Optimal. Leaf size=290 \[ -\frac{2 b^2 \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 f \left (c^2+d^2\right )}+\frac{4 (b c-a d)^3 \left (3 a c d+2 b c^2+5 b d^2\right )}{3 d^3 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{i (a-i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}}+\frac{i (a+i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}} \]
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Rubi [A] time = 1.00933, antiderivative size = 290, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {3565, 3635, 3630, 3539, 3537, 63, 208} \[ -\frac{2 b^2 \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 f \left (c^2+d^2\right )}+\frac{4 (b c-a d)^3 \left (3 a c d+2 b c^2+5 b d^2\right )}{3 d^3 f \left (c^2+d^2\right )^2 \sqrt{c+d \tan (e+f x)}}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}-\frac{i (a-i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (c-i d)^{5/2}}+\frac{i (a+i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (c+i d)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3565
Rule 3635
Rule 3630
Rule 3539
Rule 3537
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{5/2}} \, dx &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \int \frac{(a+b \tan (e+f x)) \left (\frac{1}{2} \left (4 b^3 c^2+3 a^3 c d-11 a b^2 c d+10 a^2 b d^2\right )+\frac{3}{2} 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}{2} b \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \tan ^2(e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx}{3 d \left (c^2+d^2\right )}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 (b c-a d)^3 \left (2 b c^2+3 a c d+5 b d^2\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{2 \int \frac{\frac{1}{2} \left (24 a^3 b c d^3-a^2 b^2 d^2 \left (17 c^2-19 d^2\right )+3 a^4 d^2 \left (c^2-d^2\right )-2 a b^3 c d \left (c^2+13 d^2\right )+2 b^4 \left (2 c^4+5 c^2 d^2\right )\right )+3 d^2 \left (a^2 c-b^2 c+2 a b d\right ) \left (2 a b c-a^2 d+b^2 d\right ) \tan (e+f x)-\frac{1}{2} b^2 \left (c^2+d^2\right ) \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \tan ^2(e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d^2 \left (c^2+d^2\right )^2}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 (b c-a d)^3 \left (2 b c^2+3 a c d+5 b d^2\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}-\frac{2 b^2 \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}+\frac{2 \int \frac{\frac{3}{2} d^2 \left (8 a^3 b c d-8 a b^3 c d+a^4 \left (c^2-d^2\right )-6 a^2 b^2 \left (c^2-d^2\right )+b^4 \left (c^2-d^2\right )\right )+3 d^2 \left (a^2 c-b^2 c+2 a b d\right ) \left (2 a b c-a^2 d+b^2 d\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 d^2 \left (c^2+d^2\right )^2}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 (b c-a d)^3 \left (2 b c^2+3 a c d+5 b d^2\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}-\frac{2 b^2 \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}+\frac{(a-i b)^4 \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c-i d)^2}+\frac{(a+i b)^4 \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (c+i d)^2}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 (b c-a d)^3 \left (2 b c^2+3 a c d+5 b d^2\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}-\frac{2 b^2 \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}+\frac{\left (i (a-i b)^4\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (c-i d)^2 f}-\frac{\left (i (a+i b)^4\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (c+i d)^2 f}\\ &=-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 (b c-a d)^3 \left (2 b c^2+3 a c d+5 b d^2\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}-\frac{2 b^2 \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}-\frac{(a-i b)^4 \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i c}{d}+\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(c-i d)^2 d f}-\frac{(a+i b)^4 \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i c}{d}-\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(c+i d)^2 d f}\\ &=-\frac{i (a-i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(c-i d)^{5/2} f}+\frac{i (a+i b)^4 \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(c+i d)^{5/2} f}-\frac{2 (b c-a d)^2 (a+b \tan (e+f x))^2}{3 d \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{4 (b c-a d)^3 \left (2 b c^2+3 a c d+5 b d^2\right )}{3 d^3 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}-\frac{2 b^2 \left (a d (2 b c-a d)-b^2 \left (4 c^2+3 d^2\right )\right ) \sqrt{c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}\\ \end{align*}
Mathematica [C] time = 3.45409, size = 368, normalized size = 1.27 \[ -\frac{-2 b^2 (c-i d) (c+i d) \left (9 a^2 d^2-20 a b c d+b^2 \left (8 c^2+d^2\right )\right )-12 a b d^2 \left (a^2-b^2\right ) (c+d \tan (e+f x)) \left (i (c+i d) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c+d \tan (e+f x)}{c-i d}\right )-(d+i c) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c+d \tan (e+f x)}{c+i d}\right )\right )+d^2 \left (-6 a^2 b^2 d-4 a^3 b c+a^4 d+4 a b^3 c+b^4 d\right ) \left ((d-i c) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{c+d \tan (e+f x)}{c-i d}\right )+(d+i c) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{c+d \tan (e+f x)}{c+i d}\right )\right )-6 b^2 d^2 (c-i d) (c+i d) (a+b \tan (e+f x))^2-12 b^2 d (c-i d) (c+i d) (2 b c-3 a d) (a+b \tan (e+f x))}{3 d^3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.126, size = 33851, normalized size = 116.7 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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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 \frac{\left (a + b \tan{\left (e + f x \right )}\right )^{4}}{\left (c + d \tan{\left (e + f x \right )}\right )^{\frac{5}{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{{\left (b \tan \left (f x + e\right ) + a\right )}^{4}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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