Optimal. Leaf size=425 \[ -\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(a-i b)^2 (c-i d)^{5/2} f}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(a+i b)^2 (c+i d)^{5/2} f}-\frac {b^{7/2} \left (4 a b c-9 a^2 d-5 b^2 d\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d \tan (e+f x)}}{\sqrt {b c-a d}}\right )}{\left (a^2+b^2\right )^2 (b c-a d)^{7/2} f}-\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )}{\left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}} \]
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Rubi [A]
time = 1.79, antiderivative size = 425, normalized size of antiderivative = 1.00, number of steps
used = 14, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3650, 3730,
3734, 3620, 3618, 65, 214, 3715} \begin {gather*} -\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 f \left (a^2+b^2\right ) \left (c^2+d^2\right ) (b c-a d)^2 (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {b^{7/2} \left (-9 a^2 d+4 a b c-5 b^2 d\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 )^2 (b c-a d)^{7/2}}+\frac {d \left (4 a^3 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )+4 a b^2 c d^3-\left (b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )\right )}{f \left (a^2+b^2\right ) \left (c^2+d^2\right )^2 (b c-a d)^3 \sqrt {c+d \tan (e+f x)}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (a-i b)^2 (c-i d)^{5/2}}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (a+i b)^2 (c+i d)^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 214
Rule 3618
Rule 3620
Rule 3650
Rule 3715
Rule 3730
Rule 3734
Rubi steps
\begin {align*} \int \frac {1}{(a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{5/2}} \, dx &=-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {\int \frac {\frac {1}{2} \left (-2 a b c+2 a^2 d+5 b^2 d\right )+b (b c-a d) \tan (e+f x)+\frac {5}{2} b^2 d \tan ^2(e+f x)}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2}} \, dx}{\left (a^2+b^2\right ) (b c-a d)}\\ &=-\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {2 \int \frac {-\frac {3}{4} \left (2 a^3 c d^2-4 a^2 b d \left (c^2+d^2\right )-5 b^3 d \left (c^2+d^2\right )+2 a b^2 c \left (c^2+2 d^2\right )\right )+\frac {3}{2} (b c-a d)^2 (b c+a d) \tan (e+f x)+\frac {3}{4} b \left (2 a^2 d^3+b^2 \left (3 c^2 d+5 d^3\right )\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right )}\\ &=-\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )}{\left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}-\frac {4 \int \frac {\frac {3}{8} \left (2 a^4 d^3 \left (c^2-d^2\right )+5 b^4 d \left (c^2+d^2\right )^2-2 a^3 b c d^2 \left (3 c^2+d^2\right )-2 a b^3 c \left (c^4+5 c^2 d^2+2 d^4\right )+2 a^2 b^2 d \left (3 c^4+7 c^2 d^2+2 d^4\right )\right )+\frac {3}{4} (b c-a d)^3 \left (2 a c d+b \left (c^2-d^2\right )\right ) \tan (e+f x)-\frac {3}{8} b d \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right ) \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx}{3 \left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2}\\ &=-\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )}{\left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {\left (b^4 \left (4 a b c-9 a^2 d-5 b^2 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 \left (a^2+b^2\right )^2 (b c-a d)^3}-\frac {4 \int \frac {-\frac {3}{4} (b c-a d)^3 (a c-b c-a d-b d) (a c+b c+a d-b d)+\frac {3}{2} (b c-a d)^3 (b c+a d) (a c-b d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{3 \left (a^2+b^2\right )^2 (b c-a d)^3 \left (c^2+d^2\right )^2}\\ &=-\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )}{\left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {\int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (a-i b)^2 (c-i d)^2}+\frac {\int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (a+i b)^2 (c+i d)^2}+\frac {\left (b^4 \left (4 a b c-9 a^2 d-5 b^2 d\right )\right ) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 \left (a^2+b^2\right )^2 (b c-a d)^3 f}\\ &=-\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )}{\left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}+\frac {i \text {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 (c-i d)^2 f}-\frac {i \text {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 (c+i d)^2 f}+\frac {\left (b^4 \left (4 a b c-9 a^2 d-5 b^2 d\right )\right ) \text {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 )^2 d (b c-a d)^3 f}\\ &=-\frac {b^{7/2} \left (4 a b c-9 a^2 d-5 b^2 d\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 )^2 (b c-a d)^{7/2} f}-\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )}{\left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}-\frac {\text {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 (c-i d)^2 d f}-\frac {\text {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 (c+i d)^2 d f}\\ &=-\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(a-i b)^2 (c-i d)^{5/2} f}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(a+i b)^2 (c+i d)^{5/2} f}-\frac {b^{7/2} \left (4 a b c-9 a^2 d-5 b^2 d\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 )^2 (b c-a d)^{7/2} f}-\frac {d \left (2 a^2 d^2+b^2 \left (3 c^2+5 d^2\right )\right )}{3 \left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {b^2}{\left (a^2+b^2\right ) (b c-a d) f (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (4 a^3 c d^3+4 a b^2 c d^3-4 a^2 b d^2 \left (2 c^2+d^2\right )-b^3 \left (c^4+10 c^2 d^2+5 d^4\right )\right )}{\left (a^2+b^2\right ) (b c-a d)^3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}
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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice
the leaf count of optimal. \(2536\) vs. \(2(425)=850\).
time = 6.27, size = 2536, normalized size = 5.97 \begin {gather*} \text {Result too large to show} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(5221\) vs.
\(2(387)=774\).
time = 0.66, size = 5222, normalized size = 12.29
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(5222\) |
default | \(\text {Expression too large to display}\) | \(5222\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \tan {\left (e + f x \right )}\right )^{2} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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