {3566, 3647, 3655, 6725, 63, 217, 206, 93, 208} \[ \frac{\sqrt{b} \left (15 a^2 d^2+10 a b c d+b^2 \left (-\left (c^2+8 d^2\right )\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{4 d^{3/2} f}+\frac{b^2 \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac{b (b c-9 a d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}-\frac{i (a-i b)^{5/2} \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{i (a+i b)^{5/2} \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f} \]
Antiderivative was successfully verified.
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Rule 3566
Rule 3647
Rule 3655
Rule 6725
Rule 63
Rule 217
Rule 206
Rule 93
Rule 208
Rubi steps
\begin{align*} \int (a+b \tan (e+f x))^{5/2} \sqrt{c+d \tan (e+f x)} \, dx &=\frac{b^2 \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac{\int \frac{\sqrt{c+d \tan (e+f x)} \left (\frac{1}{2} \left (-b^3 c+4 a^3 d-3 a b^2 d\right )+2 b \left (3 a^2-b^2\right ) d \tan (e+f x)-\frac{1}{2} b^2 (b c-9 a d) \tan ^2(e+f x)\right )}{\sqrt{a+b \tan (e+f x)}} \, dx}{2 d}\\ &=-\frac{b (b c-9 a d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}+\frac{b^2 \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac{\int \frac{-\frac{1}{4} b \left (b^3 c^2-8 a^3 c d+14 a b^2 c d+9 a^2 b d^2\right )+2 b d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) \tan (e+f x)+\frac{1}{4} b^2 \left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right ) \tan ^2(e+f x)}{\sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}} \, dx}{2 b d}\\ &=-\frac{b (b c-9 a d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}+\frac{b^2 \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{4} b \left (b^3 c^2-8 a^3 c d+14 a b^2 c d+9 a^2 b d^2\right )+2 b d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) x+\frac{1}{4} b^2 \left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right ) x^2}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 b d f}\\ &=-\frac{b (b c-9 a d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}+\frac{b^2 \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac{\operatorname{Subst}\left (\int \left (\frac{b^2 \left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right )}{4 \sqrt{a+b x} \sqrt{c+d x}}+\frac{2 \left (b d \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )+b d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) x\right )}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{2 b d f}\\ &=-\frac{b (b c-9 a d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}+\frac{b^2 \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac{\operatorname{Subst}\left (\int \frac{b d \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )+b d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) x}{\sqrt{a+b x} \sqrt{c+d x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{b d f}+\frac{\left (b \left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{8 d f}\\ &=-\frac{b (b c-9 a d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}+\frac{b^2 \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}+\frac{\operatorname{Subst}\left (\int \left (\frac{-b d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right )+i b d \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )}{2 (i-x) \sqrt{a+b x} \sqrt{c+d x}}+\frac{b d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right )+i b d \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )}{2 (i+x) \sqrt{a+b x} \sqrt{c+d x}}\right ) \, dx,x,\tan (e+f x)\right )}{b d f}+\frac{\left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b \tan (e+f x)}\right )}{4 d f}\\ &=-\frac{b (b c-9 a d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}+\frac{b^2 \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac{\left ((i a+b)^3 (c-i d)\right ) \operatorname{Subst}\left (\int \frac{1}{(i+x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac{\left (-b d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right )+i b d \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )\right ) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 b d f}+\frac{\left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{4 d f}\\ &=\frac{\sqrt{b} \left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{4 d^{3/2} f}-\frac{b (b c-9 a d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}+\frac{b^2 \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}-\frac{\left ((i a+b)^3 (c-i d)\right ) \operatorname{Subst}\left (\int \frac{1}{-a+i b-(-c+i d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{\left (-b d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right )+i b d \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+i b-(c+i d) x^2} \, dx,x,\frac{\sqrt{a+b \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}}\right )}{b d f}\\ &=-\frac{i (a-i b)^{5/2} \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c-i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a-i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{i (a+i b)^{5/2} \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+i d} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+i b} \sqrt{c+d \tan (e+f x)}}\right )}{f}+\frac{\sqrt{b} \left (10 a b c d+15 a^2 d^2-b^2 \left (c^2+8 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c+d \tan (e+f x)}}\right )}{4 d^{3/2} f}-\frac{b (b c-9 a d) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 d f}+\frac{b^2 \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}{2 d f}\\ \end{align*}
Mathematica [A] time = 5.95575, size = 565, normalized size = 1.68 \[ \frac{-\frac{b^{5/2} \sqrt{c-\frac{a d}{b}} \left (-15 a^2 d^2-10 a b c d+b^2 \left (c^2+8 d^2\right )\right ) \sqrt{\frac{b (c+d \tan (e+f x))}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \tan (e+f x)}}{\sqrt{b} \sqrt{c-\frac{a d}{b}}}\right )}{\sqrt{d} \sqrt{c+d \tan (e+f x)}}+\frac{4 b d \left (b \left (3 a^2 b c+a^3 d-3 a b^2 d-b^3 c\right )+\sqrt{-b^2} \left (-3 a^2 b d+a^3 c-3 a b^2 c+b^3 d\right )\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{b d}{\sqrt{-b^2}}+c} \sqrt{a+b \tan (e+f x)}}{\sqrt{\sqrt{-b^2}-a} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{\sqrt{-b^2}-a} \sqrt{\frac{b d}{\sqrt{-b^2}}+c}}-\frac{4 b d \left (b \left (3 a^2 b c+a^3 d-3 a b^2 d-b^3 c\right )-\sqrt{-b^2} \left (-3 a^2 b d+a^3 c-3 a b^2 c+b^3 d\right )\right ) \tan ^{-1}\left (\frac{\sqrt{-\frac{\sqrt{-b^2} d+b c}{b}} \sqrt{a+b \tan (e+f x)}}{\sqrt{a+\sqrt{-b^2}} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a+\sqrt{-b^2}} \sqrt{-\frac{\sqrt{-b^2} d+b c}{b}}}+2 b^4 \sqrt{a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}+b^3 (9 a d-b c) \sqrt{a+b \tan (e+f x)} \sqrt{c+d \tan (e+f x)}}{4 b^2 d f} \]
Antiderivative was successfully verified.
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Maple [F] time = 180., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c+d\tan \left ( fx+e \right ) } \left ( a+b\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}\, 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 (b \tan \left (f x + e\right ) + a\right )}^{\frac{5}{2}} \sqrt{d \tan \left (f x + e\right ) + c}\,{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(-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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Giac [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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