Optimal. Leaf size=365 \[ -\frac{2 b^{3/2} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{f \left (a^2+b^2\right ) (b c-a d)^{5/2}}+\frac{2 \left (b \left (c^2 d^2 (3 A-C)+A d^4-2 B c^3 d+c^4 C\right )-a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )\right )}{f \left (c^2+d^2\right )^2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}+\frac{2 \left (A d^2-B c d+c^2 C\right )}{3 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^{3/2}}+\frac{(A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (b+i a) (c-i d)^{5/2}}+\frac{(i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b) (c+i d)^{5/2}} \]
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Rubi [A] time = 2.46572, antiderivative size = 365, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.149, Rules used = {3649, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac{2 b^{3/2} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{f \left (a^2+b^2\right ) (b c-a d)^{5/2}}+\frac{2 \left (b \left (c^2 d^2 (3 A-C)+A d^4-2 B c^3 d+c^4 C\right )-a d^2 \left (2 c d (A-C)-B \left (c^2-d^2\right )\right )\right )}{f \left (c^2+d^2\right )^2 (b c-a d)^2 \sqrt{c+d \tan (e+f x)}}+\frac{2 \left (A d^2-B c d+c^2 C\right )}{3 f \left (c^2+d^2\right ) (b c-a d) (c+d \tan (e+f x))^{3/2}}+\frac{(A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (b+i a) (c-i d)^{5/2}}+\frac{(i A-B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b) (c+i d)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3649
Rule 3653
Rule 3539
Rule 3537
Rule 63
Rule 208
Rule 3634
Rubi steps
\begin{align*} \int \frac{A+B \tan (e+f x)+C \tan ^2(e+f x)}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}} \, dx &=\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \int \frac{-\frac{3}{2} \left (a A c d-a d (c C-B d)-A b \left (c^2+d^2\right )\right )+\frac{3}{2} (b c-a d) (B c-(A-C) d) \tan (e+f x)+\frac{3}{2} b \left (c^2 C-B c d+A d^2\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx}{3 (b c-a d) \left (c^2+d^2\right )}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{4 \int \frac{-\frac{3}{4} \left (A \left (2 a b c^3 d-a^2 d^2 \left (c^2-d^2\right )-b^2 \left (c^2+d^2\right )^2\right )+a d \left (a d \left (c^2 C-2 B c d-C d^2\right )-b \left (2 c^3 C-3 B c^2 d-B d^3\right )\right )\right )-\frac{3}{4} (b c-a d)^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right ) \tan (e+f x)+\frac{3}{4} b \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^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 c-a d)^2 \left (c^2+d^2\right )^2}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{\left (b^2 \left (A b^2-a (b B-a C)\right )\right ) \int \frac{1+\tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{\left (a^2+b^2\right ) (b c-a d)^2}+\frac{4 \int \frac{-\frac{3}{4} (b c-a d)^2 \left (a \left (c^2 C-2 B c d-C d^2-A \left (c^2-d^2\right )\right )+b \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )-\frac{3}{4} (b c-a d)^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)}{\sqrt{c+d \tan (e+f x)}} \, dx}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right )^2}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{(A-i B-C) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a-i b) (c-i d)^2}+\frac{(A+i B-C) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a+i b) (c+i d)^2}+\frac{\left (b^2 \left (A b^2-a (b B-a C)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{\left (a^2+b^2\right ) (b c-a d)^2 f}\\ &=\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}+\frac{(i A+B-i C) \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) (c-i d)^2 f}-\frac{(i (A+i B-C)) \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) (c+i d)^2 f}+\frac{\left (2 b^2 \left (A b^2-a (b B-a C)\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 )}{\left (a^2+b^2\right ) d (b c-a d)^2 f}\\ &=-\frac{2 b^{3/2} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{\left (a^2+b^2\right ) (b c-a d)^{5/2} f}+\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}-\frac{(A+i B-C) \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) (c+i d)^2 d f}+\frac{(A-i B-C) \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) d (i c+d)^2 f}\\ &=\frac{(A-i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(i a+b) (c-i d)^{5/2} f}-\frac{(A+i B-C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(i a-b) (c+i d)^{5/2} f}-\frac{2 b^{3/2} \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{\left (a^2+b^2\right ) (b c-a d)^{5/2} f}+\frac{2 \left (c^2 C-B c d+A d^2\right )}{3 (b c-a d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 \left (b \left (c^4 C-2 B c^3 d+c^2 (3 A-C) d^2+A d^4\right )-a d^2 \left (2 c (A-C) d-B \left (c^2-d^2\right )\right )\right )}{(b c-a d)^2 \left (c^2+d^2\right )^2 f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}
Mathematica [B] time = 6.26429, size = 1948, normalized size = 5.34 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.25, size = 45119, normalized size = 123.6 \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{C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A}{{\left (b \tan \left (f x + e\right ) + a\right )}{\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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