Optimal. Leaf size=473 \[ -\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac{d \left (5 a^2 C-3 a b B+3 A b^2+2 b^2 C\right ) (c+d \tan (e+f x))^{3/2}}{3 b^2 f \left (a^2+b^2\right )}-\frac{d \sqrt{c+d \tan (e+f x)} \left (-a^2 b (3 B d+5 c C)+5 a^3 C d-A b^2 (b c-a d)+a b^2 (B c+4 C d)-2 b^3 (B d+2 c C)\right )}{b^3 f \left (a^2+b^2\right )}+\frac{(b c-a d)^{3/2} \left (a^2 b^2 (2 B c-d (A+9 C))+3 a^3 b B d-5 a^4 C d-a b^3 (4 A c-7 B d-4 c C)-b^4 (5 A d+2 B c)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{7/2} f \left (a^2+b^2\right )^2}-\frac{(c-i d)^{5/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)^2}-\frac{(c+i d)^{5/2} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)^2} \]
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Rubi [A] time = 3.89577, antiderivative size = 473, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.17, Rules used = {3645, 3647, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac{d \left (5 a^2 C-3 a b B+3 A b^2+2 b^2 C\right ) (c+d \tan (e+f x))^{3/2}}{3 b^2 f \left (a^2+b^2\right )}-\frac{d \sqrt{c+d \tan (e+f x)} \left (-a^2 b (3 B d+5 c C)+5 a^3 C d-A b^2 (b c-a d)+a b^2 (B c+4 C d)-2 b^3 (B d+2 c C)\right )}{b^3 f \left (a^2+b^2\right )}+\frac{(b c-a d)^{3/2} \left (a^2 b^2 (2 B c-d (A+9 C))+3 a^3 b B d-5 a^4 C d-a b^3 (4 A c-7 B d-4 c C)-b^4 (5 A d+2 B c)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{7/2} f \left (a^2+b^2\right )^2}-\frac{(c-i d)^{5/2} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)^2}-\frac{(c+i d)^{5/2} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)^2} \]
Antiderivative was successfully verified.
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Rule 3645
Rule 3647
Rule 3653
Rule 3539
Rule 3537
Rule 63
Rule 208
Rule 3634
Rubi steps
\begin{align*} \int \frac{(c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^2} \, dx &=-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac{\int \frac{(c+d \tan (e+f x))^{3/2} \left (\frac{1}{2} \left (2 (b B-a C) \left (b c-\frac{5 a d}{2}\right )+2 A b \left (a c+\frac{5 b d}{2}\right )\right )-b ((A-C) (b c-a d)-B (a c+b d)) \tan (e+f x)+\frac{1}{2} \left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx}{b \left (a^2+b^2\right )}\\ &=\frac{\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac{2 \int \frac{\sqrt{c+d \tan (e+f x)} \left (-\frac{3}{4} \left (a \left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d^2-b c ((b B-a C) (2 b c-5 a d)+A b (2 a c+5 b d))\right )+\frac{3}{2} b^2 \left (2 a A c d-2 a c C d-A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )+b \left (c^2 C+2 B c d-C d^2\right )\right ) \tan (e+f x)-\frac{3}{4} d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \tan ^2(e+f x)\right )}{a+b \tan (e+f x)} \, dx}{3 b^2 \left (a^2+b^2\right )}\\ &=-\frac{d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt{c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac{\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac{4 \int \frac{\frac{3}{8} \left (5 a^4 C d^3+b^4 c^2 (2 B c+5 A d)-a^3 b d^2 (10 c C+3 B d)+a^2 b^2 d \left (5 c^2 C+4 B c d+(A+4 C) d^2\right )+a b^3 \left (2 A c^3-2 c^3 C-5 B c^2 d-4 A c d^2-6 c C d^2-2 B d^3\right )\right )+\frac{3}{4} b^3 \left (a A d \left (3 c^2-d^2\right )-A b \left (c^3-3 c d^2\right )+b \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3\right )-a \left (C d \left (3 c^2-d^2\right )-B \left (c^3-3 c d^2\right )\right )\right ) \tan (e+f x)+\frac{3}{8} d \left (5 a^4 C d^2-a^3 b d (10 c C+3 B d)+a b^3 \left (B c^2-12 c C d-4 B d^2\right )+2 b^4 \left (3 c^2 C+3 B c d-C d^2\right )+a^2 b^2 \left (5 c^2 C+4 B c d+4 C d^2\right )+A b^2 \left (2 a b c d+a^2 d^2-b^2 \left (c^2-2 d^2\right )\right )\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{3 b^3 \left (a^2+b^2\right )}\\ &=-\frac{d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt{c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac{\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac{4 \int \frac{-\frac{3}{4} b^3 \left (b^2 \left (A c^3-c^3 C-3 B c^2 d-3 A c d^2+3 c C d^2+B d^3\right )+a^2 \left (c^3 C+3 B c^2 d-3 c C d^2-B d^3-A \left (c^3-3 c d^2\right )\right )-2 a b \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right )-\frac{3}{4} b^3 \left (2 a b \left (A c^3-c^3 C-3 B c^2 d-3 A c d^2+3 c C d^2+B d^3\right )-a^2 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )+b^2 \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 b^3 \left (a^2+b^2\right )^2}-\frac{\left ((b c-a d)^2 \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right )\right ) \int \frac{1+\tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{2 b^3 \left (a^2+b^2\right )^2}\\ &=-\frac{d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt{c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac{\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac{\left ((A-i B-C) (c-i d)^3\right ) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a-i b)^2}+\frac{\left ((A+i B-C) (c+i d)^3\right ) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a+i b)^2}-\frac{\left ((b c-a d)^2 \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 b^3 \left (a^2+b^2\right )^2 f}\\ &=-\frac{d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt{c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac{\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac{\left ((i A+B-i C) (c-i d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (a-i b)^2 f}-\frac{\left (i (A+i B-C) (c+i d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (a+i b)^2 f}-\frac{\left ((b c-a d)^2 \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{b^3 \left (a^2+b^2\right )^2 d f}\\ &=\frac{(b c-a d)^{3/2} \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{7/2} \left (a^2+b^2\right )^2 f}-\frac{d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt{c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac{\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\frac{\left ((A-i B-C) (c-i d)^3\right ) \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 )}{(a-i b)^2 d f}-\frac{\left ((A+i B-C) (c+i d)^3\right ) \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 )}{(a+i b)^2 d f}\\ &=-\frac{(i A+B-i C) (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(a-i b)^2 f}-\frac{(B-i (A-C)) (c+i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(a+i b)^2 f}+\frac{(b c-a d)^{3/2} \left (3 a^3 b B d-5 a^4 C d-b^4 (2 B c+5 A d)-a b^3 (4 A c-4 c C-7 B d)+a^2 b^2 (2 B c-(A+9 C) d)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{7/2} \left (a^2+b^2\right )^2 f}-\frac{d \left (5 a^3 C d-A b^2 (b c-a d)-2 b^3 (2 c C+B d)-a^2 b (5 c C+3 B d)+a b^2 (B c+4 C d)\right ) \sqrt{c+d \tan (e+f x)}}{b^3 \left (a^2+b^2\right ) f}+\frac{\left (3 A b^2-3 a b B+5 a^2 C+2 b^2 C\right ) d (c+d \tan (e+f x))^{3/2}}{3 b^2 \left (a^2+b^2\right ) f}-\frac{\left (A b^2-a (b B-a C)\right ) (c+d \tan (e+f x))^{5/2}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}\\ \end{align*}
Mathematica [B] time = 6.54989, size = 6112, normalized size = 12.92 \[ \text{Result too large to show} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.258, size = 14119, normalized size = 29.9 \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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )}{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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